Chapter 2 Polynomials (Class 10 - Maths NCERT Exemplar Solutions)

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 given polynomial (whether they are positive, negative, or equal).

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Solution:

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

Comparing the given polynomial $x^2 + 99x + 127$ with the standard form $ax^2 + bx + c$, we have:

$a = 1$, $b = 99$, and $c = 127$.

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

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Analysis:

From equation (ii), we see that the product of the zeroes is positive ($127 > 0$).

If the product of two numbers is positive, then either both numbers are positive or both numbers are negative.

From equation (i), we see that the sum of the zeroes is negative ($-99 < 0$).

If the sum of two numbers (which have the same sign) is negative, then both numbers must be negative.

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Final Answer:

Since both the zeroes $\alpha$ and $\beta$ must be negative to satisfy both the sum and product conditions, the zeroes of the quadratic polynomial $x^2 + 99x + 127$ are both negative.

Thus, the correct option is (B).

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 given polynomial.

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Solution:

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

Comparing $x^2 + kx + k$ with the standard form $ax^2 + bx + c$, we have:

$a = 1, b = k, c = k$

Using the relationship between zeroes and coefficients:

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To analyze the signs of the zeroes, we consider two cases for the non-zero constant $k$:

Case I: When $k$ is positive ($k > 0$)

If $k > 0$, then from equation (ii), the product $\alpha\beta$ is positive.

A positive product implies that both zeroes must have the same sign (either both positive or both negative).

However, from equation (i), if $k > 0$, then the sum $\alpha + \beta = -k$ must be negative.

For the sum of two numbers with the same sign to be negative, both zeroes must be negative.

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Case II: When $k$ is negative ($k < 0$)

If $k < 0$, then from equation (ii), the product $\alpha\beta$ is negative.

A negative product implies that one zero is positive and the other zero is negative.

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Conclusion:

In Case I, both zeroes are negative. In Case II, one is positive and one is negative.

In neither case is it possible for both zeroes to be positive.

Therefore, the zeroes cannot both be positive.

Thus, the correct option is (A).

Given:

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

The zeroes of this polynomial are equal.

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To Find:

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

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Solution:

For a quadratic polynomial $ax^2 + bx + c$ to have equal zeroes, the discriminant ($D$) must be equal to zero.

The formula for the discriminant is:

Since the zeroes are equal, we set $D = 0$:

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Analysis of Signs:

In equation (ii), we observe the following:

1. The term $b^2$ is the square of a real number. We know that the square of any real number is always non-negative ($b^2 \geq 0$).

2. Since $c \neq 0$ (given) and for the polynomial to be quadratic $a \neq 0$, the product $ac$ cannot be zero. Therefore, $b^2$ must be positive ($b^2 > 0$).

3. This implies that the product $4ac$ must also be positive ($4ac > 0$).

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For the product of two numbers $a$ and $c$ to be positive (greater than zero), both numbers must have the same sign.

That is, either:

$\bullet$ Both $a$ and $c$ are positive ($a > 0$ and $c > 0$), OR

$\bullet$ Both $a$ and $c$ are negative ($a < 0$ and $c < 0$).

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Final Answer:

Therefore, if the zeroes of the quadratic polynomial $ax^2 + bx + c$ are equal, $c$ and $a$ must have the same sign.

Thus, the correct option is (C).

Given:

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

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

According to the question, one zero is the negative of the other, so $\alpha = -\beta$.

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To Find:

The characteristics of the linear term (the term with $x$) and the constant term $b$.

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Solution:

Comparing the given polynomial $x^2 + ax + b$ with the standard form $Ax^2 + Bx + C$, we get:

$A = 1, B = a, C = b$.

Using the relationship between zeroes and coefficients:

1. Sum of zeroes:

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

Since the coefficient of the linear term $a$ is 0, the polynomial has no linear term.

