Chapter 2 Polynomials

Given:

The quadratic polynomial is $p(x) = x^2 + 3x + k$.

One zero of the polynomial is 2.

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To Find:

The value of $k$.

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Solution:

Since 2 is a zero of the quadratic polynomial $p(x) = x^2 + 3x + k$, substituting $x = 2$ into the polynomial must make the value of the polynomial equal to zero.

So, we have:

$p(2) = (2)^2 + 3(2) + k = 0$

$4 + 6 + k = 0$

$10 + k = 0$

Subtracting 10 from both sides:

$k = -10$

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Final Answer:

The value of $k$ is $-10$.

Thus, the correct option is (B) –10.

Given:

The cubic polynomial is $p(x) = ax^3 + bx^2 + cx + d$, where $a \neq 0$.

Two of the zeroes of the polynomial are 0.

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To Find:

The third zero of the polynomial.

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Solution:

Let the three zeroes of the cubic polynomial $p(x)$ be $\alpha$, $\beta$, and $\gamma$.

According to the problem, two of the zeroes are 0. Let $\alpha = 0$ and $\beta = 0$.

The sum of the zeroes of a cubic polynomial $ax^3 + bx^2 + cx + d$ is given by the formula:

Sum of zeroes $= \alpha + \beta + \gamma = -\frac{b}{a}$

Substitute the given values of the two zeroes ($\alpha=0, \beta=0$) into the formula:

$0 + 0 + \gamma = -\frac{b}{a}$

This simplifies to:

$\gamma = -\frac{b}{a}$

Thus, the third zero is $-\frac{b}{a}$.

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Final Answer:

The third zero of the polynomial is $-\frac{b}{a}$.

Therefore, the correct option is (A) $\frac{-b}{a}$.

Given:

The quadratic polynomial is $p(x) = (k – 1) x^2 + kx + 1$.

One zero of the polynomial is $-3$.

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To Find:

The value of $k$.

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Solution:

If $-3$ is a zero of the polynomial $p(x)$, then substituting $x = -3$ into the polynomial must result in a value of 0.

So, we set $p(-3) = 0$:

$p(-3) = (k - 1)(-3)^2 + k(-3) + 1 = 0$

$(k - 1)(9) - 3k + 1 = 0$

Distribute the 9:

$9k - 9 - 3k + 1 = 0$

Combine like terms:

$(9k - 3k) + (-9 + 1) = 0$

$6k - 8 = 0$

Add 8 to both sides:

$6k = 8$

Divide both sides by 6:

$k = \frac{8}{6}$

Simplify the fraction:

$k = \frac{4}{3}$

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Final Answer:

The value of $k$ is $\frac{4}{3}$.

Thus, the correct option is (A) $\frac{4}{3}$.

Given:

The zeroes of a quadratic polynomial are $-3$ and $4$.

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To Find:

A quadratic polynomial whose zeroes are $-3$ and $4$.

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Solution:

Let the zeroes of the quadratic polynomial be $\alpha$ and $\beta$.

We are given $\alpha = -3$ and $\beta = 4$.

A quadratic polynomial with zeroes $\alpha$ and $\beta$ can be expressed in the form $p(x) = k(x^2 - (\alpha + \beta)x + \alpha\beta)$, where $k$ is any non-zero real constant.

First, calculate the sum of the zeroes:

Sum of zeroes $= \alpha + \beta = -3 + 4 = 1$

Next, calculate the product of the zeroes:

Product of zeroes $= \alpha \beta = (-3)(4) = -12$

Now, substitute these values into the general form of the polynomial:

$p(x) = k(x^2 - (1)x + (-12))$

$p(x) = k(x^2 - x - 12)$

For $k=1$, the polynomial is $x^2 - x - 12$. This polynomial has zeroes $-3$ and $4$. Let's check the options.

Option (A) is $x^2 - x + 12$. This does not match $x^2 - x - 12$.

Option (B) is $x^2 + x + 12$. This does not match $x^2 - x - 12$.

Option (C) is $\frac{x^2}{2} - \frac{x}{2} - 6$. We can factor out $\frac{1}{2}$:

$\frac{x^2}{2} - \frac{x}{2} - 6 = \frac{1}{2}(x^2 - x - 12)$

This matches the form $k(x^2 - x - 12)$ with $k = \frac{1}{2}$. Since $k$ can be any non-zero constant, this is a valid quadratic polynomial with the given zeroes.

Option (D) is $2x^2 + 2x - 24$. We can factor out 2:

$2x^2 + 2x - 24 = 2(x^2 + x - 12)$

Let's find the zeroes of $x^2 + x - 12$: $(x+4)(x-3) = 0$. The zeroes are $x = -4$ and $x = 3$. This does not match the given zeroes.

Therefore, option (C) is the correct polynomial.

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Final Answer:

A quadratic polynomial whose zeroes are $-3$ and $4$ is $\frac{x^2}{2} - \frac{x}{2} - 6$.

Thus, the correct option is (C) $\frac{x^2}{2}$ - $\frac{x}{2}$ - 6.

Given:

The quadratic polynomial is $p(x) = x^2 + (a + 1) x + b$.

The zeroes of the polynomial are $2$ and $-3$.

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To Find:

The values of $a$ and $b$.

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Solution:

Let the given zeroes be $\alpha = 2$ and $\beta = -3$.

For a quadratic polynomial $Ax^2 + Bx + C$, the sum of the zeroes is $\alpha + \beta = -\frac{B}{A}$ and the product of the zeroes is $\alpha\beta = \frac{C}{A}$.

In the given polynomial $x^2 + (a + 1) x + b$, we have $A=1$, $B=(a+1)$, and $C=b$.

Using the sum of zeroes formula:

$\alpha + \beta = -\frac{B}{A}$

$2 + (-3) = -\frac{(a + 1)}{1}$

$-1 = -(a + 1)$

$-1 = -a - 1$

Add 1 to both sides:

$-1 + 1 = -a - 1 + 1$

$0 = -a$

Multiply by -1:

$a = 0$

Using the product of zeroes formula:

$\alpha \beta = \frac{C}{A}$

$(2)(-3) = \frac{b}{1}$

$-6 = b$

So, the values are $a = 0$ and $b = -6$.

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Final Answer:

The values of $a$ and $b$ are $0$ and $-6$, respectively.

Thus, the correct option is (D) a = 0, b = – 6.

Given:

The zeroes of a polynomial are $-2$ and $5$.

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To Find:

The number of polynomials having these zeroes.

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Solution:

Let the zeroes be $\alpha = -2$ and $\beta = 5$.

For any polynomial, if $\alpha$ and $\beta$ are its zeroes, then $(x - \alpha)$ and $(x - \beta)$ are factors of the polynomial.

So, $(x - (-2))$ and $(x - 5)$ are factors. This means $(x + 2)$ and $(x - 5)$ are factors.

A polynomial having these zeroes can be written in the form $p(x) = k(x + 2)(x - 5)$, where $k$ is any non-zero real constant.

Let's expand the factors:

$(x + 2)(x - 5) = x(x - 5) + 2(x - 5)$

$= x^2 - 5x + 2x - 10$

$= x^2 - 3x - 10$

So, a polynomial with zeroes $-2$ and $5$ is of the form $p(x) = k(x^2 - 3x - 10)$.

The constant $k$ can be any real number except 0.

For example, if $k=1$, the polynomial is $x^2 - 3x - 10$.

If $k=2$, the polynomial is $2(x^2 - 3x - 10) = 2x^2 - 6x - 20$.

If $k=-1$, the polynomial is $-1(x^2 - 3x - 10) = -x^2 + 3x + 10$.

Since there are infinitely many non-zero real numbers that $k$ can take, there are infinitely many polynomials that have $-2$ and $5$ as zeroes.

Thus, the number of such polynomials is more than 3.

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Final Answer:

The number of polynomials having zeroes as $-2$ and $5$ is more than 3.

Therefore, the correct option is (D) more than 3.

Given:

The cubic polynomial is $p(x) = ax^3 + bx^2 + cx + d$, where $a \neq 0$.

One of the zeroes of the polynomial is $0$.

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To Find:

The product of the other two zeroes.

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Solution:

Let the three zeroes of the cubic polynomial $p(x)$ be $\alpha$, $\beta$, and $\gamma$.

We are given that one of the zeroes is 0. Let $\alpha = 0$.

For a cubic polynomial $ax^3 + bx^2 + cx + d$, the relationships between the coefficients and the zeroes are:

Sum of zeroes: $\alpha + \beta + \gamma = -\frac{b}{a}$

Sum of the product of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$

Product of zeroes: $\alpha\beta\gamma = -\frac{d}{a}$

We are interested in the product of the other two zeroes, which is $\beta\gamma$.

Using the second relationship and substituting $\alpha = 0$:

$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$

$(0)\beta + \beta\gamma + \gamma(0) = \frac{c}{a}$

$0 + \beta\gamma + 0 = \frac{c}{a}$

$\beta\gamma = \frac{c}{a}$

The product of the other two zeroes is $\frac{c}{a}$.

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Final Answer:

The product of the other two zeroes is $\frac{c}{a}$.

Thus, the correct option is (B) $\frac{c}{a}$.

Given:

The cubic polynomial is $p(x) = x^3 + ax^2 + bx + c$.

One zero of the polynomial is $-1$.

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To Find:

The product of the other two zeroes.

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Solution:

Let the three zeroes of the cubic polynomial $p(x)$ be $\alpha$, $\beta$, and $\gamma$.

We are given that one of the zeroes is $-1$. Let $\alpha = -1$.

For a cubic polynomial $Ax^3 + Bx^2 + Cx + D$, the sum of the product of zeroes taken two at a time is given by the formula:

$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{C}{A}$

In the given polynomial $x^3 + ax^2 + bx + c$, we have $A=1$, $B=a$, $C=b$, and $D=c$.

So, $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{b}{1} = b$.

Substitute the given zero $\alpha = -1$ into this equation:

$(-1)\beta + \beta\gamma + \gamma(-1) = b$

$-\beta + \beta\gamma - \gamma = b$

Rearrange the terms to isolate $\beta\gamma$:

$\beta\gamma - (\beta + \gamma) = b$

Now, we need to find the value of $(\beta + \gamma)$. We can use the formula for the sum of zeroes:

$\alpha + \beta + \gamma = -\frac{B}{A}$

$\alpha + \beta + \gamma = -\frac{a}{1} = -a$

Substitute $\alpha = -1$ into this equation:

$-1 + \beta + \gamma = -a$

Add 1 to both sides to find $(\beta + \gamma)$:

$\beta + \gamma = -a + 1$

Now substitute this expression for $(\beta + \gamma)$ back into the equation $\beta\gamma - (\beta + \gamma) = b$:

$\beta\gamma - (-a + 1) = b$

$\beta\gamma + a - 1 = b$

Subtract $a$ and add 1 to both sides to solve for $\beta\gamma$:

$\beta\gamma = b - a + 1$

The product of the other two zeroes is $b - a + 1$.

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Final Answer:

The product of the other two zeroes is $b - a + 1$.

Thus, the correct option is (A) b – a + 1.

Given:

The quadratic polynomial is $p(x) = x^2 + 99x + 127$.

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To Find:

The nature of the zeroes of the polynomial.

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Solution:

Let the quadratic polynomial be in the standard form $Ax^2 + Bx + C$.

In the given polynomial $x^2 + 99x + 127$, we have $A=1$, $B=99$, and $C=127$.

Let the zeroes of the polynomial be $\alpha$ and $\beta$.

For a quadratic polynomial, the sum of the zeroes is given by $\alpha + \beta = -\frac{B}{A}$.

Sum of zeroes $= \alpha + \beta = -\frac{99}{1} = -99$.

The product of the zeroes is given by $\alpha\beta = \frac{C}{A}$.

Product of zeroes $= \alpha\beta = \frac{127}{1} = 127$.

We observe the signs of the sum and product of the zeroes:

Sum of zeroes ($\alpha + \beta$) is negative ($-99 < 0$).

Product of zeroes ($\alpha\beta$) is positive ($127 > 0$).

If the product of two numbers is positive, the numbers must have the same sign (either both positive or both negative).

If the sum of two numbers is negative, and they have the same sign, then both numbers must be negative.

If both zeroes were positive, their sum would be positive, which is not the case.

If one zero were positive and the other negative, their product would be negative, which is not the case.

Therefore, both zeroes must be negative.

We can also check the discriminant $\Delta = B^2 - 4AC$ to confirm the nature of the roots:

$\Delta = (99)^2 - 4(1)(127)$

$\Delta = 9801 - 508$

$\Delta = 9293$

Since $\Delta = 9293 > 0$, the zeroes are real and distinct. This confirms that the zeroes are not equal (option D is incorrect).

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Final Answer:

Since the product of the zeroes is positive and the sum of the zeroes is negative, both zeroes must be negative.

Thus, the correct option is (B) both negative.

Given:

The quadratic polynomial is $p(x) = x^2 + kx + k$, where $k \neq 0$.

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To Find:

The nature of the zeroes of the polynomial.

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Solution:

Let the quadratic polynomial be in the standard form $Ax^2 + Bx + C$.

In the given polynomial $x^2 + kx + k$, we have $A=1$, $B=k$, and $C=k$.

Let the zeroes of the polynomial be $\alpha$ and $\beta$.

Using the relationships between the coefficients and the zeroes:

Sum of zeroes: $\alpha + \beta = -\frac{B}{A} = -\frac{k}{1} = -k$

Product of zeroes: $\alpha\beta = \frac{C}{A} = \frac{k}{1} = k$

Now let's analyze the options:

(A) cannot both be positive: If both zeroes $\alpha$ and $\beta$ are positive, then their sum $\alpha + \beta$ must be positive, and their product $\alpha\beta$ must also be positive.

$\alpha + \beta = -k > 0 \implies k < 0$

$\alpha\beta = k > 0 \implies k > 0$

The conditions $k < 0$ and $k > 0$ are contradictory. Therefore, the zeroes cannot both be positive.

(B) cannot both be negative: If both zeroes $\alpha$ and $\beta$ are negative, then their sum $\alpha + \beta$ must be negative, and their product $\alpha\beta$ must be positive.

$\alpha + \beta = -k < 0 \implies k > 0$

$\alpha\beta = k > 0 \implies k > 0$

The condition $k > 0$ is consistent. For example, if $k=4$, the polynomial is $x^2 + 4x + 4 = (x+2)^2$, and the zeroes are $-2$ and $-2$, which are both negative. Therefore, the zeroes can both be negative.

(C) are always unequal: The zeroes are unequal if the discriminant $\Delta = B^2 - 4AC > 0$. The zeroes are equal if $\Delta = 0$.

$\Delta = k^2 - 4(1)(k) = k^2 - 4k = k(k-4)$

For the zeroes to be equal, $\Delta = 0$, which means $k(k-4) = 0$. Since $k \neq 0$ is given, this implies $k-4 = 0$, so $k=4$. If $k=4$, the zeroes are equal. Thus, the zeroes are not always unequal.

(D) are always equal: As shown above, the zeroes are equal only when $k=4$ (and $k \neq 0$). For other values of $k$ (e.g., $k=-1$, $\Delta = (-1)^2 - 4(-1) = 1+4=5 > 0$, zeroes are unequal), the zeroes are not equal. Thus, the zeroes are not always equal.

