Chapter 2 Polynomials (Additional Questions)

A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. The general form of a polynomial in one variable $x$ is $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $a_i$ are coefficients and $n$ is a non-negative integer.

Let's examine each given expression:

(A) $x^3 + \sqrt{x} - 4$ can be written as $x^3 + x^{1/2} - 4$. The term $x^{1/2}$ has an exponent $\frac{1}{2}$, which is not a non-negative integer. Therefore, this expression is not a polynomial.

(B) $\frac{1}{x} + x^2 + 2$ can be written as $x^{-1} + x^2 + 2$. The term $x^{-1}$ has an exponent $-1$, which is a negative integer. Therefore, this expression is not a polynomial.

(C) $y^4 - 5y^{1/2} + 1$. The term $y^{1/2}$ has an exponent $\frac{1}{2}$, which is not a non-negative integer. Therefore, this expression is not a polynomial.

(D) $2z^5 - 3z^2 + \frac{1}{2}$. The terms $2z^5$, $-3z^2$, and $\frac{1}{2}$ have variables raised to the powers 5, 2, and 0 (as $\frac{1}{2} = \frac{1}{2} z^0$), respectively. These exponents (5, 2, 0) are all non-negative integers. The coefficients ($2, -3, \frac{1}{2}$) are constants. The operations involved are subtraction and addition. This expression fits the definition of a polynomial.

Based on the analysis, the expression $2z^5 - 3z^2 + \frac{1}{2}$ is a polynomial.

The correct option is (D) $2z^5 - 3z^2 + \frac{1}{2}$.

The degree of a polynomial is the highest power of the variable in the polynomial. For a polynomial in one variable, such as $5x^4 - 3x^5 + 7x - 1$, we look at the exponents of the variable in each term.

The given polynomial is $5x^4 - 3x^5 + 7x - 1$. We can rewrite this polynomial in standard form (descending powers of $x$) as $-3x^5 + 5x^4 + 7x^1 - 1x^0$.

Let's identify the terms and their corresponding degrees:

Term 1: $5x^4$, the power of $x$ is 4.

Term 2: $-3x^5$, the power of $x$ is 5.

Term 3: $7x$, which is $7x^1$, the power of $x$ is 1.

Term 4: $-1$, which is $-1x^0$, the power of $x$ is 0.

The degrees of the terms are 4, 5, 1, and 0. The highest power among these is 5.

Therefore, the degree of the polynomial $5x^4 - 3x^5 + 7x - 1$ is 5.

The correct option is (B) 5.

We are given the polynomial $p(x) = x^2 - 3x + 2$. We need to find the value of $p(-1)$. This means we need to substitute $x = -1$ into the expression for $p(x)$.

Substitute $x = -1$ into $p(x)$:

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

Now, evaluate the expression:

$(-1)^2 = (-1) \times (-1) = 1$

$-3(-1) = -3 \times (-1) = 3$

So, the expression becomes:

$p(-1) = 1 + 3 + 2$

$p(-1) = 6$

The value of $p(-1)$ is 6.

The correct option is (B) 6.

A linear polynomial is a polynomial of degree 1. Its general form is given by $p(x) = ax + b$, where $a$ and $b$ are real numbers and $a \neq 0$.

A zero of a polynomial $p(x)$ is a value of $x$ for which $p(x) = 0$.

To find the zero(s) of a linear polynomial $p(x) = ax + b$, we set $p(x) = 0$ and solve for $x$:

$ax + b = 0$

Subtract $b$ from both sides:

$ax = -b$

Since $a \neq 0$, we can divide by $a$:

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

Since $a$ and $b$ are real numbers and $a \neq 0$, the value $-\frac{b}{a}$ is a unique real number. This means that a linear polynomial has one and only one zero.

Therefore, a linear polynomial has exactly one zero.

The correct option is (A) Exactly one.

A zero of a polynomial $p(x)$ is a value of $x$ for which the value of the polynomial is 0. To find the zero(s) of the polynomial $p(x) = 5x - 10$, we need to set $p(x)$ equal to zero and solve the resulting equation for $x$.

Set the polynomial equal to 0:

$p(x) = 0$

$5x - 10 = 0$

Now, we solve this linear equation for $x$. Add 10 to both sides of the equation:

$5x - 10 + 10 = 0 + 10$

$5x = 10$

Divide both sides by 5:

$\frac{5x}{5} = \frac{10}{5}$

$x = 2$

The value of $x$ for which the polynomial $p(x) = 5x - 10$ becomes zero is $x = 2$. Therefore, the zero of the polynomial is 2.