2. Product of zeroes:

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

Since $\alpha$ is a zero and for the zeroes to be distinct and non-zero, $\alpha^2$ will always be a positive value. Therefore, $-\alpha^2$ must be negative.

This implies that the constant term $b$ is negative.

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Final Answer:

The polynomial has no linear term and the constant term is negative.

Thus, the correct option is (A).

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Alternate Solution:

Consider a quadratic polynomial where zeroes are 2 and -2 (one is the negative of the other).

The polynomial is formed as:

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

$p(x) = x^2 - 4$

Comparing this with $x^2 + ax + b$:

$\bullet$ Here, $a = 0$, so there is no linear term (no $x$ term).

$\bullet$ Here, $b = -4$, which is a negative constant term.

This confirms that option (A) is correct.

To Find:

Identify which graph among the given options does not represent a quadratic polynomial.

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Solution:

We know that the graph of a quadratic polynomial of the form $p(x) = ax^2 + bx + c$ ($a \neq 0$) is always a parabola.

A parabola can open either upwards (if $a > 0$) or downwards (if $a < 0$).

Furthermore, a quadratic polynomial can have at most two zeroes, which means its graph can intersect the $x$-axis at most at two points.

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Analysis of Options:

$\bullet$ Option (A): This graph is a parabola opening upwards and does not intersect the $x$-axis. It represents a quadratic polynomial with no real zeroes. Thus, it is a quadratic graph.

$\bullet$ Option (B): This graph is a parabola opening downwards and touches the $x$-axis at exactly one point. It represents a quadratic polynomial with two equal real zeroes. Thus, it is a quadratic graph.

$\bullet$ Option (C): This graph is a parabola opening downwards and intersects the $x$-axis at two distinct points. It represents a quadratic polynomial with two distinct real zeroes. Thus, it is a quadratic graph.

$\bullet$ Option (D): This graph intersects the $x$-axis at three distinct points. A polynomial that intersects the $x$-axis at three points must be at least a cubic polynomial (degree 3). Since its shape is not a parabola and it has more than two zeroes, it is not the graph of a quadratic polynomial.

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Final Answer:

The graph in Option (D) does not represent a quadratic polynomial.

Thus, the correct option is (D).

Given:

Divisor $g(x) = 2x + 3$

Proposed remainder $r(x) = x - 1$

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To Find:

Whether $x - 1$ can be the remainder when a polynomial $p(x)$ is divided by $2x + 3$.

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Solution / Justification:

According to the Division Algorithm for polynomials, if $p(x)$ and $g(x)$ are two polynomials with $g(x) \neq 0$, then we can find polynomials $q(x)$ and $r(x)$ such that:

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

where the remainder $r(x)$ must satisfy the following condition:

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In the given problem:

The divisor is $g(x) = 2x + 3$.

The proposed remainder is $r(x) = x - 1$.

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Comparison:

Comparing the degrees, we find that:

$\text{degree}(r(x)) = \text{degree}(g(x))$

This contradicts the condition of the division algorithm stated in equation (i), which requires the degree of the remainder to be strictly less than the degree of the divisor.

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Final Answer:

No, $x - 1$ cannot be the remainder on division of a polynomial $p(x)$ by $2x + 3$, because the degree of the remainder is not less than the degree of the divisor.

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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Answer:

The given statement is True.

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Justification:

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

According to the given condition, both zeroes are negative:

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

1. Sum of zeroes:

Since both $\alpha$ and $\beta$ are negative, their sum $(\alpha + \beta)$ must also be negative.

For the ratio $\frac{b}{a}$ to be positive, both $a$ and $b$ must have the same sign (either both positive or both negative).

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2. Product of zeroes:

Since both $\alpha$ and $\beta$ are negative, their product $(\alpha \cdot \beta)$ must be positive.

For the ratio $\frac{c}{a}$ to be positive, both $a$ and $c$ must have the same sign (either both positive or both negative).

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Conclusion:

From results (ii) and (iv), we have seen that $b$ has the same sign as $a$, and $c$ also has the same sign as $a$.