From the analysis, only option (A) is always true for $k \neq 0$.

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Final Answer:

Based on the sum and product of zeroes, the zeroes of the polynomial $x^2 + kx + k$ (where $k \neq 0$) cannot both be positive.

Thus, the correct option is (A) cannot both be positive.

Given:

The quadratic polynomial is $ax^2 + bx + c$, where $c \neq 0$.

The zeroes of the polynomial are equal.

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To Find:

The relationship between the signs of the coefficients $a$ and $c$ or $b$ and $c$.

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Solution:

For a quadratic polynomial in the form $Ax^2 + Bx + C$, the zeroes are equal if and only if the discriminant, $\Delta$, is equal to zero.

The discriminant is given by the formula $\Delta = B^2 - 4AC$.

In the given polynomial $ax^2 + bx + c$, we have $A = a$, $B = b$, and $C = c$.

Since the zeroes are equal, the discriminant must be zero:

$\Delta = b^2 - 4ac = 0$

This equation can be rewritten as:

$b^2 = 4ac$

We know that the square of any real number is non-negative, so $b^2 \ge 0$.

Thus, we must have $4ac \ge 0$.

Dividing by 4 (which is a positive number), we get:

$ac \ge 0$

Since the polynomial is quadratic, $a \neq 0$. We are also given that $c \neq 0$.

Because $a \neq 0$ and $c \neq 0$, their product $ac$ cannot be zero.

Therefore, the condition $ac \ge 0$ combined with $ac \neq 0$ implies that $ac > 0$.

The product of two non-zero numbers is positive if and only if the numbers have the same sign.

Thus, $a$ and $c$ must have the same sign.

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Final Answer:

If the zeroes of the quadratic polynomial $ax^2 + bx + c$, $c \neq 0$ are equal, then $a$ and $c$ have the same sign.

Thus, the correct option is (C) c and a have the same sign.

Given:

The quadratic polynomial is $p(x) = x^2 + ax + b$.

One zero of the polynomial is the negative of the other.

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To Find:

The characteristics of the polynomial based on the given information.

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Solution:

Let the two zeroes of the quadratic polynomial $p(x) = x^2 + ax + b$ be $\alpha$ and $\beta$.

According to the given information, one zero is the negative of the other. We can write this as $\beta = -\alpha$.

For a quadratic polynomial in the form $x^2 + ax + b$, the sum of the zeroes is given by the coefficient of the $x$ term with the sign changed, and the product of the zeroes is given by the constant term.

Sum of zeroes $= \alpha + \beta = -a$

Product of zeroes $= \alpha \beta = b$

Substitute $\beta = -\alpha$ into the sum of zeroes equation:

$\alpha + (-\alpha) = -a$

$0 = -a$

This implies that $a = 0$.

Substitute $\beta = -\alpha$ into the product of zeroes equation:

$\alpha (-\alpha) = b$

$-\alpha^2 = b$

So, $b = -\alpha^2$.

Since the polynomial has real coefficients, its zeroes must be real or a complex conjugate pair. As one zero is the negative of the other, if one is complex, say $c+di$, the other must be $-(c+di) = -c-di$. For them to be a conjugate pair, $-c-di$ must be equal to $c-di$, which implies $c=0$. So, the zeroes must be purely imaginary or real.

If the zeroes are real, then $\alpha$ is a real number. The square of any real number is non-negative, i.e., $\alpha^2 \ge 0$.

Therefore, $b = -\alpha^2 \le 0$.

The polynomial has the form $p(x) = x^2 + (0)x + b = x^2 + b$.

The coefficient of the linear term ($x$ term) is $a$, which we found to be 0. Thus, the polynomial has no linear term.

The constant term is $b$, which we found to be $b = -\alpha^2 \le 0$. This means the constant term is either negative (if the zeroes are non-zero) or zero (if the zeroes are both 0). If the zeroes are 0 and 0, the polynomial is $x^2$, which has no linear term and a constant term of 0.

Let's evaluate the given options:

(A) has no linear term and the constant term is negative. This holds true when the zeroes are non-zero (e.g., $x^2 - 4$, zeroes are 2, -2).

(B) has no linear term and the constant term is positive. This is false because $b \le 0$.

(C) can have a linear term but the constant term is negative. This is false because $a=0$ always, so it cannot have a linear term.

(D) can have a linear term but the constant term is positive. This is false because $a=0$ always and $b \le 0$.

Given the options, option (A) is the most fitting description, assuming the typical case where the zeroes are non-zero and distinct (e.g., $\alpha \neq 0$). In this case, $b = -\alpha^2 < 0$, so the constant term is negative.

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Final Answer:

The polynomial always has no linear term. The constant term is $b = -\alpha^2$. If the zeroes are non-zero, the constant term is negative. Considering the options, the most appropriate description is that it has no linear term and the constant term is negative.

Thus, the correct option is (A) has no linear term and the constant term is negative.

Given:

Four graphs are provided, representing possible curves in the xy-plane.

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To Find:

Which of the given graphs is not the graph of a quadratic polynomial.

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Solution:

The graph of a quadratic polynomial $p(x) = ax^2 + bx + c$, where $a \neq 0$, is always a parabola.

A parabola is a U-shaped curve that opens either upwards (if $a > 0$) or downwards (if $a < 0$).

The number of zeroes of a quadratic polynomial is equal to the number of times its graph intersects the x-axis. A quadratic polynomial can have at most two zeroes (real or complex).

Therefore, the graph of a quadratic polynomial can intersect the x-axis at most two times.

Let's examine each graph:

Graph (A): This graph is a parabola opening upwards. It intersects the x-axis at two distinct points. This is a valid graph for a quadratic polynomial with two distinct real zeroes.

Graph (B): This graph is a parabola opening downwards. It touches the x-axis at exactly one point. This is a valid graph for a quadratic polynomial with exactly one real zero (or two equal real zeroes).

Graph (C): This graph is a curve that intersects the x-axis at three distinct points. A polynomial of degree $n$ can intersect the x-axis at most $n$ times. Since this curve intersects the x-axis at three points, it must represent a polynomial of degree at least 3. Therefore, it cannot be the graph of a quadratic polynomial (which has degree 2).

Graph (D): This graph is a parabola opening upwards. It does not intersect the x-axis. This is a valid graph for a quadratic polynomial with no real zeroes (two complex conjugate zeroes).

Based on the number of intersections with the x-axis, only Graph (C) does not fit the properties of a quadratic polynomial's graph.

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Final Answer:

The graph of a quadratic polynomial is a parabola which intersects the x-axis at most twice. Graph (C) intersects the x-axis at three points.

Thus, the graph which is not the graph of a quadratic polynomial is shown in (C).

Given:

A polynomial $p(x)$ is divided by the polynomial $d(x) = 2x + 3$.

The potential remainder is given as $r(x) = x - 1$.

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Justify:

Determine if $x-1$ can be the remainder when $p(x)$ is divided by $2x+3$.

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Solution:

According to the Division Algorithm for polynomials, when a polynomial $p(x)$ is divided by a non-zero polynomial $d(x)$, we get a quotient $q(x)$ and a remainder $r(x)$ such that:

$p(x) = d(x) \cdot q(x) + r(x)$

where $r(x) = 0$ or $\text{degree of } r(x) < \text{degree of } d(x)$.

In this case, the divisor is $d(x) = 2x + 3$. The degree of the divisor is the highest power of $x$ in $d(x)$, which is $1$.

Degree of divisor, $\text{deg}(2x + 3) = 1$.

The potential remainder is $r(x) = x - 1$. The degree of this potential remainder is the highest power of $x$ in $r(x)$, which is $1$.

Degree of potential remainder, $\text{deg}(x - 1) = 1$.

For $x-1$ to be a valid remainder when dividing by $2x+3$, the condition $\text{degree of } r(x) < \text{degree of } d(x)$ must be satisfied.

Comparing the degrees, we have $\text{deg}(x - 1) = 1$ and $\text{deg}(2x + 3) = 1$.

Since $1$ is not strictly less than $1$ ($1 \not< 1$), the condition $\text{degree of } r(x) < \text{degree of } d(x)$ is not satisfied.

Therefore, $x-1$ cannot be the remainder when dividing a polynomial $p(x)$ by $2x+3$.

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Justification:

No, $x-1$ cannot be the remainder. The degree of the remainder must always be strictly less than the degree of the divisor. The degree of the divisor $2x+3$ is 1, and the degree of the potential remainder $x-1$ is also 1. Since the degree of the potential remainder is not less than the degree of the divisor, it cannot be the remainder.

Given:

The statement: "If the zeroes of a quadratic polynomial $ax^2 + bx + c$ are both negative, then $a$, $b$ and $c$ all have the same sign."

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Justify:

Determine if the given statement is True or False.

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Solution:

Let the quadratic polynomial be $p(x) = ax^2 + bx + c$, where $a \neq 0$.

Let the zeroes of the polynomial be $\alpha$ and $\beta$.

According to the given information, both zeroes are negative, which means $\alpha < 0$ and $\beta < 0$.

We know the following relationships between the zeroes and the coefficients of a quadratic polynomial:

Sum of zeroes: $\alpha + \beta = -\frac{b}{a}$

Product of zeroes: $\alpha\beta = \frac{c}{a}$

Since $\alpha < 0$ and $\beta < 0$, the sum of the zeroes $\alpha + \beta$ must be negative.

$\alpha + \beta < 0$

Therefore, $-\frac{b}{a} < 0$. This implies $\frac{b}{a} > 0$.

For $\frac{b}{a}$ to be positive, $a$ and $b$ must have the same sign.

Also, since $\alpha < 0$ and $\beta < 0$, the product of the zeroes $\alpha\beta$ must be positive.

$\alpha\beta > 0$

Therefore, $\frac{c}{a} > 0$.

For $\frac{c}{a}$ to be positive, $a$ and $c$ must have the same sign.

Since $a$ and $b$ have the same sign, and $a$ and $c$ have the same sign, it follows that $a$, $b$, and $c$ must all have the same sign.

For example, consider the polynomial $x^2 + 5x + 6$. Its zeroes are $-2$ and $-3$, which are both negative. Here, $a=1$, $b=5$, and $c=6$, all of which are positive (same sign).

Consider the polynomial $-x^2 - 5x - 6$. Its zeroes are also $-2$ and $-3$. Here, $a=-1$, $b=-5$, and $c=-6$, all of which are negative (same sign).

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Final Answer:

The statement is True. If the zeroes of a quadratic polynomial $ax^2 + bx + c$ are both negative, then $a$, $b$, and $c$ all have the same sign.

Part (i):

No, $x^2 - 1$ cannot be the quotient.

Justification:

According to the Division Algorithm, for polynomials $p(x)$ (dividend), $d(x)$ (divisor), $q(x)$ (quotient), and $r(x)$ (remainder), we have $p(x) = d(x) \cdot q(x) + r(x)$, where $\text{deg}(r(x)) < \text{deg}(d(x))$.

If $p(x) = d(x) \cdot q(x)$ (i.e., $r(x) = 0$), then $\text{deg}(p(x)) = \text{deg}(d(x)) + \text{deg}(q(x))$.

If $r(x) \neq 0$, then $\text{deg}(p(x)) = \text{deg}(d(x) \cdot q(x)) = \text{deg}(d(x)) + \text{deg}(q(x))$ because $\text{deg}(d(x) \cdot q(x)) > \text{deg}(r(x))$.

In this problem, the dividend is $p(x) = x^6 + 2x^3 + x - 1$, so $\text{deg}(p(x)) = 6$.

The divisor $d(x)$ is a polynomial of degree 5, so $\text{deg}(d(x)) = 5$.

The proposed quotient is $q(x) = x^2 - 1$, so $\text{deg}(q(x)) = 2$.

According to the Division Algorithm, the degree of the dividend should be the sum of the degrees of the divisor and the quotient:

$\text{deg}(p(x)) = \text{deg}(d(x)) + \text{deg}(q(x))$

Substituting the given degrees:

$6 = 5 + 2$

$6 = 7$

This is a contradiction. The degrees do not satisfy the property of polynomial division.

Therefore, $x^2 - 1$ cannot be the quotient when dividing a polynomial of degree 6 by a polynomial of degree 5.

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Part (ii):

Given dividend $p(x) = ax^2 + bx + c$ and divisor $d(x) = px^3 + qx^2 + rx + s$, with $p \neq 0$.

The degree of the dividend is $\text{deg}(p(x)) = 2$ (assuming $a \neq 0$; if $a=0$ but $b \neq 0$, deg=1; if $a=b=0$ but $c \neq 0$, deg=0; if $a=b=c=0$, $p(x)$ is the zero polynomial). We are given $p \neq 0$, so the degree of the divisor is $\text{deg}(d(x)) = 3$.

Justification:

According to the Division Algorithm, when the degree of the dividend is less than the degree of the divisor ($\text{deg}(p(x)) < \text{deg}(d(x))$), the quotient is the zero polynomial, and the remainder is the dividend itself.

In this case, the degree of the dividend (at most 2) is less than the degree of the divisor (3).

Thus, the quotient is $q(x) = 0$ and the remainder is $r(x) = p(x) = ax^2 + bx + c$.

Quotient: $0$

Remainder: $ax^2 + bx + c$

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Part (iii):

Given that on division of polynomial $p(x)$ by polynomial $g(x)$, the quotient is zero.

Relation between degrees:

The relation between the degrees of $p(x)$ and $g(x)$ is $\text{deg}(p(x)) < \text{deg}(g(x))$.

Justification:

According to the Division Algorithm, $p(x) = g(x) \cdot q(x) + r(x)$, where $\text{deg}(r(x)) < \text{deg}(g(x))$ or $r(x) = 0$.

If the quotient $q(x) = 0$, the algorithm becomes $p(x) = g(x) \cdot 0 + r(x)$, which means $p(x) = r(x)$.

The condition $\text{deg}(r(x)) < \text{deg}(g(x))$ must still hold for the remainder (unless $p(x)=0$).

Since $p(x) = r(x)$, their degrees are the same: $\text{deg}(p(x)) = \text{deg}(r(x))$.

Combining $\text{deg}(p(x)) = \text{deg}(r(x))$ and $\text{deg}(r(x)) < \text{deg}(g(x))$, we get $\text{deg}(p(x)) < \text{deg}(g(x))$.

This is the standard condition under which the quotient of a polynomial division is 0.

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Part (iv):

Given that on division of a non-zero polynomial $p(x)$ by polynomial $g(x)$, the remainder is zero.

Relation between degrees:

The relation between the degrees of $p(x)$ and $g(x)$ is $\text{deg}(p(x)) \ge \text{deg}(g(x))$.

Justification:

According to the Division Algorithm, $p(x) = g(x) \cdot q(x) + r(x)$.

If the remainder $r(x) = 0$, the algorithm becomes $p(x) = g(x) \cdot q(x)$.

Since $p(x)$ is a non-zero polynomial, neither $g(x)$ nor $q(x)$ can be the zero polynomial.

For any two non-zero polynomials, the degree of their product is the sum of their degrees:

$\text{deg}(p(x)) = \text{deg}(g(x) \cdot q(x)) = \text{deg}(g(x)) + \text{deg}(q(x))$.