We can verify this by substituting $x=2$ into $p(x)$: $p(2) = 5(2) - 10 = 10 - 10 = 0$.

The correct option is (A) 2.

The graph of a polynomial equation $y = p(x)$ shows the relationship between the input $x$ and the output $y$. The shape of the graph depends on the degree of the polynomial.

We are given the polynomial equation $y = ax + b$, where $a \neq 0$. This is the equation of a linear polynomial because the highest power of the variable $x$ is 1.

The equation $y = ax + b$ is in the form $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept. In this case, $m=a$ and $c=b$.

The graph of any equation of the form $y = mx + c$ is a straight line.

The condition $a \neq 0$ ensures that the line is not horizontal (which would be the case if $a=0$, resulting in $y=b$, a constant polynomial).

Therefore, the graph of a linear polynomial $y = ax + b$ (where $a \neq 0$) is a straight line.

The correct option is (B) Straight line.

For a quadratic polynomial of the form $ax^2 + bx + c$, where $a$, $b$, and $c$ are coefficients and $a \neq 0$, there is a relationship between the zeros (roots) of the polynomial and its coefficients.

Let the zeros of the quadratic polynomial $ax^2 + bx + c$ be $\alpha$ and $\beta$. According to Vieta's formulas (or the relationships between zeros and coefficients), the sum and product of the zeros are given by:

The sum of the zeros ($\alpha + \beta$) is equal to the negative of the coefficient of $x$ divided by the coefficient of $x^2$.

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

The product of the zeros ($\alpha \beta$) is equal to the constant term divided by the coefficient of $x^2$.

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

The question asks for the sum of the zeros, which is $\alpha + \beta$. From the relationships, we know that $\alpha + \beta = -\frac{b}{a}$.

Comparing this with the given options, we find that option (B) matches our result.

The correct option is (B) $-\frac{b}{a}$.

For a quadratic polynomial in the standard form $ax^2 + bx + c$, where $a$, $b$, and $c$ are coefficients and $a \neq 0$, there are well-defined relationships between the zeros (also known as roots) of the polynomial and its coefficients. Let the zeros of the polynomial be $\alpha$ and $\beta$.

The sum of the zeros ($\alpha + \beta$) is given by the formula:

$\alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{b}{a}$

The product of the zeros ($\alpha \cdot \beta$) is given by the formula:

$\alpha \cdot \beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{c}{a}$

The question asks for the product of the zeros, $\alpha \cdot \beta$. According to the relationship between zeros and coefficients, the product of the zeros is $\frac{c}{a}$.

Comparing this result with the given options, we find that option (A) matches our result.

The correct option is (A) $\frac{c}{a}$.

For a quadratic polynomial in the standard form $ax^2 + bx + c$, where $a$, $b$, and $c$ are coefficients and $a \neq 0$, the sum of its zeros (roots) is given by a specific formula related to the coefficients.

The given polynomial is $x^2 - 5x + 6$. We need to compare this polynomial with the standard form $ax^2 + bx + c$ to identify the coefficients $a$, $b$, and $c$.

Comparing $x^2 - 5x + 6$ with $ax^2 + bx + c$:

Coefficient of $x^2$ is $a$. In the given polynomial, the coefficient of $x^2$ is 1 (since $x^2 = 1 \cdot x^2$). So, $a = 1$.

Coefficient of $x$ is $b$. In the given polynomial, the coefficient of $x$ is -5. So, $b = -5$.

The constant term is $c$. In the given polynomial, the constant term is 6. So, $c = 6$.

The formula for the sum of the zeros of a quadratic polynomial $ax^2 + bx + c$ is:

Sum of zeros $= -\frac{b}{a}$

Now, substitute the values of $a$ and $b$ that we found into the formula:

Sum of zeros $= -\frac{-5}{1}$

Sum of zeros $= -(-5)$

Sum of zeros $= 5$

Thus, the sum of the zeros of the quadratic polynomial $x^2 - 5x + 6$ is 5.

The correct option is (A) 5.

For a quadratic polynomial in the standard form $ax^2 + bx + c$, where $a$, $b$, and $c$ are real coefficients and $a \neq 0$, there are specific relationships between the zeros (roots) of the polynomial and its coefficients. Let the zeros be $\alpha$ and $\beta$.

The given polynomial is $2x^2 - 7x + 3$. We compare this with the standard form $ax^2 + bx + c$ to find the values of $a$, $b$, and $c$.