Therefore, $a, b,$ and $c$ all have the same sign.

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Alternate Solution:

Consider a quadratic polynomial with negative zeroes, say $-2$ and $-3$.

The polynomial is $p(x) = (x + 2)(x + 3) = x^2 + 5x + 6$.

Here, $a = 1, b = 5, c = 6$. All are positive (same sign).

If we multiply the entire polynomial by $-1$, we get $-x^2 - 5x - 6$.

Here, $a = -1, b = -5, c = -6$. All are negative (same sign).

In both cases, $a, b,$ and $c$ share the same sign.

(i) Solution:

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

Justification:

According to the division algorithm for polynomials, we have the following relationship between degrees:

Given that:

$\text{Degree of dividend } p(x) = 6$

$\text{Degree of divisor } g(x) = 5$

Substituting these values into equation (i):

$6 = 5 + \text{Degree}(q(x))$

$\text{Degree}(q(x)) = 6 - 5 = 1$

However, the given quotient is $x^2 - 1$, which has a degree of 2. Since the calculated degree is 1 and the given degree is 2, the statement is not possible.

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(ii) Solution:

Quotient = 0

Remainder = $ax^2 + bx + c$

Justification:

In polynomial division, if the degree of the dividend is less than the degree of the divisor, the division cannot proceed further.

Degree of dividend $(ax^2 + bx + c) = 2$

Degree of divisor $(px^3 + qx^2 + rx + s) = 3$

Since $2 < 3$, the dividend itself becomes the remainder and the quotient is 0.

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(iii) Solution:

The relation is $\text{Degree}(p(x)) < \text{Degree}(g(x))$.

Justification:

By division algorithm: $p(x) = g(x) \cdot q(x) + r(x)$.

If the quotient $q(x) = 0$, then the expression becomes:

$p(x) = g(x) \cdot 0 + r(x)$

$p(x) = r(x)$

We know that the degree of the remainder $r(x)$ is always strictly less than the degree of the divisor $g(x)$.

Therefore, $\text{Degree}(p(x)) < \text{Degree}(g(x))$.

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(iv) Solution:

The relation is $\text{Degree}(p(x)) \ge \text{Degree}(g(x))$.

Justification:

If the remainder is zero, then $g(x)$ is a factor of $p(x)$. Thus:

$p(x) = g(x) \cdot q(x)$

$\text{Degree}(p(x)) = \text{Degree}(g(x)) + \text{Degree}(q(x))$

Since $p(x)$ is a non-zero polynomial, $q(x)$ must have a degree of at least 0 (i.e., it is at least a constant). Therefore, the degree of the dividend must be greater than or equal to the degree of the divisor.

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(v) Solution:

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

Justification:

For the quadratic polynomial $x^2 + kx + k$ to have equal zeroes, its discriminant ($D$) must be zero.

Comparing with $ax^2 + bx + c$, we have $a = 1, b = k, c = k$.

This gives two possible values for $k$:

$k = 0$ or $k = 4$

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

Since 4 is an even integer and 0 is not greater than 1, there is no odd integer $k > 1$ that satisfies the condition for equal zeroes.

(i) Solution: False

Justification:

Let the zeroes of $ax^2 + bx + c$ be $\alpha$ and $\beta$. Given that $\alpha > 0$ and $\beta > 0$.

Since both zeroes are positive, their sum $\alpha + \beta$ is positive. From (i), $-\frac{b}{a} > 0$, which implies $\frac{b}{a} < 0$. This means $a$ and $b$ must have opposite signs.

From (ii), since the product is positive, $\frac{c}{a} > 0$, meaning $a$ and $c$ must have the same sign. Thus, $a, b,$ and $c$ cannot all have the same sign.

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(ii) Solution: False

Justification:

A quadratic polynomial can intersect the $x$-axis at exactly one point if it has two equal real zeroes. In such cases, the vertex of the parabola touches the $x$-axis.