Since $q(x)$ is a non-zero polynomial, its degree $\text{deg}(q(x))$ is a non-negative integer ($\ge 0$).

Therefore, $\text{deg}(p(x))$ must be greater than or equal to $\text{deg}(g(x))$ because $\text{deg}(p(x)) = \text{deg}(g(x)) + (\text{a non-negative integer})$.

This means that $g(x)$ is a factor of $p(x)$.

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Part (v):

Given the quadratic polynomial $x^2 + kx + k$, where $k$ is an odd integer and $k > 1$.

Question: Can this polynomial have equal zeroes?

Justification:

A quadratic polynomial $Ax^2 + Bx + C$ has equal zeroes if and only if its discriminant $\Delta = B^2 - 4AC$ is equal to zero.

For the polynomial $x^2 + kx + k$, we have $A = 1$, $B = k$, and $C = k$.

The discriminant is $\Delta = k^2 - 4(1)(k) = k^2 - 4k$.

For equal zeroes, we must have $\Delta = 0$:

$k^2 - 4k = 0$

Factor out $k$:

$k(k - 4) = 0$

This equation holds if $k = 0$ or $k - 4 = 0 \implies k = 4$.

So, the quadratic polynomial $x^2 + kx + k$ has equal zeroes only when $k=0$ or $k=4$.

The question asks if this is possible for some odd integer $k > 1$.

The values $k=0$ and $k=4$ are the only possibilities for equal zeroes.

Neither 0 nor 4 is an odd integer greater than 1.

Therefore, the quadratic polynomial $x^2 + kx + k$ cannot have equal zeroes for any odd integer $k > 1$.

No, it cannot have equal zeroes for some odd integer $k > 1$.

Part (i):

The statement is False.

Justification:

Let the quadratic polynomial be $p(x) = ax^2 + bx + c$, with zeroes $\alpha$ and $\beta$. We are given that $\alpha > 0$ and $\beta > 0$.

The sum of the zeroes is $\alpha + \beta = -\frac{b}{a}$. Since both zeroes are positive, their sum is positive: $\alpha + \beta > 0$.

So, $-\frac{b}{a} > 0$, which implies $\frac{b}{a} < 0$. This means $a$ and $b$ have opposite signs.

The product of the zeroes is $\alpha\beta = \frac{c}{a}$. Since both zeroes are positive, their product is positive: $\alpha\beta > 0$.

So, $\frac{c}{a} > 0$. This means $a$ and $c$ have the same sign.

If $a$ and $c$ have the same sign, and $a$ and $b$ have opposite signs, then $b$ and $c$ must also have opposite signs.

For example, if $a > 0$, then $c > 0$ and $b < 0$. In this case, $a, b, c$ do not all have the same sign.

Consider the polynomial $x^2 - 5x + 6$. Its zeroes are $x=2$ and $x=3$, which are both positive. Here, $a=1$, $b=-5$, and $c=6$. The signs are $+, -, +$, which are not all the same.

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Part (ii):

The statement is False.

Justification:

The graph of a quadratic polynomial is a parabola. A parabola can intersect the x-axis at zero points (no real zeroes), exactly one point (equal real zeroes, where the parabola touches the x-axis), or exactly two points (two distinct real zeroes).

If the graph of a polynomial intersects the x-axis at only one point, it could be a quadratic polynomial with equal real zeroes (i.e., the discriminant is zero). For instance, the graph of $p(x) = x^2$ intersects the x-axis only at $x=0$. This is a quadratic polynomial.

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Part (iii):

The statement is True.

Justification:

A polynomial graph intersecting the x-axis at exactly two points means the polynomial has exactly two real zeroes. A quadratic polynomial can have exactly two real zeroes (when the discriminant is positive). However, other types of polynomials can also intersect the x-axis at exactly two points.

For example, consider the cubic polynomial $p(x) = (x-1)^2(x+2)$. The zeroes are $x=1$ (with multiplicity 2) and $x=-2$ (with multiplicity 1). The graph of this polynomial touches the x-axis at $x=1$ and crosses the x-axis at $x=-2$. Thus, it intersects the x-axis at exactly two points (at $x=1$ and $x=-2$), but it is a cubic polynomial, not a quadratic one.

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Part (iv):

The statement is True.

Justification:

Let the cubic polynomial be $p(x) = ax^3 + bx^2 + cx + d$, where $a \neq 0$. Let the zeroes be $\alpha, \beta, \gamma$. We are given that two of the zeroes are zero. Let $\alpha = 0$ and $\beta = 0$.

The product of the zeroes is $\alpha\beta\gamma = -\frac{d}{a}$. Substituting the given zeroes: $(0)(0)\gamma = -\frac{d}{a} \implies 0 = -\frac{d}{a}$. Since $a \neq 0$, this implies $d = 0$. The constant term is $d$. So, the constant term is zero.

The sum of the products of the zeroes taken two at a time is $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$. Substituting the given zeroes: $(0)(0) + (0)\gamma + \gamma(0) = \frac{c}{a} \implies 0 = \frac{c}{a}$. Since $a \neq 0$, this implies $c = 0$. The coefficient of the linear term ($x$ term) is $c$. So, the linear term is zero.

Therefore, if two of the zeroes of a cubic polynomial are zero, the polynomial must be of the form $ax^3 + bx^2 + (0)x + 0 = ax^3 + bx^2$. This polynomial does not have a linear term (coefficient of $x$ is 0) and does not have a constant term (constant term is 0).

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Part (v):

The statement is True.

Justification:

Let the cubic polynomial be $p(x) = ax^3 + bx^2 + cx + d$, where $a \neq 0$. Let the zeroes be $\alpha, \beta, \gamma$. We are given that all zeroes are negative, i.e., $\alpha < 0$, $\beta < 0$, and $\gamma < 0$.

Sum of zeroes: $\alpha + \beta + \gamma = -\frac{b}{a}$. Since all zeroes are negative, their sum is negative. So, $-\frac{b}{a} < 0 \implies \frac{b}{a} > 0$. This means $a$ and $b$ have the same sign.

Sum of products of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$. Since the product of two negative numbers is positive, each term $\alpha\beta, \beta\gamma, \gamma\alpha$ is positive. Their sum is also positive. So, $\frac{c}{a} > 0$. This means $a$ and $c$ have the same sign.

Product of zeroes: $\alpha\beta\gamma = -\frac{d}{a}$. Since the product of three negative numbers is negative, their product is negative. So, $-\frac{d}{a} < 0 \implies \frac{d}{a} > 0$. This means $a$ and $d$ have the same sign.

Since $b$, $c$, and $d$ all have the same sign as $a$, it means $a, b, c, d$ all have the same sign.

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Part (vi):

The statement is False.

Justification:

Let the cubic polynomial be $p(x) = x^3 + ax^2 - bx + c$. Let the zeroes be $\alpha, \beta, \gamma$. We are given that all zeroes are positive, i.e., $\alpha > 0$, $\beta > 0$, and $\gamma > 0$.

The polynomial is $1 \cdot x^3 + a \cdot x^2 + (-b) \cdot x + c$. Comparing with the standard form $Ax^3 + Bx^2 + Cx + D$, we have $A=1, B=a, C=-b, D=c$.

Sum of zeroes: $\alpha + \beta + \gamma = -\frac{B}{A} = -\frac{a}{1} = -a$. Since $\alpha, \beta, \gamma > 0$, their sum is positive. So, $-a > 0 \implies a < 0$. $a$ is negative.

Sum of products of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{C}{A} = \frac{-b}{1} = -b$. Since $\alpha, \beta, \gamma > 0$, the pairwise products $\alpha\beta, \beta\gamma, \gamma\alpha$ are positive. Their sum is positive. So, $-b > 0 \implies b < 0$. $b$ is negative.

Product of zeroes: $\alpha\beta\gamma = -\frac{D}{A} = -\frac{c}{1} = -c$. Since $\alpha, \beta, \gamma > 0$, their product is positive. So, $-c > 0 \implies c < 0$. $c$ is negative.

Thus, if all three zeroes are positive, then $a$, $b$, and $c$ are all negative.

The statement says that at least one of $a$, $b$, and $c$ is non-negative (i.e., $\ge 0$). Since we found that $a < 0$, $b < 0$, and $c < 0$, none of them are non-negative. This contradicts the statement.

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Part (vii):

The statement is False.

Justification:

A quadratic polynomial $Ax^2 + Bx + C$ has equal zeroes if and only if its discriminant $\Delta = B^2 - 4AC$ is equal to zero.

For the polynomial $kx^2 + x + k$, we have $A=k$, $B=1$, and $C=k$. For it to be a quadratic polynomial, $k \neq 0$.

The discriminant is $\Delta = (1)^2 - 4(k)(k) = 1 - 4k^2$.

For equal zeroes, we set the discriminant to zero:

$1 - 4k^2 = 0$

$1 = 4k^2$

$k^2 = \frac{1}{4}$

Taking the square root of both sides:

$k = \pm\sqrt{\frac{1}{4}}$

$k = \pm \frac{1}{2}$

The values of $k$ for which the polynomial has equal zeroes are $k = \frac{1}{2}$ and $k = -\frac{1}{2}$. Both of these values are non-zero, so the polynomial remains quadratic in both cases.

The statement claims that the *only* value of $k$ is $\frac{1}{2}$. This is incorrect because $k = -\frac{1}{2}$ also results in equal zeroes.

Given:

The quadratic polynomial is $p(x) = x^2 + \frac{1}{6} x - 2$.

---

To Find:

The zeroes of the polynomial and verify the relation between its coefficients and zeroes.

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Solution:

To find the zeroes of the polynomial, we set $p(x) = 0$:

$x^2 + \frac{1}{6} x - 2 = 0$

To clear the fraction, multiply the entire equation by 6:

$6 \left( x^2 + \frac{1}{6} x - 2 \right) = 6(0)$

$6x^2 + x - 12 = 0$

Now, we solve this quadratic equation by factoring. We look for two numbers that multiply to $(6)(-12) = -72$ and add up to 1 (the coefficient of $x$). These numbers are 9 and -8.

Rewrite the middle term ($x$) as $9x - 8x$:

$6x^2 + 9x - 8x - 12 = 0$

Group the terms and factor:

$(6x^2 + 9x) + (-8x - 12) = 0$

Factor out the common terms from each group:

$3x(2x + 3) - 4(2x + 3) = 0$

Factor out the common binomial $(2x + 3)$:

$(3x - 4)(2x + 3) = 0$

Set each factor equal to zero to find the zeroes:

$3x - 4 = 0 \implies 3x = 4 \implies x = \frac{4}{3}$

$2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}$

The zeroes of the polynomial are $\alpha = \frac{4}{3}$ and $\beta = -\frac{3}{2}$.

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Verification of the relation between coefficients and zeroes:

The given polynomial is $p(x) = x^2 + \frac{1}{6} x - 2$. Comparing this with the standard form $Ax^2 + Bx + C$, we have $A = 1$, $B = \frac{1}{6}$, and $C = -2$.

Relationship 1: Sum of zeroes

The sum of zeroes is $\alpha + \beta$. According to the relationship, $\alpha + \beta = -\frac{B}{A}$.

Calculate the sum of the found zeroes:

$\alpha + \beta = \frac{4}{3} + \left(-\frac{3}{2}\right) = \frac{4}{3} - \frac{3}{2}$

Find a common denominator (6):

$\frac{4 \times 2}{3 \times 2} - \frac{3 \times 3}{2 \times 3} = \frac{8}{6} - \frac{9}{6} = \frac{8 - 9}{6} = -\frac{1}{6}$

Calculate $-\frac{B}{A}$ from the coefficients:

$-\frac{B}{A} = -\frac{1/6}{1} = -\frac{1}{6}$

Since $-\frac{1}{6} = -\frac{1}{6}$, the sum of zeroes relation is verified.

Relationship 2: Product of zeroes

The product of zeroes is $\alpha \beta$. According to the relationship, $\alpha \beta = \frac{C}{A}$.

Calculate the product of the found zeroes:

$\alpha \beta = \left(\frac{4}{3}\right) \times \left(-\frac{3}{2}\right) = -\frac{4 \times \cancel{3}}{\cancel{3} \times 2} = -\frac{4}{2} = -2$

Calculate $\frac{C}{A}$ from the coefficients:

$\frac{C}{A} = \frac{-2}{1} = -2$

Since $-2 = -2$, the product of zeroes relation is verified.

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Final Answer:

The zeroes of the polynomial $x^2 + \frac{1}{6} x - 2$ are $\frac{4}{3}$ and $-\frac{3}{2}$.

The relations between the coefficients and the zeroes are verified.

Given:

The quadratic polynomial is $p(x) = 4x^2 - 3x - 1$.

---

To Find:

The zeroes of the polynomial by factorisation method and verify the relation between its coefficients and zeroes.

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Solution:

To find the zeroes of the polynomial, we set $p(x) = 0$:

$4x^2 - 3x - 1 = 0$

We will use the factorisation method by splitting the middle term. We need to find two numbers whose product is equal to the product of the coefficient of $x^2$ and the constant term ($4 \times -1 = -4$), and whose sum is equal to the coefficient of the $x$ term ($-3$). The numbers are $-4$ and $1$, since $(-4) \times (1) = -4$ and $(-4) + 1 = -3$.

Rewrite the middle term ($-3x$) as $-4x + x$:

$4x^2 - 4x + x - 1 = 0$

Group the terms and factor common terms from each group:

$(4x^2 - 4x) + (x - 1) = 0$

$4x(x - 1) + 1(x - 1) = 0$

Factor out the common binomial factor $(x - 1)$:

$(x - 1)(4x + 1) = 0$

Set each factor equal to zero to find the zeroes:

$x - 1 = 0 \implies x = 1$

$4x + 1 = 0 \implies 4x = -1 \implies x = -\frac{1}{4}$

The zeroes of the polynomial are $\alpha = 1$ and $\beta = -\frac{1}{4}$.

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Verification of the relation between coefficients and zeroes:

The given polynomial is $p(x) = 4x^2 - 3x - 1$. Comparing this with the standard form $Ax^2 + Bx + C$, we have $A = 4$, $B = -3$, and $C = -1$.

Let the zeroes be $\alpha = 1$ and $\beta = -\frac{1}{4}$.

Relationship 1: Sum of zeroes

Sum of zeroes $= \alpha + \beta = 1 + \left(-\frac{1}{4}\right) = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$.

From coefficients, sum of zeroes $= -\frac{B}{A} = -\frac{-3}{4} = \frac{3}{4}$.

Since $\frac{3}{4} = \frac{3}{4}$, the sum of zeroes relation is verified.

Relationship 2: Product of zeroes

Product of zeroes $= \alpha \beta = (1) \times \left(-\frac{1}{4}\right) = -\frac{1}{4}$.

From coefficients, product of zeroes $= \frac{C}{A} = \frac{-1}{4} = -\frac{1}{4}$.

Since $-\frac{1}{4} = -\frac{1}{4}$, the product of zeroes relation is verified.

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Final Answer:

The zeroes of the polynomial $4x^2 - 3x - 1$ are $1$ and $-\frac{1}{4}$.

The relations between the coefficients and the zeroes are verified.