For example, consider $p(x) = x^2 - 2x + 1 = (x-1)^2$. This graph intersects the $x$-axis only at the point $(1, 0)$.

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(iii) Solution: True

Justification:

While a quadratic polynomial can have exactly two points of intersection, higher-degree polynomials can also behave this way. For example, a cubic polynomial can have one point where it crosses the $x$-axis and another where it only touches the axis.

Consider $p(x) = x^2(x - 1)$. This is a cubic polynomial that intersects the $x$-axis at exactly two points: $(0, 0)$ and $(1, 0)$.

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(iv) Solution: True

Justification:

Let the zeroes of the cubic polynomial be $\alpha, \beta,$ and $\gamma$. Given $\alpha = 0$ and $\beta = 0$.

The cubic polynomial is given by:

$p(x) = k(x - \alpha)(x - \beta)(x - \gamma)$

$p(x) = k(x - 0)(x - 0)(x - \gamma)$

$p(x) = kx^2(x - \gamma) = kx^3 - k\gamma x^2$

In this polynomial, the terms involving $x$ (linear term) and the constant term are both zero. Therefore, it has no linear or constant terms.

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(v) Solution: True

Justification:

Let $p(x) = ax^3 + bx^2 + cx + d$ have negative zeroes $\alpha, \beta, \gamma$.

$\alpha + \beta + \gamma = -\frac{b}{a} \implies \text{Sum is negative, so } \frac{b}{a} > 0$ ($a, b$ same sign).

$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \implies \text{Sum of products is positive, so } \frac{c}{a} > 0$ ($a, c$ same sign).

$\alpha\beta\gamma = -\frac{d}{a} \implies \text{Product is negative, so } \frac{d}{a} > 0$ ($a, d$ same sign).

Since $b, c,$ and $d$ all have the same sign as $a$, all coefficients and the constant term have the same sign.

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(vi) Solution: False

Justification:

Let the zeroes be $\alpha, \beta, \gamma > 0$. For $x^3 + ax^2 - bx + c$:

Sum of zeroes $= -a = \alpha + \beta + \gamma > 0 \implies a < 0$ (negative).

Sum of products $= -b = \alpha\beta + \beta\gamma + \gamma\alpha > 0 \implies b < 0$ (negative).

Product of zeroes $= -c = \alpha\beta\gamma > 0 \implies c < 0$ (negative).

Since $a, b,$ and $c$ are all negative, the statement that "at least one of them is non-negative" is False.

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(vii) Solution: False

Justification:

For $kx^2 + x + k$ to have equal zeroes, the discriminant $D$ must be zero.

$D = (1)^2 - 4(k)(k) = 0$

$1 - 4k^2 = 0$

$4k^2 = 1$

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

Since $k$ can be both $+\frac{1}{2}$ and $-\frac{1}{2}$, the statement that $\frac{1}{2}$ is the "only" value is False.

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 $\beta = -\frac{3}{2}$.

---

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.

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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 $\beta = -\frac{1}{4}$.

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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.

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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 $\beta = \frac{\sqrt{2}}{4}$.

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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.

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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.

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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 $\beta = \sqrt{2}$.

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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.

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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 $\beta = \sqrt{3}$.

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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.

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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.

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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.

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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.

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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.

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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.

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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:

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

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

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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:

Dividend polynomial $p(x) = x^3 + 2x^2 + kx + 3$

Divisor $g(x) = x - 3$

Remainder $R = 21$

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To Find:

1. The value of $k$.

2. The quotient $q(x)$ when $p(x)$ is divided by $x - 3$.

3. The zeroes of the polynomial $f(x) = x^3 + 2x^2 + kx - 18$.

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Solution:

Step 1: Finding the value of $k$

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

Here, the divisor is $x - 3$, so $a = 3$.

Substituting $x = 3$ in $p(x)$:

$(3)^3 + 2(3)^2 + k(3) + 3 = 21$

$27 + 2(9) + 3k + 3 = 21$

$27 + 18 + 3k + 3 = 21$

$48 + 3k = 21$

$3k = 21 - 48$

$3k = -27$

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Step 2: Finding the quotient $q(x)$

Substituting $k = -9$ in $p(x)$, we get $p(x) = x^3 + 2x^2 - 9x + 3$.

Now, we perform long division of $p(x)$ by $x - 3$:

$\begin{array}{r} x^2+5x+6\phantom{x^3)} \\ x-3{\overline{\smash{\big)}\,x^3+2x^2-9x+3\phantom{)}}} \\ \underline{-~\phantom{(}(x^3-3x^2)\phantom{-9x+3)}} \\ 0+5x^2-9x+3\phantom{)} \\ \underline{-~\phantom{()}(5x^2-15x)\phantom{+3)}} \\ 0+6x+3\phantom{)} \\ \underline{-~\phantom{()}(6x-18)\phantom{)}} \\ 0+21\phantom{)} \end{array}$

Thus, the quotient is $q(x) = x^2 + 5x + 6$.

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Step 3: Finding the zeroes of $f(x) = x^3 + 2x^2 + kx - 18$

Substituting $k = -9$, the polynomial becomes:

To find the zeroes, we factorise $f(x)$ by grouping terms:

$f(x) = x^2(x + 2) - 9(x + 2)$

$f(x) = (x + 2)(x^2 - 9)$

Using the identity $a^2 - b^2 = (a - b)(a + b)$:

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

To find the zeroes, set $f(x) = 0$:

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

This gives $x = -2, x = 3,$ and $x = -3$.

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Final Answer:

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

The quotient is $x^2 + 5x + 6$.

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

General Formula:

A quadratic polynomial whose sum of zeroes $(\alpha + \beta)$ and product of zeroes $(\alpha\beta)$ are given can be written as:

$p(x) = k[x^2 - (\text{Sum of zeroes})x + (\text{Product of zeroes})]$

Where $k$ is a non-zero real constant used to remove fractions.

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

The polynomial is:

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

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

To clear the denominator, let $k = 3$:

Factorisation:

We split the middle term. We need two numbers whose sum is 8 and product is $3 \times 4 = 12$. The numbers are 6 and 2.

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

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

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

For zeroes, $3x + 2 = 0 \implies x = \frac{-2}{3}$ or $x + 2 = 0 \implies x = -2$.

Zeroes: $-\frac{2}{3}, -2$

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

The polynomial is:

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

To clear the denominator, let $k = 16$:

Factorisation:

We need two numbers whose sum is $-42$ and product is $16 \times 5 = 80$. The numbers are $-40$ and $-2$.

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

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

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

For zeroes, $8x - 1 = 0 \implies x = \frac{1}{8}$ or $2x - 5 = 0 \implies x = \frac{5}{2}$.

Zeroes: $\frac{1}{8}, \frac{5}{2}$

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

The polynomial is:

Factorisation:

We need two numbers whose sum is $2\sqrt{3}$ and product is $-9$. Let the numbers be $3\sqrt{3}$ and $-\sqrt{3}$.

$(3\sqrt{3}) \times (-\sqrt{3}) = -3 \times 3 = -9$

$x^2 + 3\sqrt{3}x - \sqrt{3}x - 9 = 0$

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

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

For zeroes, $x - \sqrt{3} = 0 \implies x = \sqrt{3}$ or $x + 3\sqrt{3} = 0 \implies x = -3\sqrt{3}$.

Zeroes: $\sqrt{3}, -3\sqrt{3}$

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

The polynomial is:

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

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

To clear the denominator, let $k = 2\sqrt{5}$:

Factorisation:

We need two numbers whose sum is 3 and product is $2\sqrt{5} \times (-\sqrt{5}) = -10$. The numbers are 5 and $-2$.