Given:

The quadratic polynomial is $p(x) = 3x^2 + 4x - 4$.

---

To Find:

The zeroes of the polynomial by factorisation method and verify the relation between its coefficients and zeroes.

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Solution:

To find the zeroes of the polynomial, we set $p(x) = 0$:

$3x^2 + 4x - 4 = 0$

We use the factorisation method by splitting the middle term. We need to find two numbers whose product is equal to the product of the coefficient of $x^2$ and the constant term ($3 \times -4 = -12$), and whose sum is equal to the coefficient of the $x$ term ($4$). The numbers are $6$ and $-2$, since $6 \times (-2) = -12$ and $6 + (-2) = 4$.

Rewrite the middle term ($4x$) as $6x - 2x$:

$3x^2 + 6x - 2x - 4 = 0$

Group the terms and factor common terms from each group:

$(3x^2 + 6x) + (-2x - 4) = 0$

$3x(x + 2) - 2(x + 2) = 0$

Factor out the common binomial factor $(x + 2)$:

$(x + 2)(3x - 2) = 0$

Set each factor equal to zero to find the zeroes:

$x + 2 = 0 \implies x = -2$

$3x - 2 = 0 \implies 3x = 2 \implies x = \frac{2}{3}$

The zeroes of the polynomial are $\alpha = -2$ and $\beta = \frac{2}{3}$.

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Verification of the relation between coefficients and zeroes:

The given polynomial is $p(x) = 3x^2 + 4x - 4$. Comparing this with the standard form $Ax^2 + Bx + C$, we have $A = 3$, $B = 4$, and $C = -4$.

Let the zeroes be $\alpha = -2$ and $\beta = \frac{2}{3}$.

Relationship 1: Sum of zeroes

Sum of zeroes $= \alpha + \beta = -2 + \frac{2}{3} = -\frac{6}{3} + \frac{2}{3} = \frac{-6 + 2}{3} = -\frac{4}{3}$.

From coefficients, sum of zeroes $= -\frac{B}{A} = -\frac{4}{3}$.

Since $-\frac{4}{3} = -\frac{4}{3}$, the sum of zeroes relation is verified.

Relationship 2: Product of zeroes

Product of zeroes $= \alpha \beta = (-2) \times \left(\frac{2}{3}\right) = -\frac{2 \times 2}{3} = -\frac{4}{3}$.

From coefficients, product of zeroes $= \frac{C}{A} = \frac{-4}{3} = -\frac{4}{3}$.

Since $-\frac{4}{3} = -\frac{4}{3}$, the product of zeroes relation is verified.

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Final Answer:

The zeroes of the polynomial $3x^2 + 4x - 4$ are $-2$ and $\frac{2}{3}$.

The relations between the coefficients and the zeroes are verified.

Given:

The quadratic polynomial is $p(t) = 5t^2 + 12t + 7$.

---

To Find:

The zeroes of the polynomial by factorisation method and verify the relation between its coefficients and zeroes.

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Solution:

To find the zeroes of the polynomial, we set $p(t) = 0$:

$5t^2 + 12t + 7 = 0$

We use the factorisation method by splitting the middle term. We need to find two numbers whose product is equal to the product of the coefficient of $t^2$ and the constant term ($5 \times 7 = 35$), and whose sum is equal to the coefficient of the $t$ term ($12$). The numbers are $5$ and $7$, since $5 \times 7 = 35$ and $5 + 7 = 12$.

Rewrite the middle term ($12t$) as $5t + 7t$:

$5t^2 + 5t + 7t + 7 = 0$

Group the terms and factor common terms from each group:

$(5t^2 + 5t) + (7t + 7) = 0$

$5t(t + 1) + 7(t + 1) = 0$

Factor out the common binomial factor $(t + 1)$:

$(t + 1)(5t + 7) = 0$

Set each factor equal to zero to find the zeroes:

$t + 1 = 0 \implies t = -1$

$5t + 7 = 0 \implies 5t = -7 \implies t = -\frac{7}{5}$

The zeroes of the polynomial are $\alpha = -1$ and $\beta = -\frac{7}{5}$.

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Verification of the relation between coefficients and zeroes:

The given polynomial is $p(t) = 5t^2 + 12t + 7$. Comparing this with the standard form $At^2 + Bt + C$, we have $A = 5$, $B = 12$, and $C = 7$.

Let the zeroes be $\alpha = -1$ and $\beta = -\frac{7}{5}$.

Relationship 1: Sum of zeroes

Sum of zeroes $= \alpha + \beta = -1 + \left(-\frac{7}{5}\right) = -1 - \frac{7}{5} = -\frac{5}{5} - \frac{7}{5} = \frac{-5 - 7}{5} = -\frac{12}{5}$.

From coefficients, sum of zeroes $= -\frac{B}{A} = -\frac{12}{5}$.

Since $-\frac{12}{5} = -\frac{12}{5}$, the sum of zeroes relation is verified.

Relationship 2: Product of zeroes

Product of zeroes $= \alpha \beta = (-1) \times \left(-\frac{7}{5}\right) = \frac{7}{5}$.

From coefficients, product of zeroes $= \frac{C}{A} = \frac{7}{5}$.

Since $\frac{7}{5} = \frac{7}{5}$, the product of zeroes relation is verified.

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Final Answer:

The zeroes of the polynomial $5t^2 + 12t + 7$ are $-1$ and $-\frac{7}{5}$.

The relations between the coefficients and the zeroes are verified.

Given:

The polynomial is $p(t) = t^3 – 2t^2 – 15t$.

---

To Find:

The zeroes of the polynomial by factorisation method and verify the relation between its coefficients and zeroes.

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Solution:

To find the zeroes of the polynomial, we set $p(t) = 0$:

$t^3 – 2t^2 – 15t = 0$

First, factor out the common term $t$ from all terms:

$t(t^2 – 2t – 15) = 0$

Now, we factor the quadratic expression inside the parentheses, $t^2 – 2t – 15$. We look for two numbers that multiply to $-15$ and add up to $-2$. These numbers are $-5$ and $3$.

Rewrite the quadratic expression:

$t^2 – 5t + 3t – 15$

Group the terms and factor:

$(t^2 – 5t) + (3t – 15)$

$t(t – 5) + 3(t – 5)$

Factor out the common binomial $(t – 5)$:

$(t – 5)(t + 3)$

So the polynomial becomes:

$t(t – 5)(t + 3) = 0$

Set each factor equal to zero to find the zeroes:

$t = 0$

$t – 5 = 0 \implies t = 5$

$t + 3 = 0 \implies t = -3$

The zeroes of the polynomial are $\alpha = 0$, $\beta = 5$, and $\gamma = -3$.

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Verification of the relation between coefficients and zeroes:

The given polynomial is $p(t) = t^3 – 2t^2 – 15t$. We can write it as $p(t) = 1 \cdot t^3 + (-2) \cdot t^2 + (-15) \cdot t + 0$.

Comparing this with the standard form of a cubic polynomial $At^3 + Bt^2 + Ct + D$, we have $A = 1$, $B = -2$, $C = -15$, and $D = 0$.

Let the zeroes be $\alpha = 0$, $\beta = 5$, and $\gamma = -3$.

Relationship 1: Sum of zeroes

Sum of zeroes $= \alpha + \beta + \gamma = 0 + 5 + (-3) = 2$.

From coefficients, sum of zeroes $= -\frac{B}{A} = -\frac{-2}{1} = 2$.

Since $2 = 2$, the sum of zeroes relation is verified.

Relationship 2: Sum of the product of zeroes taken two at a time

Sum of products $= \alpha\beta + \beta\gamma + \gamma\alpha = (0)(5) + (5)(-3) + (-3)(0) = 0 - 15 + 0 = -15$.

From coefficients, sum of products $= \frac{C}{A} = \frac{-15}{1} = -15$.

Since $-15 = -15$, the sum of products relation is verified.

Relationship 3: Product of zeroes

Product of zeroes $= \alpha\beta\gamma = (0)(5)(-3) = 0$.

From coefficients, product of zeroes $= -\frac{D}{A} = -\frac{0}{1} = 0$.

Since $0 = 0$, the product of zeroes relation is verified.

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Final Answer:

The zeroes of the polynomial $t^3 – 2t^2 – 15t$ are $0$, $5$, and $-3$.

The relations between the coefficients and the zeroes are verified.

Given:

The quadratic polynomial is $p(x) = 2x^2 + \frac{7}{2} x + \frac{3}{4}$.

---

To Find:

The zeroes of the polynomial by factorisation method and verify the relation between its coefficients and zeroes.

---

Solution:

To find the zeroes of the polynomial, we set $p(x) = 0$:

$2x^2 + \frac{7}{2} x + \frac{3}{4} = 0$

To clear the fractions, multiply the entire equation by the least common multiple of the denominators (2 and 4), which is 4:

$4 \left( 2x^2 + \frac{7}{2} x + \frac{3}{4} \right) = 4(0)$

$8x^2 + 14x + 3 = 0$

Now, we solve this quadratic equation by factoring. We need to find two numbers whose product is equal to the product of the coefficient of $x^2$ and the constant term ($8 \times 3 = 24$), and whose sum is equal to the coefficient of the $x$ term ($14$). The numbers are $12$ and $2$, since $12 \times 2 = 24$ and $12 + 2 = 14$.

Rewrite the middle term ($14x$) as $12x + 2x$:

$8x^2 + 12x + 2x + 3 = 0$

Group the terms and factor common terms from each group:

$(8x^2 + 12x) + (2x + 3) = 0$

$4x(2x + 3) + 1(2x + 3) = 0$

Factor out the common binomial factor $(2x + 3)$:

$(2x + 3)(4x + 1) = 0$

Set each factor equal to zero to find the zeroes:

$2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}$

$4x + 1 = 0 \implies 4x = -1 \implies x = -\frac{1}{4}$

The zeroes of the polynomial are $\alpha = -\frac{3}{2}$ and $\beta = -\frac{1}{4}$.

---

Verification of the relation between coefficients and zeroes:

The given polynomial is $p(x) = 2x^2 + \frac{7}{2} x + \frac{3}{4}$. Comparing this with the standard form $Ax^2 + Bx + C$, we have $A = 2$, $B = \frac{7}{2}$, and $C = \frac{3}{4}$.

Let the zeroes be $\alpha = -\frac{3}{2}$ and $\beta = -\frac{1}{4}$.

Relationship 1: Sum of zeroes

Sum of zeroes $= \alpha + \beta = -\frac{3}{2} + \left(-\frac{1}{4}\right) = -\frac{3}{2} - \frac{1}{4}$.

Find a common denominator (4):

$-\frac{3 \times 2}{2 \times 2} - \frac{1}{4} = -\frac{6}{4} - \frac{1}{4} = \frac{-6 - 1}{4} = -\frac{7}{4}$.

From coefficients, sum of zeroes $= -\frac{B}{A} = -\frac{7/2}{2} = -\frac{7}{2 \times 2} = -\frac{7}{4}$.

Since $-\frac{7}{4} = -\frac{7}{4}$, the sum of zeroes relation is verified.

Relationship 2: Product of zeroes

Product of zeroes $= \alpha \beta = \left(-\frac{3}{2}\right) \times \left(-\frac{1}{4}\right) = \frac{(-3) \times (-1)}{2 \times 4} = \frac{3}{8}$.

From coefficients, product of zeroes $= \frac{C}{A} = \frac{3/4}{2} = \frac{3}{4 \times 2} = \frac{3}{8}$.

Since $\frac{3}{8} = \frac{3}{8}$, the product of zeroes relation is verified.

---

Final Answer:

The zeroes of the polynomial $2x^2 + \frac{7}{2} x + \frac{3}{4}$ are $-\frac{3}{2}$ and $-\frac{1}{4}$.

The relations between the coefficients and the zeroes are verified.

Given:

The quadratic polynomial is $p(x) = 4x^2 + 5 \sqrt{2}x - 3$.

---

To Find:

The zeroes of the polynomial by factorisation method and verify the relation between its coefficients and zeroes.

---

Solution:

To find the zeroes of the polynomial, we set $p(x) = 0$:

$4x^2 + 5 \sqrt{2}x - 3 = 0$

We use the factorisation method by splitting the middle term. We need to find two numbers whose product is equal to the product of the coefficient of $x^2$ and the constant term ($4 \times -3 = -12$), and whose sum is equal to the coefficient of the $x$ term ($5\sqrt{2}$).

Let the two numbers be $p$ and $q$. We need $pq = -12$ and $p+q = 5\sqrt{2}$.

Since the middle term involves $\sqrt{2}$, let's consider numbers involving $\sqrt{2}$. Note that $-12 = -6 \times 2 = -6 \times (\sqrt{2})^2$.

We look for factors of $-12$. Consider pairs that can form $5\sqrt{2}$ when combined with $\sqrt{2}$. The numbers $6\sqrt{2}$ and $-\sqrt{2}$ satisfy the conditions:

Product: $(6\sqrt{2})(-\sqrt{2}) = -6 \times (\sqrt{2})^2 = -6 \times 2 = -12$.

Sum: $6\sqrt{2} + (-\sqrt{2}) = 6\sqrt{2} - \sqrt{2} = 5\sqrt{2}$.

Rewrite the middle term ($5\sqrt{2}x$) as $6\sqrt{2}x - \sqrt{2}x$:

$4x^2 + 6\sqrt{2}x - \sqrt{2}x - 3 = 0$

Group the terms and factor common terms from each group:

$(4x^2 + 6\sqrt{2}x) + (-\sqrt{2}x - 3) = 0$

From the first group, factor out $2x$:

$2x(2x + 3\sqrt{2})$

From the second group, factor out $-\frac{\sqrt{2}}{2}$. Note that $-3 = -\frac{\sqrt{2}}{2} \times \frac{6}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \times \frac{6\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} \times 3\sqrt{2}$:

$-\frac{\sqrt{2}}{2}(2x + 3\sqrt{2})$

So the equation becomes:

$2x(2x + 3\sqrt{2}) - \frac{\sqrt{2}}{2}(2x + 3\sqrt{2}) = 0$

Factor out the common binomial factor $(2x + 3\sqrt{2})$:

$(2x + 3\sqrt{2})(2x - \frac{\sqrt{2}}{2}) = 0$

Set each factor equal to zero to find the zeroes:

$2x + 3\sqrt{2} = 0 \implies 2x = -3\sqrt{2} \implies x = -\frac{3\sqrt{2}}{2}$

$2x - \frac{\sqrt{2}}{2} = 0 \implies 2x = \frac{\sqrt{2}}{2} \implies x = \frac{\sqrt{2}}{4}$

The zeroes of the polynomial are $\alpha = -\frac{3\sqrt{2}}{2}$ and $\beta = \frac{\sqrt{2}}{4}$.

---

Verification of the relation between coefficients and zeroes:

The given polynomial is $p(x) = 4x^2 + 5 \sqrt{2}x - 3$. Comparing this with the standard form $Ax^2 + Bx + C$, we have $A = 4$, $B = 5\sqrt{2}$, and $C = -3$.

Let the zeroes be $\alpha = -\frac{3\sqrt{2}}{2}$ and $\beta = \frac{\sqrt{2}}{4}$.