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

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

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

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

For zeroes, $\sqrt{5}x - 1 = 0 \implies x = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$ or $2x + \sqrt{5} = 0 \implies x = -\frac{\sqrt{5}}{2}$.

Zeroes: $\frac{\sqrt{5}}{5}, -\frac{\sqrt{5}}{2}$

Given:

Cubic polynomial: $x^3 - 6x^2 + 3x + 10$

Form of zeroes: $a, a + b, a + 2b$

---

To Find:

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

---

Solution:

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

Comparing $x^3 - 6x^2 + 3x + 10$ with the standard form $Ax^3 + Bx^2 + Cx + D$, we have:

$A = 1, B = -6, C = 3, D = 10$

We know the relationship between the zeroes and the coefficients:

1. Sum of zeroes:

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

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

$3a + 3b = 6$

From equation (i), we can say that the middle zero $\beta = 2$.

Also, $b = 2 - a$.

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2. Product of zeroes:

$\alpha \cdot \beta \cdot \gamma = -\frac{D}{A}$

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

Substituting $(a + b) = 2$ from equation (i):

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

Substitute $b = 2 - a$ into equation (ii):

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

$a[a + 4 - 2a] = -5$

$a(4 - a) = -5$

$4a - a^2 = -5$

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

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Solving the quadratic equation for $a$:

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

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

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

So, $a = 5$ or $a = -1$.

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Finding the values of $b$ and the zeroes:

Case I: When $a = 5$

$b = 2 - a = 2 - 5 = -3$

Zeroes are: $a = 5$, $a + b = 2$, $a + 2b = 5 + 2(-3) = -1$.

Case II: When $a = -1$

$b = 2 - a = 2 - (-1) = 3$

Zeroes are: $a = -1$, $a + b = 2$, $a + 2b = -1 + 2(3) = 5$.

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Final Answer:

The values of $a$ and $b$ are $a = 5, b = -3$ or $a = -1, b = 3$.

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

Given:

Cubic polynomial: $p(x) = 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}$

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

---

To Find:

The other two zeroes of the given cubic polynomial.

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Solution:

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

To find the other zeroes, we divide the cubic polynomial $6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}$ by $(x - \sqrt{2})$ using the long division method.

$\begin{array}{r} 6x^2+7\sqrt{2}x+4\phantom{6x^3+\sqrt{2})} \\ x-\sqrt{2} \ {\overline{\smash{\big)}\,6x^3+\sqrt{2}x^2-10x-4\sqrt{2}\phantom{)}}} \\ \underline{-~\phantom{(}(6x^3-6\sqrt{2}x^2)\phantom{-10x-4\sqrt{2})}} \\ 0+7\sqrt{2}x^2-10x-4\sqrt{2}\phantom{)} \\ \underline{-~\phantom{()}(7\sqrt{2}x^2-14x)\phantom{-4\sqrt{2}}} \\ 0+4x-4\sqrt{2}\phantom{)} \\ \underline{-~\phantom{()}(4x-4\sqrt{2})} \\ 0+0\phantom{)} \end{array}$

The other two zeroes of $p(x)$ are the roots of the quadratic quotient $q(x) = 0$.

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Finding zeroes of the quotient:

To find the zeroes, we solve $6x^2 + 7\sqrt{2}x + 4 = 0$ by splitting the middle term.

We need two numbers whose sum is $7\sqrt{2}$ and whose product is $6 \times 4 = 24$.

The numbers are $4\sqrt{2}$ and $3\sqrt{2}$ (Since $4\sqrt{2} + 3\sqrt{2} = 7\sqrt{2}$ and $4\sqrt{2} \times 3\sqrt{2} = 12 \times 2 = 24$).

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

Taking common factors:

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

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

Now, equating each factor to zero:

Case I:

Case II:

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Final Answer:

The other two zeroes of the given cubic polynomial are $-\frac{\sqrt{2}}{2}$ and $-\frac{2\sqrt{2}}{3}$.

Given:

The quadratic polynomial $g(x) = x^2 + 2x + k$ is a factor of the quartic polynomial $p(x) = 2x^4 + x^3 - 14x^2 + 5x + 6$.

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To Find:

1. The value of $k$.

2. All zeroes of the quadratic polynomial $x^2 + 2x + k$.

3. All zeroes of the quartic polynomial $2x^4 + x^3 - 14x^2 + 5x + 6$.

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Solution:

Since $x^2 + 2x + k$ is a factor of $2x^4 + x^3 - 14x^2 + 5x + 6$, the remainder obtained on dividing the quartic polynomial by the quadratic polynomial must be zero.

Let us perform the long division:

$\begin{array}{r} 2x^2-3x+(-8-2k)\phantom{2x^4)} \\ x^2+2x+k{\overline{\smash{\big)}\,2x^4+x^3-14x^2+5x+6\phantom{)}}} \\ \underline{-~\phantom{(}(2x^4+4x^3+2kx^2)\phantom{-14x^2)}} \\ 0-3x^3-(14+2k)x^2+5x+6\phantom{)} \\ \underline{-~\phantom{()}(-3x^3-6x^2-3kx)\phantom{+6)}} \\ 0+(-8-2k)x^2+(5+3k)x+6\phantom{)} \\ \underline{-~\phantom{()}((-8-2k)x^2+(-16-4k)x+(-8k-2k^2))} \\ 0+(21+7k)x+(6+8k+2k^2)\phantom{)} \end{array}$

For $x^2 + 2x + k$ to be a factor, the remainder must be zero. Therefore, the coefficients of the remainder must be zero:

We can verify this value of $k$ with the constant term of the remainder:

$2k^2 + 8k + 6 = 2(-3)^2 + 8(-3) + 6 = 18 - 24 + 6 = 0$.

Thus, the value of $k = -3$.

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Finding Zeroes of the Quadratic Polynomial:

Substituting $k = -3$, the quadratic polynomial is $x^2 + 2x - 3$.

To find its zeroes, we set $x^2 + 2x - 3 = 0$:

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

$x(x + 3) - 1(x + 3) = 0$

$(x - 1)(x + 3) = 0$

So, the zeroes of the quadratic polynomial are $1$ and $-3$.

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Finding All Zeroes of the Quartic Polynomial:

The quartic polynomial can be written as the product of the divisor and the quotient. From the long division, the quotient is $2x^2 - 3x + (-8 - 2k)$.

Substituting $k = -3$ in the quotient:

$Q(x) = 2x^2 - 3x + (-8 - 2(-3))$

$Q(x) = 2x^2 - 3x + (-8 + 6)$

$Q(x) = 2x^2 - 3x - 2$

Now, we find the zeroes of this quotient by factorisation:

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

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

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

This gives $x = -\frac{1}{2}$ and $x = 2$.

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Final Answer:

The value of $k$ is $-3$.

The zeroes of the quadratic polynomial are $1$ and $-3$.

The zeroes of the quartic polynomial are $1, -3, 2,$ and $-\frac{1}{2}$.

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 the given polynomial.

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To Find:

All the zeroes of the cubic polynomial.

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Solution:

Since $(x - \sqrt{5})$ is a factor of $p(x)$, by the Factor Theorem, $\sqrt{5}$ is one of the zeroes of the polynomial.