Relationship 1: Sum of zeroes

Sum of zeroes $= \alpha + \beta = -\frac{3\sqrt{2}}{2} + \frac{\sqrt{2}}{4}$.

Find a common denominator (4):

$-\frac{3\sqrt{2} \times 2}{2 \times 2} + \frac{\sqrt{2}}{4} = -\frac{6\sqrt{2}}{4} + \frac{\sqrt{2}}{4} = \frac{-6\sqrt{2} + \sqrt{2}}{4} = \frac{-5\sqrt{2}}{4}$.

From coefficients, sum of zeroes $= -\frac{B}{A} = -\frac{5\sqrt{2}}{4}$.

Since $\frac{-5\sqrt{2}}{4} = \frac{-5\sqrt{2}}{4}$, the sum of zeroes relation is verified.

Relationship 2: Product of zeroes

Product of zeroes $= \alpha \beta = \left(-\frac{3\sqrt{2}}{2}\right) \times \left(\frac{\sqrt{2}}{4}\right)$.

$= -\frac{3 \times (\sqrt{2} \times \sqrt{2})}{2 \times 4} = -\frac{3 \times 2}{8} = -\frac{6}{8} = -\frac{3}{4}$.

From coefficients, product of zeroes $= \frac{C}{A} = \frac{-3}{4}$.

Since $-\frac{3}{4} = -\frac{3}{4}$, the product of zeroes relation is verified.

---

Final Answer:

The zeroes of the polynomial $4x^2 + 5 \sqrt{2}x - 3$ are $-\frac{3\sqrt{2}}{2}$ and $\frac{\sqrt{2}}{4}$.

The relations between the coefficients and the zeroes are verified.

Given:

The quadratic polynomial is $p(s) = 2s^2 – (1 + 2\sqrt{2})s + \sqrt{2}$.

---

To Find:

The zeroes of the polynomial by factorisation method and verify the relation between its coefficients and zeroes.

---

Solution:

To find the zeroes of the polynomial, we set $p(s) = 0$:

$2s^2 – (1 + 2\sqrt{2})s + \sqrt{2} = 0$

We use the factorisation method by splitting the middle term. The middle term is $-(1 + 2\sqrt{2})s = -s - 2\sqrt{2}s$.

We need two numbers whose product is equal to the product of the coefficient of $s^2$ and the constant term ($2 \times \sqrt{2} = 2\sqrt{2}$), and whose sum is equal to the coefficient of the $s$ term ($-(1 + 2\sqrt{2})$ or $-1 - 2\sqrt{2}$). The numbers are $-1$ and $-2\sqrt{2}$.

Rewrite the polynomial by splitting the middle term:

$2s^2 - s - 2\sqrt{2}s + \sqrt{2} = 0$

Group the terms and factor common terms from each group:

$(2s^2 - s) + (-2\sqrt{2}s + \sqrt{2}) = 0$

$s(2s - 1) - \sqrt{2}(2s - 1) = 0$

Factor out the common binomial factor $(2s - 1)$:

$(2s - 1)(s - \sqrt{2}) = 0$

Set each factor equal to zero to find the zeroes:

$2s - 1 = 0 \implies 2s = 1 \implies s = \frac{1}{2}$

$s - \sqrt{2} = 0 \implies s = \sqrt{2}$

The zeroes of the polynomial are $\alpha = \frac{1}{2}$ and $\beta = \sqrt{2}$.

---

Verification of the relation between coefficients and zeroes:

The given polynomial is $p(s) = 2s^2 – (1 + 2\sqrt{2})s + \sqrt{2}$. Comparing this with the standard form $As^2 + Bs + C$, we have $A = 2$, $B = -(1 + 2\sqrt{2})$, and $C = \sqrt{2}$.

Let the zeroes be $\alpha = \frac{1}{2}$ and $\beta = \sqrt{2}$.

Relationship 1: Sum of zeroes

Sum of zeroes $= \alpha + \beta = \frac{1}{2} + \sqrt{2}$.

From coefficients, sum of zeroes $= -\frac{B}{A} = -\frac{-(1 + 2\sqrt{2})}{2} = \frac{1 + 2\sqrt{2}}{2} = \frac{1}{2} + \frac{2\sqrt{2}}{2} = \frac{1}{2} + \sqrt{2}$.

Since $\frac{1}{2} + \sqrt{2} = \frac{1}{2} + \sqrt{2}$, the sum of zeroes relation is verified.

Relationship 2: Product of zeroes

Product of zeroes $= \alpha \beta = \left(\frac{1}{2}\right)(\sqrt{2}) = \frac{\sqrt{2}}{2}$.

From coefficients, product of zeroes $= \frac{C}{A} = \frac{\sqrt{2}}{2}$.

Since $\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$, the product of zeroes relation is verified.

---

Final Answer:

The zeroes of the polynomial $2s^2 – (1 + 2\sqrt{2})s + \sqrt{2}$ are $\frac{1}{2}$ and $\sqrt{2}$.

The relations between the coefficients and the zeroes are verified.

Given:

The quadratic polynomial is $p(v) = v^2 + 4\sqrt{3}v - 15$.

---

To Find:

The zeroes of the polynomial by factorisation method and verify the relation between its coefficients and zeroes.

---

Solution:

To find the zeroes of the polynomial, we set $p(v) = 0$:

$v^2 + 4\sqrt{3}v - 15 = 0$

We use the factorisation method by splitting the middle term. We need to find two numbers whose product is equal to the product of the coefficient of $v^2$ and the constant term ($1 \times -15 = -15$), and whose sum is equal to the coefficient of the $v$ term ($4\sqrt{3}$).

Let the two numbers be $p$ and $q$. We need $pq = -15$ and $p+q = 4\sqrt{3}$.

Consider numbers involving $\sqrt{3}$. We know that $(\sqrt{3})^2 = 3$. So, $-15 = -5 \times 3 = -5 \times (\sqrt{3})^2$. We can look for numbers of the form $a\sqrt{3}$ and $b\sqrt{3}$. Their product is $(a\sqrt{3})(b\sqrt{3}) = 3ab$. So, $3ab = -15 \implies ab = -5$. Their sum is $a\sqrt{3} + b\sqrt{3} = (a+b)\sqrt{3}$. We need $(a+b)\sqrt{3} = 4\sqrt{3}$, which implies $a+b = 4$. We need two numbers $a$ and $b$ such that their product is $-5$ and their sum is $4$. These numbers are $5$ and $-1$. Thus, the numbers for splitting the middle term are $5\sqrt{3}$ and $-\sqrt{3}$.

Rewrite the middle term ($4\sqrt{3}v$) as $5\sqrt{3}v - \sqrt{3}v$:

$v^2 + 5\sqrt{3}v - \sqrt{3}v - 15 = 0$

Group the terms and factor common terms from each group:

$(v^2 + 5\sqrt{3}v) + (-\sqrt{3}v - 15) = 0$

$v(v + 5\sqrt{3}) - \sqrt{3}(v + 5\sqrt{3}) = 0$

Factor out the common binomial factor $(v + 5\sqrt{3})$:

$(v + 5\sqrt{3})(v - \sqrt{3}) = 0$

Set each factor equal to zero to find the zeroes:

$v + 5\sqrt{3} = 0 \implies v = -5\sqrt{3}$

$v - \sqrt{3} = 0 \implies v = \sqrt{3}$

The zeroes of the polynomial are $\alpha = -5\sqrt{3}$ and $\beta = \sqrt{3}$.

---

Verification of the relation between coefficients and zeroes:

The given polynomial is $p(v) = v^2 + 4\sqrt{3}v - 15$. Comparing this with the standard form $Av^2 + Bv + C$, we have $A = 1$, $B = 4\sqrt{3}$, and $C = -15$.

Let the zeroes be $\alpha = -5\sqrt{3}$ and $\beta = \sqrt{3}$.

Relationship 1: Sum of zeroes

Sum of zeroes $= \alpha + \beta = -5\sqrt{3} + \sqrt{3} = (-5 + 1)\sqrt{3} = -4\sqrt{3}$.

From coefficients, sum of zeroes $= -\frac{B}{A} = -\frac{4\sqrt{3}}{1} = -4\sqrt{3}$.

Since $-4\sqrt{3} = -4\sqrt{3}$, the sum of zeroes relation is verified.

Relationship 2: Product of zeroes

Product of zeroes $= \alpha \beta = (-5\sqrt{3})(\sqrt{3}) = -5 \times (\sqrt{3})^2 = -5 \times 3 = -15$.

From coefficients, product of zeroes $= \frac{C}{A} = \frac{-15}{1} = -15$.

Since $-15 = -15$, the product of zeroes relation is verified.

---

Final Answer:

The zeroes of the polynomial $v^2 + 4\sqrt{3}v - 15$ are $-5\sqrt{3}$ and $\sqrt{3}$.

The relations between the coefficients and the zeroes are verified.

Given:

The quadratic polynomial is $p(y) = y^2 + \frac{3}{2}\sqrt{5}y - 5$.

---

To Find:

The zeroes of the polynomial by factorisation method and verify the relation between its coefficients and zeroes.

---

Solution:

To find the zeroes of the polynomial, we set $p(y) = 0$:

$y^2 + \frac{3}{2}\sqrt{5}y - 5 = 0$

To clear the fraction, multiply the entire equation by 2:

$2 \left( y^2 + \frac{3}{2}\sqrt{5}y - 5 \right) = 2(0)$

$2y^2 + 3\sqrt{5}y - 10 = 0$

Now, we solve this quadratic equation by factoring. We need to find two numbers whose product is equal to the product of the coefficient of $y^2$ and the constant term ($2 \times -10 = -20$), and whose sum is equal to the coefficient of the $y$ term ($3\sqrt{5}$).

Let the two numbers be $p$ and $q$. We need $pq = -20$ and $p+q = 3\sqrt{5}$.

Consider numbers involving $\sqrt{5}$. We know $(\sqrt{5})^2 = 5$. So, $-20 = -4 \times 5 = -4 \times (\sqrt{5})^2$. We look for numbers of the form $a\sqrt{5}$ and $b\sqrt{5}$. Their product is $(a\sqrt{5})(b\sqrt{5}) = 5ab$. So, $5ab = -20 \implies ab = -4$. Their sum is $a\sqrt{5} + b\sqrt{5} = (a+b)\sqrt{5}$. We need $(a+b)\sqrt{5} = 3\sqrt{5}$, which implies $a+b = 3$. We need two numbers $a$ and $b$ such that their product is $-4$ and their sum is $3$. These numbers are $4$ and $-1$. Thus, the numbers for splitting the middle term are $4\sqrt{5}$ and $-\sqrt{5}$.

Rewrite the middle term ($3\sqrt{5}y$) as $4\sqrt{5}y - \sqrt{5}y$:

$2y^2 + 4\sqrt{5}y - \sqrt{5}y - 10 = 0$

Group the terms and factor common terms from each group:

$(2y^2 + 4\sqrt{5}y) + (-\sqrt{5}y - 10) = 0$

$2y(y + 2\sqrt{5}) - \sqrt{5}(y + 2\sqrt{5}) = 0$

Factor out the common binomial factor $(y + 2\sqrt{5})$:

$(y + 2\sqrt{5})(2y - \sqrt{5}) = 0$

Set each factor equal to zero to find the zeroes:

$y + 2\sqrt{5} = 0 \implies y = -2\sqrt{5}$

$2y - \sqrt{5} = 0 \implies 2y = \sqrt{5} \implies y = \frac{\sqrt{5}}{2}$

The zeroes of the polynomial are $\alpha = -2\sqrt{5}$ and $\beta = \frac{\sqrt{5}}{2}$.

---

Verification of the relation between coefficients and zeroes:

The given polynomial is $p(y) = y^2 + \frac{3}{2}\sqrt{5}y - 5$. Comparing this with the standard form $Ay^2 + By + C$, we have $A = 1$, $B = \frac{3}{2}\sqrt{5}$, and $C = -5$.

Let the zeroes be $\alpha = -2\sqrt{5}$ and $\beta = \frac{\sqrt{5}}{2}$.

Relationship 1: Sum of zeroes

Sum of zeroes $= \alpha + \beta = -2\sqrt{5} + \frac{\sqrt{5}}{2}$.

Find a common denominator (2):

$-\frac{2\sqrt{5} \times 2}{2} + \frac{\sqrt{5}}{2} = -\frac{4\sqrt{5}}{2} + \frac{\sqrt{5}}{2} = \frac{-4\sqrt{5} + \sqrt{5}}{2} = \frac{-3\sqrt{5}}{2}$.

From coefficients, sum of zeroes $= -\frac{B}{A} = -\frac{3\sqrt{5}/2}{1} = -\frac{3\sqrt{5}}{2}$.

Since $-\frac{3\sqrt{5}}{2} = -\frac{3\sqrt{5}}{2}$, the sum of zeroes relation is verified.

Relationship 2: Product of zeroes

Product of zeroes $= \alpha \beta = (-2\sqrt{5}) \times \left(\frac{\sqrt{5}}{2}\right)$.

$= -\frac{2\sqrt{5} \times \sqrt{5}}{2} = -\frac{2 \times 5}{2} = -\frac{10}{2} = -5$.

From coefficients, product of zeroes $= \frac{C}{A} = \frac{-5}{1} = -5$.

Since $-5 = -5$, the product of zeroes relation is verified.

---

Final Answer:

The zeroes of the polynomial $y^2 + \frac{3}{2}\sqrt{5}y - 5$ are $-2\sqrt{5}$ and $\frac{\sqrt{5}}{2}$.

The relations between the coefficients and the zeroes are verified.

Given:

The quadratic polynomial is $p(y) = 7y^2 – \frac{11}{3}y - \frac{2}{3}$.

---

To Find:

The zeroes of the polynomial by factorisation method and verify the relation between its coefficients and zeroes.

---

Solution:

To find the zeroes of the polynomial, we set $p(y) = 0$:

$7y^2 – \frac{11}{3}y - \frac{2}{3} = 0$

To clear the fractions, multiply the entire equation by the least common multiple of the denominators (3), which is 3:

$3 \left( 7y^2 – \frac{11}{3}y - \frac{2}{3} \right) = 3(0)$

$21y^2 - 11y - 2 = 0$

Now, we solve this quadratic equation by factoring. We need to find two numbers whose product is equal to the product of the coefficient of $y^2$ and the constant term ($21 \times -2 = -42$), and whose sum is equal to the coefficient of the $y$ term ($-11$). The numbers are $-14$ and $3$, since $(-14) \times (3) = -42$ and $(-14) + 3 = -11$.

Rewrite the middle term ($-11y$) as $-14y + 3y$:

$21y^2 - 14y + 3y - 2 = 0$

Group the terms and factor common terms from each group:

$(21y^2 - 14y) + (3y - 2) = 0$

$7y(3y - 2) + 1(3y - 2) = 0$

Factor out the common binomial factor $(3y - 2)$:

$(3y - 2)(7y + 1) = 0$

Set each factor equal to zero to find the zeroes:

$3y - 2 = 0 \implies 3y = 2 \implies y = \frac{2}{3}$

$7y + 1 = 0 \implies 7y = -1 \implies y = -\frac{1}{7}$

The zeroes of the polynomial are $\alpha = \frac{2}{3}$ and $\beta = -\frac{1}{7}$.