To find the other zeroes, we divide the polynomial $p(x)$ by the factor $(x - \sqrt{5})$ using long division:

$\begin{array}{r} x^2-2\sqrt{5}x+3\phantom{x^3-3\sqrt{5})} \\ x-\sqrt{2}{\overline{\smash{\big)}\,x^3-3\sqrt{5}x^2+13x-3\sqrt{5}\phantom{)}}} \\ \underline{-~\phantom{(}(x^3-\sqrt{5}x^2)\phantom{-3\sqrt{5}x^2+13x)}} \\ 0-2\sqrt{5}x^2+13x-3\sqrt{5}\phantom{)} \\ \underline{-~\phantom{()}(-2\sqrt{5}x^2+10x)\phantom{-3\sqrt{5}}} \\ 0+3x-3\sqrt{5}\phantom{)} \\ \underline{-~\phantom{()}(3x-3\sqrt{5})} \\ 0+0\phantom{)} \end{array}$

The quotient obtained is a quadratic polynomial $q(x) = x^2 - 2\sqrt{5}x + 3$.

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The other zeroes of $p(x)$ are the zeroes of the quadratic polynomial $q(x)$.

Setting $q(x) = 0$:

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

Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = -2\sqrt{5}$, and $c = 3$.

$x = \frac{-(-2\sqrt{5}) \pm \sqrt{(-2\sqrt{5})^2 - 4(1)(3)}}{2(1)}$

$x = \frac{2\sqrt{5} \pm \sqrt{20 - 12}}{2}$

$x = \frac{2\sqrt{5} \pm \sqrt{8}}{2}$

$x = \frac{2\sqrt{5} \pm 2\sqrt{2}}{2}$

Dividing numerator and denominator by 2:

$x = \sqrt{5} \pm \sqrt{2}$

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Final Answer:

The zeroes of the cubic polynomial are $\sqrt{5}$, $\sqrt{5} + \sqrt{2}$, and $\sqrt{5} - \sqrt{2}$.

Given:

Polynomial $q(x) = x^3 + 2x^2 + a$

Polynomial $p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b$

The zeroes of $q(x)$ are also the zeroes of $p(x)$.

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To Find:

1. The values of $a$ and $b$.

2. The zeroes of $p(x)$ that are not the zeroes of $q(x)$.

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Solution:

If the zeroes of $q(x)$ are also the zeroes of $p(x)$, then the polynomial $q(x)$ must be a factor of the polynomial $p(x)$. This means when we divide $p(x)$ by $q(x)$, the remainder must be zero.

Let us perform the long division of $p(x)$ by $q(x)$:

$\begin{array}{r} x^2-3x+2\phantom{x^3+2x^2+a)} \\ x^3+2x^2+a{\overline{\smash{\big)}\,x^5-x^4-4x^3+3x^2+3x+b\phantom{)}}} \\ \underline{-~\phantom{(}(x^5+2x^4+ax^2)\phantom{-4x^3--)}} \\ -3x^4-4x^3+(3-a)x^2+3x+b\phantom{)} \\ \underline{-~\phantom{()}(-3x^4-6x^3-3ax)} \\ 2x^3+(3-a)x^2+(3+3a)x+b\phantom{)} \\ \underline{-~\phantom{()}(2x^3+4x^2+2a)} \\ (-1-a)x^2+(3+3a)x+(b-2a)\phantom{)} \end{array}$

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For $q(x)$ to be a factor of $p(x)$, the remainder must be identically zero for all values of $x$. Therefore, the coefficients of the remainder must be zero.

From the coefficient of $x^2$:

From the constant term of the remainder:

Substituting $a = -1$ in equation (ii):

$b - 2(-1) = 0$

$b + 2 = 0$

We can verify the coefficient of $x$: $3 + 3a = 3 + 3(-1) = 0$. This is consistent.

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Finding zeroes of $p(x)$ that are not zeroes of $q(x)$:

The zeroes of $p(x)$ that are not zeroes of $q(x)$ are the zeroes of the quotient polynomial obtained from the division.

The quotient is $g(x) = x^2 - 3x + 2$.

To find its zeroes, we set $g(x) = 0$ and factorise it by splitting the middle term:

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

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

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

This gives $x = 1$ and $x = 2$.

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Final Answer:

The values are $a = -1$ and $b = -2$.

The zeroes of $p(x)$ which are not zeroes of $q(x)$ are $1$ and $2$.