---

Verification of the relation between coefficients and zeroes:

The given polynomial is $p(y) = 7y^2 – \frac{11}{3}$y - $\frac{2}{3}$. Comparing this with the standard form $Ay^2 + By + C$, we have $A = 7$, $B = -\frac{11}{3}$, and $C = -\frac{2}{3}$.

Let the zeroes be $\alpha = \frac{2}{3}$ and $\beta = -\frac{1}{7}$.

Relationship 1: Sum of zeroes

Sum of zeroes $= \alpha + \beta = \frac{2}{3} + \left(-\frac{1}{7}\right) = \frac{2}{3} - \frac{1}{7}$.

Find a common denominator (21):

$\frac{2 \times 7}{3 \times 7} - \frac{1 \times 3}{7 \times 3} = \frac{14}{21} - \frac{3}{21} = \frac{14 - 3}{21} = \frac{11}{21}$.

From coefficients, sum of zeroes $= -\frac{B}{A} = -\frac{-11/3}{7} = \frac{11/3}{7} = \frac{11}{3 \times 7} = \frac{11}{21}$.

Since $\frac{11}{21} = \frac{11}{21}$, the sum of zeroes relation is verified.

Relationship 2: Product of zeroes

Product of zeroes $= \alpha \beta = \left(\frac{2}{3}\right) \times \left(-\frac{1}{7}\right) = \frac{2 \times (-1)}{3 \times 7} = -\frac{2}{21}$.

From coefficients, product of zeroes $= \frac{C}{A} = \frac{-2/3}{7} = \frac{-2}{3 \times 7} = -\frac{2}{21}$.

Since $-\frac{2}{21} = -\frac{2}{21}$, the product of zeroes relation is verified.

---

Final Answer:

The zeroes of the polynomial $7y^2 – \frac{11}{3}y - \frac{2}{3}$ are $\frac{2}{3}$ and $-\frac{1}{7}$.

The relations between the coefficients and the zeroes are verified.

Given:

Sum of the zeroes of a quadratic polynomial $= \sqrt{2}$.

Product of the zeroes of the same quadratic polynomial $= -\frac{3}{2}$.

---

To Find:

A quadratic polynomial with the given sum and product of zeroes, and its zeroes.

---

Solution:

Let the sum of the zeroes be $S$ and the product of the zeroes be $P$.

We are given $S = \sqrt{2}$ and $P = -\frac{3}{2}$.

A quadratic polynomial whose sum of zeroes is $S$ and product of zeroes is $P$ can be written in the form:

$p(x) = k(x^2 - Sx + P)$

where $k$ is any non-zero real constant.

Substituting the given values of $S$ and $P$:

$p(x) = k\left(x^2 - (\sqrt{2})x + \left(-\frac{3}{2}\right)\right)$

$p(x) = k\left(x^2 - \sqrt{2}x - \frac{3}{2}\right)$

We can choose any non-zero value for $k$. A common choice is $k=1$ or a value that eliminates fractions. Let's choose $k=2$ to eliminate the fraction:

$p(x) = 2\left(x^2 - \sqrt{2}x - \frac{3}{2}\right)$

$p(x) = 2x^2 - 2\sqrt{2}x - 3$

This is one quadratic polynomial that satisfies the given conditions.

---

Finding the zeroes of the polynomial:

To find the zeroes of $p(x) = 2x^2 - 2\sqrt{2}x - 3$, we set $p(x) = 0$ and solve for $x$ using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

Here, $A = 2$, $B = -2\sqrt{2}$, and $C = -3$.

Calculate the discriminant $\Delta = B^2 - 4AC$:

$\Delta = (-2\sqrt{2})^2 - 4(2)(-3)$

$\Delta = (4 \times 2) + 24$

$\Delta = 8 + 24$

$\Delta = 32$

Now, use the quadratic formula to find the zeroes:

$x = \frac{-(-2\sqrt{2}) \pm \sqrt{32}}{2(2)}$

$x = \frac{2\sqrt{2} \pm \sqrt{16 \times 2}}{4}$

$x = \frac{2\sqrt{2} \pm 4\sqrt{2}}{4}$

The two zeroes are:

$x_1 = \frac{2\sqrt{2} + 4\sqrt{2}}{4} = \frac{6\sqrt{2}}{4} = \frac{3\sqrt{2}}{2}$

$x_2 = \frac{2\sqrt{2} - 4\sqrt{2}}{4} = \frac{-2\sqrt{2}}{4} = -\frac{\sqrt{2}}{2}$

---

Final Answer:

A quadratic polynomial with the given sum and product of zeroes is $2x^2 - 2\sqrt{2}x - 3$.

The zeroes of this polynomial are $\frac{3\sqrt{2}}{2}$ and $-\frac{\sqrt{2}}{2}$.

(Note: Another possible polynomial is $x^2 - \sqrt{2}x - \frac{3}{2}$. Both polynomials have the same zeroes).

Given:

Polynomial $p(x) = x^3 + 2x^2 + kx + 3$.

Divisor $d(x) = x - 3$.

Remainder $r = 21$ when $p(x)$ is divided by $d(x)$.

The second polynomial is $q(x) = x^3 + 2x^2 + kx - 18$.

---

To Find:

The value of $k$ and the quotient when $p(x)$ is divided by $x-3$.

The zeroes of the polynomial $x^3 + 2x^2 + kx - 18$ (using the found value of $k$).

---

Solution:

Part 1: Find k and the quotient for $p(x)$

According to the Remainder Theorem, if a polynomial $p(x)$ is divided by $(x - a)$, the remainder is $p(a)$.

In this case, the divisor is $x - 3$, so $a = 3$. The remainder is given as 21.

Therefore, we must have $p(3) = 21$.

Substitute $x = 3$ into the polynomial $p(x) = x^3 + 2x^2 + kx + 3$:

$p(3) = (3)^3 + 2(3)^2 + k(3) + 3$

$p(3) = 27 + 2(9) + 3k + 3$

$p(3) = 27 + 18 + 3k + 3$

$p(3) = 48 + 3k$

Now, equate this to the given remainder:

$48 + 3k = 21$

$3k = 21 - 48$

$3k = -27$

$k = \frac{-27}{3}$

$k = -9$

So, the value of $k$ is $-9$.

Now, substitute $k = -9$ into the polynomial $p(x)$:

$p(x) = x^3 + 2x^2 - 9x + 3$

To find the quotient when $p(x)$ is divided by $x - 3$, we can perform polynomial long division or synthetic division. Using synthetic division with $a = 3$:

$\begin{array}{c|cccc} 3 & 1 & 2 & -9 & 3 \\ & & 3 & 15 & 18 \\ \hline & 1 & 5 & 6 & 21 \\ \end{array}$

The numbers in the bottom row represent the coefficients of the quotient and the remainder. The coefficients of the quotient are $1$, $5$, and $6$. Since the degree of the dividend is 3 and the degree of the divisor is 1, the degree of the quotient is $3 - 1 = 2$.

Thus, the quotient is $x^2 + 5x + 6$ and the remainder is $21$, which matches the given information.

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Part 2: Find the zeroes of $x^3 + 2x^2 + kx - 18$

Substitute the found value of $k = -9$ into the second polynomial $q(x) = x^3 + 2x^2 + kx - 18$:

$q(x) = x^3 + 2x^2 - 9x - 18$

We know that $p(x) = x^3 + 2x^2 - 9x + 3$ and $q(x) = x^3 + 2x^2 - 9x - 18$.

Notice that $p(x) = q(x) + 21$.

From the Division Algorithm for $p(x)$, we have $p(x) = (x-3)(x^2 + 5x + 6) + 21$.

Substituting $p(x)$ into the relationship between $p(x)$ and $q(x)$:

$q(x) + 21 = (x-3)(x^2 + 5x + 6) + 21$

$q(x) = (x-3)(x^2 + 5x + 6)$

This shows that $(x-3)$ is a factor of $q(x)$, and the other factor (the quotient when $q(x)$ is divided by $x-3$) is $x^2 + 5x + 6$.

The zeroes of $q(x)$ are the values of $x$ for which $q(x) = 0$.

$(x-3)(x^2 + 5x + 6) = 0$

One zero is found by setting the first factor to zero:

$x - 3 = 0 \implies x = 3$

The other zeroes are found by setting the quadratic factor to zero and solving:

$x^2 + 5x + 6 = 0$

Factor the quadratic: We look for two numbers that multiply to 6 and add up to 5. These numbers are 2 and 3.

$(x + 2)(x + 3) = 0$

Set each factor equal to zero:

$x + 2 = 0 \implies x = -2$

$x + 3 = 0 \implies x = -3$

The zeroes of the cubic polynomial $q(x) = x^3 + 2x^2 - 9x - 18$ are $3$, $-2$, and $-3$.

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Final Answer:

The value of $k$ is $-9$.

The quotient when $x^3 + 2x^2 - 9x + 3$ is divided by $x - 3$ is $x^2 + 5x + 6$.

The zeroes of the cubic polynomial $x^3 + 2x^2 - 9x - 18$ are $3$, $-2$, and $-3$.

Part (i): Sum = $\frac{-8}{3}$, Product = $\frac{4}{3}$

Given: $S = \frac{-8}{3}$, $P = \frac{4}{3}$.

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To Find: Quadratic polynomial and its zeroes.

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Solution:

Polynomial form: $p(x) = k(x^2 - Sx + P)$. Choose $k=3$ to clear fractions.

$p(x) = 3\left(x^2 - \left(\frac{-8}{3}\right)x + \frac{4}{3}\right) = 3x^2 + 8x + 4$

Find zeroes: Set $p(x) = 0$ and factorise $3x^2 + 8x + 4$.

Product $= (3)(4) = 12$, Sum $= 8$. Numbers are $6, 2$.

$3x^2 + 6x + 2x + 4 = 0$

$3x(x + 2) + 2(x + 2) = 0$

$(x + 2)(3x + 2) = 0$

Zeroes: $x + 2 = 0 \implies x = -2$; $3x + 2 = 0 \implies x = -\frac{2}{3}$.

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Verification: Zeroes $\alpha = -2, \beta = -\frac{2}{3}$. Polynomial $3x^2 + 8x + 4 \implies A=3, B=8, C=4$.

Sum: $\alpha + \beta = -2 - \frac{2}{3} = -\frac{8}{3}$. Match: $-\frac{B}{A} = -\frac{8}{3}$.

Product: $\alpha \beta = (-2)(-\frac{2}{3}) = \frac{4}{3}$. Match: $\frac{C}{A} = \frac{4}{3}$.

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Final Answer: Polynomial $3x^2 + 8x + 4$. Zeroes $-2, -\frac{2}{3}$.

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Part (ii): Sum = $\frac{21}{8}$, Product = $\frac{5}{16}$

Given: $S = \frac{21}{8}$, $P = \frac{5}{16}$.

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To Find: Quadratic polynomial and its zeroes.

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Solution:

Polynomial form: $p(x) = k(x^2 - Sx + P)$. Choose $k=16$ to clear fractions.

$p(x) = 16\left(x^2 - \frac{21}{8}x + \frac{5}{16}\right) = 16x^2 - 42x + 5$

Find zeroes: Set $p(x) = 0$ and factorise $16x^2 - 42x + 5$.

Product $= (16)(5) = 80$, Sum $= -42$. Numbers are $-40, -2$.

$16x^2 - 40x - 2x + 5 = 0$

$8x(2x - 5) - 1(2x - 5) = 0$

$(2x - 5)(8x - 1) = 0$

Zeroes: $2x - 5 = 0 \implies x = \frac{5}{2}$; $8x - 1 = 0 \implies x = \frac{1}{8}$.

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Verification: Zeroes $\alpha = \frac{5}{2}, \beta = \frac{1}{8}$. Polynomial $16x^2 - 42x + 5 \implies A=16, B=-42, C=5$.

Sum: $\alpha + \beta = \frac{5}{2} + \frac{1}{8} = \frac{21}{8}$. Match: $-\frac{B}{A} = -\frac{-42}{16} = \frac{42}{16} = \frac{21}{8}$.

Product: $\alpha \beta = (\frac{5}{2})(\frac{1}{8}) = \frac{5}{16}$. Match: $\frac{C}{A} = \frac{5}{16}$.

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Final Answer: Polynomial $16x^2 - 42x + 5$. Zeroes $\frac{5}{2}, \frac{1}{8}$.

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Part (iii): Sum = $-2\sqrt{3}$, Product = - 9

Given: $S = -2\sqrt{3}$, $P = -9$.

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To Find: Quadratic polynomial and its zeroes.

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Solution:

Polynomial form: $p(x) = k(x^2 - Sx + P)$. Choose $k=1$.

$p(x) = x^2 - (-2\sqrt{3})x + (-9) = x^2 + 2\sqrt{3}x - 9$

Find zeroes: Set $p(x) = 0$ and factorise $x^2 + 2\sqrt{3}x - 9$.

Product $= (1)(-9) = -9$, Sum $= 2\sqrt{3}$. Numbers are $3\sqrt{3}, -\sqrt{3}$ (product $3\sqrt{3}(-\sqrt{3}) = -9$, sum $3\sqrt{3}-\sqrt{3}=2\sqrt{3}$). *Correction: The numbers are $3\sqrt{3}$ and $-\sqrt{3}$ based on the equation $x^2 + 2\sqrt{3}x - 9 = 0$. However, the sum is $-S = -(-2\sqrt{3})=2\sqrt{3}$. So numbers should add to $2\sqrt{3}$ and multiply to $-9$. The numbers are $3\sqrt{3}$ and $-\sqrt{3}$. Sum $= 2\sqrt{3}$, Product $=-9$. Wait, sum is $-S$, not $S$. $x^2 - Sx + P$. $S = -2\sqrt{3}$. So $x^2 - (-2\sqrt{3})x + (-9) = x^2 + 2\sqrt{3}x - 9$. Need product -9 and sum $2\sqrt{3}$. Numbers are $3\sqrt{3}$ and $-\sqrt{3}$.* No, the sum of zeroes is $-S$. If zeroes are $\alpha, \beta$, $\alpha+\beta = S = -2\sqrt{3}$, $\alpha\beta = P = -9$. Need numbers that add to $-2\sqrt{3}$ and multiply to $-9$. These are $-3\sqrt{3}$ and $\sqrt{3}$. $(-3\sqrt{3})+\sqrt{3}=-2\sqrt{3}$, $(-3\sqrt{3})(\sqrt{3})=-9$. Let's use these zeroes to factor the polynomial $x^2 + 2\sqrt{3}x - 9$. This polynomial is derived from sum $-S = -(-2\sqrt{3}) = 2\sqrt{3}$ and product $P=-9$. Need numbers that multiply to -9 and add to $2\sqrt{3}$. These are $3\sqrt{3}$ and $-\sqrt{3}$. Let's restart this part.

Part (iii): Sum = $-2\sqrt{3}$, Product = - 9

Given: Sum of zeroes, $S = -2\sqrt{3}$. Product of zeroes, $P = -9$.

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To Find: Quadratic polynomial and its zeroes by factorisation.

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Solution:

Polynomial form: $p(x) = k(x^2 - Sx + P)$. Choose $k=1$.

$p(x) = x^2 - (-2\sqrt{3})x + (-9) = x^2 + 2\sqrt{3}x - 9$

Find zeroes: Set $p(x) = 0$ and factorise $x^2 + 2\sqrt{3}x - 9$.

We need two numbers whose product is $(1)(-9) = -9$ and whose sum is $2\sqrt{3}$. Let the zeroes be $\alpha, \beta$. Then $\alpha+\beta = -2\sqrt{3}$ and $\alpha\beta = -9$. *Mistake in prompt sample or my understanding of how this question is phrased.* "sum and product *respectively* of the zeroes are as given". So the sum *is* $-2\sqrt{3}$ and the product *is* $-9$. Let the polynomial be $Ax^2+Bx+C$. Sum $= -B/A = -2\sqrt{3}$, Product $= C/A = -9$. If we choose $A=1$, then $B = 2\sqrt{3}$ and $C=-9$. The polynomial is $x^2 + 2\sqrt{3}x - 9$. The zeroes $\alpha, \beta$ satisfy $\alpha+\beta = -2\sqrt{3}$ and $\alpha\beta = -9$. We need to factor $x^2 + 2\sqrt{3}x - 9$. We look for two numbers that multiply to $-9$ and add to $2\sqrt{3}$. These numbers are $3\sqrt{3}$ and $-\sqrt{3}$... Oh, that's correct for the *coefficients* but not for the sum and product. The formula is $x^2 - (\text{sum})x + (\text{product})$. So the polynomial is $x^2 - (-2\sqrt{3})x + (-9) = x^2 + 2\sqrt{3}x - 9$. Its zeroes $\alpha, \beta$ satisfy $\alpha+\beta = -2\sqrt{3}$ and $\alpha\beta = -9$. We need to find two numbers that sum to $-2\sqrt{3}$ and multiply to $-9$. These numbers are $-3\sqrt{3}$ and $\sqrt{3}$.

Rewrite the middle term ($2\sqrt{3}x$). Product $=-9$, Sum $= 2\sqrt{3}$. The numbers are $3\sqrt{3}$ and $-\sqrt{3}$. Okay, the polynomial is $x^2 - (-2\sqrt{3})x + (-9) = x^2 + 2\sqrt{3}x - 9$. We need to factor $x^2 + 2\sqrt{3}x - 9$. We look for two numbers that multiply to -9 and add to $2\sqrt{3}$. Let the zeroes be $a, b$. $a+b = -2\sqrt{3}$ and $ab = -9$. The polynomial with these zeroes is $x^2 - (\text{sum})x + (\text{product}) = x^2 - (-2\sqrt{3})x + (-9) = x^2 + 2\sqrt{3}x - 9$. The zeroes of this polynomial are the numbers that sum to $2\sqrt{3}$ and multiply to $-9$. These numbers are $3\sqrt{3}$ and $-\sqrt{3}$.

Using these zeroes to factor $x^2 + 2\sqrt{3}x - 9$ is difficult by splitting the middle term in the standard way without knowing the zeroes beforehand. Let's use the quadratic formula to find the zeroes first, then factor.

$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4(1)(-9)}}{2(1)}$

$x = \frac{-2\sqrt{3} \pm \sqrt{12 + 36}}{2} = \frac{-2\sqrt{3} \pm \sqrt{48}}{2} = \frac{-2\sqrt{3} \pm \sqrt{16 \times 3}}{2} = \frac{-2\sqrt{3} \pm 4\sqrt{3}}{2}$

Zeroes: $x_1 = \frac{-2\sqrt{3} + 4\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$; $x_2 = \frac{-2\sqrt{3} - 4\sqrt{3}}{2} = \frac{-6\sqrt{3}}{2} = -3\sqrt{3}$.

The zeroes are $\alpha = \sqrt{3}$ and $\beta = -3\sqrt{3}$.

Factorisation using these zeroes: $x^2 + 2\sqrt{3}x - 9 = (x - \sqrt{3})(x - (-3\sqrt{3})) = (x - \sqrt{3})(x + 3\sqrt{3})$.

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Verification: Zeroes $\alpha = \sqrt{3}, \beta = -3\sqrt{3}$. Polynomial $x^2 + 2\sqrt{3}x - 9 \implies A=1, B=2\sqrt{3}, C=-9$.

Sum: $\alpha + \beta = \sqrt{3} + (-3\sqrt{3}) = -2\sqrt{3}$. Match: $-\frac{B}{A} = -\frac{2\sqrt{3}}{1} = -2\sqrt{3}$.

Product: $\alpha \beta = (\sqrt{3})(-3\sqrt{3}) = -3 \times 3 = -9$. Match: $\frac{C}{A} = \frac{-9}{1} = -9$.

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Final Answer: Polynomial $x^2 + 2\sqrt{3}x - 9$. Zeroes $\sqrt{3}, -3\sqrt{3}$.

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Part (iv): Sum = $\frac{-3}{2\sqrt{5}}$, Product = $-\frac{1}{2}$

Given: $S = \frac{-3}{2\sqrt{5}}$, $P = -\frac{1}{2}$.

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To Find: Quadratic polynomial and its zeroes.

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Solution:

Polynomial form: $p(x) = k(x^2 - Sx + P)$. Choose $k=2\sqrt{5}$ to clear fractions and radical.

$p(x) = 2\sqrt{5}\left(x^2 - \left(\frac{-3}{2\sqrt{5}}\right)x + \left(-\frac{1}{2}\right)\right) = 2\sqrt{5}x^2 + 3x - \sqrt{5}$

Find zeroes: Set $p(x) = 0$ and factorise $2\sqrt{5}x^2 + 3x - \sqrt{5}$.

Product $= (2\sqrt{5})(-\sqrt{5}) = -10$, Sum $= 3$. Numbers are $5, -2$.

$2\sqrt{5}x^2 + 5x - 2x - \sqrt{5} = 0$

$\sqrt{5}x(2x + \sqrt{5}) - 1(2x + \sqrt{5}) = 0$

$(2x + \sqrt{5})(\sqrt{5}x - 1) = 0$

Zeroes: $2x + \sqrt{5} = 0 \implies x = -\frac{\sqrt{5}}{2}$; $\sqrt{5}x - 1 = 0 \implies x = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$.

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Verification: Zeroes $\alpha = -\frac{\sqrt{5}}{2}, \beta = \frac{\sqrt{5}}{5}$. Polynomial $2\sqrt{5}x^2 + 3x - \sqrt{5} \implies A=2\sqrt{5}, B=3, C=-\sqrt{5}$.

Sum: $\alpha + \beta = -\frac{\sqrt{5}}{2} + \frac{\sqrt{5}}{5} = \frac{-5\sqrt{5}+2\sqrt{5}}{10} = \frac{-3\sqrt{5}}{10}$. Match: $-\frac{B}{A} = -\frac{3}{2\sqrt{5}} = -\frac{3\sqrt{5}}{10}$.

Product: $\alpha \beta = (-\frac{\sqrt{5}}{2})(\frac{\sqrt{5}}{5}) = -\frac{5}{10} = -\frac{1}{2}$. Match: $\frac{C}{A} = \frac{-\sqrt{5}}{2\sqrt{5}} = -\frac{1}{2}$.

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Final Answer: Polynomial $2\sqrt{5}x^2 + 3x - \sqrt{5}$. Zeroes $-\frac{\sqrt{5}}{2}, \frac{\sqrt{5}}{5}$.

Given:

The cubic polynomial is $p(x) = x^3 – 6x^2 + 3x + 10$.

The zeroes of the polynomial are of the form $a$, $a+b$, $a+2b$ for some real numbers $a$ and $b$. These zeroes form an arithmetic progression (AP).

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To Find:

The values of $a$ and $b$, and the zeroes of the polynomial.

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Solution:

Let the zeroes of the polynomial be $\alpha = a$, $\beta = a+b$, and $\gamma = a+2b$.

The given cubic polynomial is $x^3 – 6x^2 + 3x + 10$. Comparing this with the standard form $Ax^3 + Bx^2 + Cx + D$, we have $A=1$, $B=-6$, $C=3$, and $D=10$.

We use the relationships between the coefficients and the zeroes of a cubic polynomial:

1. Sum of zeroes: $\alpha + \beta + \gamma = -\frac{B}{A}$

$(a) + (a+b) + (a+2b) = -\frac{-6}{1}$

$3a + 3b = 6$

Dividing by 3:

This equation tells us that the middle zero ($a+b$) is equal to 2.

2. Product of zeroes: $\alpha\beta\gamma = -\frac{D}{A}$

$(a)(a+b)(a+2b) = -\frac{10}{1}$

Using equation (1), substitute $a+b = 2$ into the product:

$a(2)(a+2b) = -10$

$2a(a+2b) = -10$

Divide by 2:

$a(a+2b) = -5$

Now, from equation (1), we have $b = 2-a$. Substitute this into the equation $a(a+2b) = -5$:

$a(a+2(2-a)) = -5$

$a(a+4-2a) = -5$

$a(4-a) = -5$

$4a - a^2 = -5$

Rearrange into a standard quadratic form:

$a^2 - 4a - 5 = 0$

Factor the quadratic equation:

$(a-5)(a+1) = 0$

This gives two possible values for $a$:

$a - 5 = 0 \implies a = 5$

$a + 1 = 0 \implies a = -1$

Now, we find the corresponding value of $b$ for each value of $a$ using equation (1), $a+b=2$ (or $b=2-a$).

Case 1: If $a = 5$

$5 + b = 2 \implies b = 2 - 5 = -3$

The zeroes are $a$, $a+b$, $a+2b$ which are $5$, $5+(-3)=2$, $5+2(-3)=-1$. The set of zeroes is $\{5, 2, -1\}$.

Case 2: If $a = -1$

$-1 + b = 2 \implies b = 2 - (-1) = 3$

The zeroes are $a$, $a+b$, $a+2b$ which are $-1$, $-1+3=2$, $-1+2(3)=5$. The set of zeroes is $\{-1, 2, 5\}$.

In both cases, the set of zeroes is $\{-1, 2, 5\}$.

The possible values for $a$ and $b$ depend on which zero is considered the first term of the AP ($a$) and the common difference ($b$). The two possibilities derived from the polynomial coefficients are:

Possibility A: $a = -1$ and $b = 3$. The AP is $(-1, 2, 5)$.

Possibility B: $a = 5$ and $b = -3$. The AP is $(5, 2, -1)$.

Both APs consist of the same set of numbers: $\{-1, 2, 5\}$.

Let's verify these zeroes with the original polynomial $p(x) = x^3 – 6x^2 + 3x + 10$.

$p(-1) = (-1)^3 - 6(-1)^2 + 3(-1) + 10 = -1 - 6(1) - 3 + 10 = -1 - 6 - 3 + 10 = -10 + 10 = 0$. So -1 is a zero.

$p(2) = (2)^3 - 6(2)^2 + 3(2) + 10 = 8 - 6(4) + 6 + 10 = 8 - 24 + 6 + 10 = 24 - 24 = 0$. So 2 is a zero.

$p(5) = (5)^3 - 6(5)^2 + 3(5) + 10 = 125 - 6(25) + 15 + 10 = 125 - 150 + 15 + 10 = 150 - 150 = 0$. So 5 is a zero.

The zeroes of the polynomial are indeed $-1$, $2$, and $5$.

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Final Answer:

There are two possible pairs of values for $a$ and $b$ such that the zeroes form an arithmetic progression:

Case 1: $a = -1$ and $b = 3$. The zeroes are $-1$, $2$, and $5$.

Case 2: $a = 5$ and $b = -3$. The zeroes are $5$, $2$, and $-1$.

In both cases, the set of zeroes of the given polynomial is $\{-1, 2, 5\}$.

Given:

The cubic polynomial is $p(x) = 6x^3 + \sqrt{2}x^2 – 10x – 4\sqrt{2}$.

One zero of the polynomial is $\sqrt{2}$.

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To Find:

The other two zeroes of the polynomial.

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Solution:

Since $\sqrt{2}$ is a zero of the polynomial $p(x)$, by the Factor Theorem, $(x - \sqrt{2})$ is a factor of $p(x)$.

We can divide $p(x)$ by $(x - \sqrt{2})$ using synthetic division to find the other factor, which will be a quadratic polynomial.

The coefficients of the polynomial $6x^3 + \sqrt{2}x^2 – 10x – 4\sqrt{2}$ are $6$, $\sqrt{2}$, $-10$, and $-4\sqrt{2}$. The root we are dividing by is $\sqrt{2}$.

$\begin{array}{c|cccc} \sqrt{2} & 6 & \sqrt{2} & -10 & -4\sqrt{2} \\ & & 6\sqrt{2} & (\sqrt{2})(7\sqrt{2})=14 & (\sqrt{2})(4)=4\sqrt{2} \\ \hline & 6 & \sqrt{2}+6\sqrt{2} & -10+14 & -4\sqrt{2}+4\sqrt{2} \\ \hline & 6 & 7\sqrt{2} & 4 & 0 \\ \end{array}$

The remainder is 0, which confirms that $\sqrt{2}$ is a zero.

The coefficients of the quotient polynomial are $6$, $7\sqrt{2}$, and $4$. Since the dividend was degree 3 and the divisor was degree 1, the quotient is degree $3 - 1 = 2$.

The quotient polynomial is $q(x) = 6x^2 + 7\sqrt{2}x + 4$.

The other two zeroes of the cubic polynomial are the zeroes of this quadratic polynomial $q(x)$. We find the zeroes by setting $q(x) = 0$ and solving using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

For $q(x) = 6x^2 + 7\sqrt{2}x + 4$, we have $A=6$, $B=7\sqrt{2}$, and $C=4$.

First, calculate the discriminant $\Delta = B^2 - 4AC$:

$\Delta = (7\sqrt{2})^2 - 4(6)(4)$

$\Delta = (49 \times 2) - 96$

$\Delta = 98 - 96$

$\Delta = 2$

Now, use the quadratic formula:

$x = \frac{-(7\sqrt{2}) \pm \sqrt{2}}{2(6)}$

$x = \frac{-7\sqrt{2} \pm \sqrt{2}}{12}$

The two other zeroes are:

$x_1 = \frac{-7\sqrt{2} + \sqrt{2}}{12} = \frac{-6\sqrt{2}}{12} = -\frac{\sqrt{2}}{2}$

$x_2 = \frac{-7\sqrt{2} - \sqrt{2}}{12} = \frac{-8\sqrt{2}}{12} = -\frac{2\sqrt{2}}{3}$

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Final Answer:

The other two zeroes of the cubic polynomial $6x^3 + \sqrt{2}x^2 – 10x – 4\sqrt{2}$ are $-\frac{\sqrt{2}}{2}$ and $-\frac{2\sqrt{2}}{3}$.

Given:

Dividend polynomial $p(x) = 2x^4 + x^3 – 14x^2 + 5x + 6$.

Divisor polynomial $d(x) = x^2 + 2x + k$.

$d(x)$ is a factor of $p(x)$.

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To Find:

The value of $k$.

All the zeroes of the polynomials $d(x)$ (with the found $k$) and $p(x)$.

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Solution:

If $x^2 + 2x + k$ is a factor of $2x^4 + x^3 – 14x^2 + 5x + 6$, then the remainder obtained by dividing $p(x)$ by $d(x)$ must be zero.

We perform polynomial long division:

$\begin{array}{r} 2x^2\phantom{+0x}-3x+(2-2k) \phantom{+0} \\ x^2+2x+k{\overline{\smash{\big)}\,2x^4+x^3-14x^2+5x+6\phantom{)}}}\\ \underline{-~\phantom{(}(2x^4+4x^3+2kx^2)\phantom{-b)}}\\ -3x^3+(-14-2k)x^2+5x\phantom{)} \\ \underline{-~\phantom{()}(-3x^3-6x^2-3kx)}\\ (-14-2k+6)x^2+(5+3k)x+6\phantom{)} \\ (-8-2k)x^2+(5+3k)x+6\phantom{)} \\ \underline{-~\phantom{()}((-8-2k)x^2+2(-8-2k)x+k(-8-2k))}\\ (5+3k-2(-8-2k))x+(6-k(-8-2k))\phantom{)} \\ (5+3k+16+4k)x+(6+8k+2k^2)\phantom{)} \\ (21+7k)x+(2k^2+8k+6)\phantom{)} \end{array}$

The remainder is $(21+7k)x + (2k^2 + 8k + 6)$.

Since the divisor is a factor, the remainder must be zero for all values of $x$. This means the coefficient of $x$ and the constant term in the remainder must both be equal to zero.

Equating the coefficient of $x$ to zero:

$7k = -21$

$k = \frac{-21}{7}$

$k = -3$

Equating the constant term to zero:

Substitute $k = -3$ into equation (2) to verify:

$2(-3)^2 + 8(-3) + 6 = 2(9) - 24 + 6 = 18 - 24 + 6 = 0$.

Since both conditions are satisfied, the value of $k$ is indeed $-3$.

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The divisor polynomial is $d(x) = x^2 + 2x + (-3) = x^2 + 2x - 3$.

The quotient polynomial from the long division, with $k=-3$, is $q(x) = 2x^2 - 3x + (2 - 2(-3)) = 2x^2 - 3x + (2+6) = 2x^2 - 3x + 8$.

The dividend polynomial $p(x)$ can be written as $p(x) = d(x) \cdot q(x) = (x^2 + 2x - 3)(2x^2 - 3x + 8)$.

The zeroes of $p(x)$ are the combined zeroes of $d(x)$ and $q(x)$.

Zeroes of $d(x) = x^2 + 2x - 3$:

Set $d(x) = 0$ and factorise:

$x^2 + 2x - 3 = 0$

Find two numbers that multiply to $-3$ and add to $2$. These are $3$ and $-1$.

$(x + 3)(x - 1) = 0$

Setting each factor to zero:

$x + 3 = 0 \implies x = -3$

$x - 1 = 0 \implies x = 1$

The zeroes of $d(x)$ are $1$ and $-3$.

Zeroes of $q(x) = 2x^2 - 3x + 8$:

Set $q(x) = 0$ and solve using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

For $2x^2 - 3x + 8 = 0$, we have $A=2$, $B=-3$, $C=8$.

Calculate the discriminant $\Delta = B^2 - 4AC$:

$\Delta = (-3)^2 - 4(2)(8) = 9 - 64 = -55$

Since $\Delta < 0$, the quadratic equation has no real roots. The zeroes are complex conjugates.

Using the formula:

$x = \frac{-(-3) \pm \sqrt{-55}}{2(2)} = \frac{3 \pm \sqrt{55}i}{4}$

The zeroes of $q(x)$ are $\frac{3 + \sqrt{55}i}{4}$ and $\frac{3 - \sqrt{55}i}{4}$.

Zeroes of $p(x) = 2x^4 + x^3 – 14x^2 + 5x + 6$:

The zeroes of $p(x)$ are the combined zeroes of its factors $d(x)$ and $q(x)$.

The zeroes of $p(x)$ are $1$, $-3$, $\frac{3 + \sqrt{55}i}{4}$, and $\frac{3 - \sqrt{55}i}{4}$.

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Final Answer:

The value of $k$ is $-3$.

The zeroes of the polynomial $x^2 + 2x + k$ (which is $x^2 + 2x - 3$) are $1$ and $-3$.

The zeroes of the polynomial $2x^4 + x^3 – 14x^2 + 5x + 6$ are $1$, $-3$, $\frac{3 + \sqrt{55}i}{4}$, and $\frac{3 - \sqrt{55}i}{4}$.

Given:

The cubic polynomial is $p(x) = x^3 – 3\sqrt{5}x^2 + 13x – 3\sqrt{5}$.

$(x - \sqrt{5})$ is a factor of $p(x)$.

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To Find:

All the zeroes of the polynomial $p(x)$.

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Solution:

Since $(x - \sqrt{5})$ is a factor of the polynomial $p(x)$, by the Factor Theorem, $x = \sqrt{5}$ is a zero of $p(x)$.

To find the other zeroes, we can divide the polynomial $p(x)$ by the factor $(x - \sqrt{5})$. We can use synthetic division.

The coefficients of the polynomial $x^3 – 3\sqrt{5}x^2 + 13x – 3\sqrt{5}$ are $1$, $-3\sqrt{5}$, $13$, and $-3\sqrt{5}$. The root we are dividing by is $\sqrt{5}$.

$\begin{array}{c|cccc} \sqrt{5} & 1 & -3\sqrt{5} & 13 & -3\sqrt{5} \\ & & 1 \cdot \sqrt{5}=\sqrt{5} & (\sqrt{5})(-2\sqrt{5}) = -10 & (\sqrt{5})(3) = 3\sqrt{5} \\ \hline & 1 & -3\sqrt{5}+\sqrt{5} & 13-10 & -3\sqrt{5}+3\sqrt{5} \\ \hline & 1 & -2\sqrt{5} & 3 & 0 \\ \end{array}$

The remainder is 0, which confirms that $(x - \sqrt{5})$ is a factor.

The numbers in the bottom row, except the last one, are the coefficients of the quotient polynomial. The dividend was degree 3 and the divisor was degree 1, so the quotient is degree $3 - 1 = 2$.

The quotient polynomial is $q(x) = 1x^2 - 2\sqrt{5}x + 3 = x^2 - 2\sqrt{5}x + 3$.

The zeroes of the cubic polynomial $p(x)$ are $\sqrt{5}$ and the zeroes of the quadratic polynomial $q(x)$.

To find the zeroes of $q(x) = x^2 - 2\sqrt{5}x + 3$, we set $q(x) = 0$ and solve for $x$. We can use the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

For $x^2 - 2\sqrt{5}x + 3 = 0$, we have $A=1$, $B=-2\sqrt{5}$, and $C=3$.

First, calculate the discriminant $\Delta = B^2 - 4AC$:

$\Delta = (-2\sqrt{5})^2 - 4(1)(3)$

$\Delta = (4 \times 5) - 12$

$\Delta = 20 - 12$

$\Delta = 8$

Now, use the quadratic formula to find the zeroes:

$x = \frac{-(-2\sqrt{5}) \pm \sqrt{8}}{2(1)}$

$x = \frac{2\sqrt{5} \pm \sqrt{4 \times 2}}{2}$

$x = \frac{2\sqrt{5} \pm 2\sqrt{2}}{2}$

Factor out 2 from the numerator:

$x = \frac{2(\sqrt{5} \pm \sqrt{2})}{2}$

Cancel out the 2:

$x = \sqrt{5} \pm \sqrt{2}$

The two other zeroes are $\sqrt{5} + \sqrt{2}$ and $\sqrt{5} - \sqrt{2}$.

The zeroes of the cubic polynomial $p(x)$ are the zeroes of its factors $(x - \sqrt{5})$ and $(x^2 - 2\sqrt{5}x + 3)$.

The zeroes are $\sqrt{5}$, $\sqrt{5} + \sqrt{2}$, and $\sqrt{5} - \sqrt{2}$.

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Final Answer:

The zeroes of the polynomial $x^3 – 3\sqrt{5}x^2 + 13x – 3\sqrt{5}$ are $\sqrt{5}$, $\sqrt{5} + \sqrt{2}$, and $\sqrt{5} - \sqrt{2}$.

Given:

Two polynomials: $q(x) = x^3 + 2x^2 + a$ and $p(x) = x^5 – x^4 – 4x^3 + 3x^2 + 3x + b$.

The zeroes of $q(x)$ are also zeroes of $p(x)$.

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To Find:

The values of $a$ and $b$.

All the zeroes of $q(x)$ and $p(x)$.

Which zeroes of $p(x)$ are not zeroes of $q(x)$.

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Solution:

Since the zeroes of $q(x)$ are also zeroes of $p(x)$, this implies that $q(x)$ must be a factor of $p(x)$. Therefore, when $p(x)$ is divided by $q(x)$, the remainder must be zero.

We perform polynomial long division of $p(x) = x^5 – x^4 – 4x^3 + 3x^2 + 3x + b$ by $q(x) = x^3 + 2x^2 + a$.

Upon dividing $p(x)$ by $q(x)$, we get a quotient and a remainder. The process is as follows:

Divide $x^5$ by $x^3$ to get $x^2$. Multiply $x^2$ by $x^3 + 2x^2 + a$ to get $x^5 + 2x^4 + ax^2$. Subtract this from $p(x)$: $(x^5 – x^4 – 4x^3 + 3x^2 + 3x + b) - (x^5 + 2x^4 + ax^2) = -3x^4 - 4x^3 + (3-a)x^2 + 3x + b$.

Divide $-3x^4$ by $x^3$ to get $-3x$. Multiply $-3x$ by $x^3 + 2x^2 + a$ to get $-3x^4 - 6x^3 - 3ax$. Subtract this from the previous result: $(-3x^4 - 4x^3 + (3-a)x^2 + 3x + b) - (-3x^4 - 6x^3 - 3ax) = 2x^3 + (3-a)x^2 + (3+3a)x + b$.

Divide $2x^3$ by $x^3$ to get $2$. Multiply $2$ by $x^3 + 2x^2 + a$ to get $2x^3 + 4x^2 + 2a$. Subtract this from the previous result: $(2x^3 + (3-a)x^2 + (3+3a)x + b) - (2x^3 + 4x^2 + 2a) = (-1-a)x^2 + (3+3a)x + (b-2a)$.

The quotient is $x^2 - 3x + 2$ and the remainder is $(-1-a)x^2 + (3+3a)x + (b-2a)$.

For the remainder to be zero for all values of $x$, the coefficients of the remainder polynomial must all be zero.

Equating the coefficient of $x^2$ to zero:

From (i), $a = -1$.

Equating the coefficient of $x$ to zero:

From (ii), $3a = -3 \implies a = -1$. This is consistent with the value from (i).

Equating the constant term to zero:

Substitute $a = -1$ into (iii):

$b - 2(-1) = 0$

$b + 2 = 0$

$b = -2$

So, the values are $a = -1$ and $b = -2$.

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Now we find the zeroes of the two polynomials using the found values of $a$ and $b$.

The polynomial $q(x)$ becomes $q(x) = x^3 + 2x^2 + (-1) = x^3 + 2x^2 - 1$.

To find the zeroes of $q(x)$, we look for values of $x$ such that $q(x) = 0$. We can test for rational roots $\pm 1$.

$q(1) = (1)^3 + 2(1)^2 - 1 = 1 + 2 - 1 = 2 \neq 0$.

$q(-1) = (-1)^3 + 2(-1)^2 - 1 = -1 + 2(1) - 1 = -1 + 2 - 1 = 0$.

Since $q(-1) = 0$, $x = -1$ is a zero of $q(x)$. By the Factor Theorem, $(x - (-1)) = (x+1)$ is a factor of $q(x)$.

We divide $q(x)$ by $(x+1)$ to find the other factors. Using synthetic division:

$\begin{array}{c|cccc} -1 & 1 & 2 & 0 & -1 \\ & & -1 & -1 & 1 \\ \hline & 1 & 1 & -1 & 0 \\ \end{array}$

The quotient is $x^2 + x - 1$.

So, $q(x) = (x+1)(x^2 + x - 1)$.

The zeroes of $q(x)$ are the roots of $x+1 = 0$ and $x^2 + x - 1 = 0$.

$x+1 = 0 \implies x = -1$.

For $x^2 + x - 1 = 0$, using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ with $A=1, B=1, C=-1$:

$x = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.

The zeroes of $q(x)$ are $-1$, $\frac{-1 + \sqrt{5}}{2}$, and $\frac{-1 - \sqrt{5}}{2}$.

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The polynomial $p(x)$ becomes $p(x) = x^5 – x^4 – 4x^3 + 3x^2 + 3x + (-2) = x^5 – x^4 – 4x^3 + 3x^2 + 3x - 2$.

We know that $p(x) = q(x) \cdot (\text{quotient from division})$. The quotient we found earlier was $x^2 - 3x + 2$ (when $a=-1$).

So, $p(x) = (x^3 + 2x^2 - 1)(x^2 - 3x + 2)$.

The zeroes of $p(x)$ are the combined zeroes of $q(x)$ and the quotient $x^2 - 3x + 2$.

We already found the zeroes of $q(x)$: $-1$, $\frac{-1 + \sqrt{5}}{2}$, and $\frac{-1 - \sqrt{5}}{2}$.

Now, find the zeroes of the quadratic quotient $x^2 - 3x + 2$ by setting it to zero and factorising:

$x^2 - 3x + 2 = 0$

$(x - 1)(x - 2) = 0$

Setting each factor to zero:

$x - 1 = 0 \implies x = 1$

$x - 2 = 0 \implies x = 2$

The zeroes of the quotient $x^2 - 3x + 2$ are $1$ and $2$.

The zeroes of $p(x)$ are the collection of zeroes from $q(x)$ and the quotient:

Zeroes of $p(x)$: $-1$, $\frac{-1 + \sqrt{5}}{2}$, $\frac{-1 - \sqrt{5}}{2}$, $1$, $2$.

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The zeroes of $p(x)$ that are not zeroes of $q(x)$ are the zeroes of the quotient $x^2 - 3x + 2$ that are not in the set of zeroes of $q(x)$.

The zeroes of the quotient are $1$ and $2$. We need to check if $1$ or $2$ are zeroes of $q(x) = x^3 + 2x^2 - 1$.

$q(1) = (1)^3 + 2(1)^2 - 1 = 1 + 2 - 1 = 2 \neq 0$. So $1$ is not a zero of $q(x)$.

$q(2) = (2)^3 + 2(2)^2 - 1 = 8 + 8 - 1 = 15 \neq 0$. So $2$ is not a zero of $q(x)$.

Thus, the zeroes of $p(x)$ that are not zeroes of $q(x)$ are $1$ and $2$.

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Final Answer:

The values of $a$ and $b$ are $a = -1$ and $b = -2$.

The zeroes of $q(x) = x^3 + 2x^2 - 1$ are $-1$, $\frac{-1 + \sqrt{5}}{2}$, and $\frac{-1 - \sqrt{5}}{2}$.

The zeroes of $p(x) = x^5 – x^4 – 4x^3 + 3x^2 + 3x - 2$ are $-1$, $\frac{-1 + \sqrt{5}}{2}$, $\frac{-1 - \sqrt{5}}{2}$, $1$, and $2$.

The zeroes of $p(x)$ that are not zeroes of $q(x)$ are $1$ and $2$.