Comparing, we get:

$a = 2$

$b = -7$

$c = 3$

The product of the zeros ($\alpha \cdot \beta$) of a quadratic polynomial $ax^2 + bx + c$ is given by the formula:

Product of zeros $= \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{c}{a}$

Substitute the values of $c$ and $a$ we found into this formula:

Product of zeros $= \frac{3}{2}$

Thus, the product of the zeros of the quadratic polynomial $2x^2 - 7x + 3$ is $\frac{3}{2}$.

The correct option is (C) $\frac{3}{2}$.

A quadratic polynomial can be expressed in terms of its zeros. If $\alpha$ and $\beta$ are the zeros of a quadratic polynomial, then the polynomial can be written in the form $k(x - \alpha)(x - \beta)$, where $k$ is a non-zero constant.

Expanding the expression $k(x - \alpha)(x - \beta)$, we get:

$p(x) = k(x^2 - \beta x - \alpha x + \alpha \beta)$

$p(x) = k(x^2 - (\alpha + \beta)x + \alpha \beta)$

We are given that the sum of the zeros is 4 and the product of the zeros is 1.

Sum of zeros: $\alpha + \beta = 4$

Product of zeros: $\alpha \beta = 1$

Substitute these values into the polynomial form:

$p(x) = k(x^2 - (4)x + (1))$

$p(x) = k(x^2 - 4x + 1)$

Since the question asks for *a* quadratic polynomial, we can choose a simple value for $k$. Let's take $k=1$.

For $k=1$, the polynomial is $p(x) = 1 \cdot (x^2 - 4x + 1) = x^2 - 4x + 1$.

Now, let's check the given options:

(A) $x^2 + 4x + 1$: Sum of zeros $= -\frac{4}{1} = -4$. Product of zeros $= \frac{1}{1} = 1$. (Incorrect sum)

(B) $x^2 - 4x + 1$: Sum of zeros $= -\frac{-4}{1} = 4$. Product of zeros $= \frac{1}{1} = 1$. (Correct sum and product)

(C) $x^2 + x + 4$: Sum of zeros $= -\frac{1}{1} = -1$. Product of zeros $= \frac{4}{1} = 4$. (Incorrect sum and product)

(D) $x^2 - x + 4$: Sum of zeros $= -\frac{-1}{1} = 1$. Product of zeros $= \frac{4}{1} = 4$. (Incorrect sum and product)

The polynomial $x^2 - 4x + 1$ has a sum of zeros equal to 4 and a product of zeros equal to 1.

The correct option is (B) $x^2 - 4x + 1$.

The Division Algorithm for polynomials states that for any two polynomials $p(x)$ and $g(x)$, where $g(x)$ is a non-zero polynomial, there exist unique polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that:

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

This algorithm provides a specific condition on the remainder $r(x)$. The condition is that either the remainder $r(x)$ is the zero polynomial ($r(x) = 0$), or the degree of the remainder polynomial $r(x)$ is strictly less than the degree of the divisor polynomial $g(x)$.

In mathematical notation, this condition is stated as:

$r(x) = 0$ or $\text{degree}(r(x)) < \text{degree}(g(x))$

Let's consider the given options in light of this condition:

(A) degree of $r(x) <$ degree of $g(x)$. This matches the second part of the condition. If $r(x) = 0$, its degree is often taken as $-\infty$, which is less than the degree of any non-zero polynomial $g(x)$. So this option correctly captures the relationship when a remainder exists and is implicitly understood to cover the $r(x)=0$ case by the convention on the degree of the zero polynomial.

(B) degree of $r(x) >$ degree of $g(x)$. This is incorrect. The remainder's degree must be smaller than the divisor's degree.

(C) degree of $r(x) =$ degree of $g(x)$. This is incorrect. If the degrees were equal or the remainder's degree was greater, further division would be possible.

(D) $r(x) = 0$ always. This is incorrect. The remainder $r(x)$ is zero only when $g(x)$ is a factor of $p(x)$ (i.e., the division is exact).

Therefore, the correct condition on the remainder $r(x)$ is that its degree is less than the degree of $g(x)$, or $r(x)$ is the zero polynomial.

The option that best represents this condition is (A).

The correct option is (A) degree of $r(x) <$ degree of $g(x)$.

We are asked to find the remainder when the polynomial $p(x) = x^2 - 4$ is divided by the polynomial $g(x) = x - 2$. We can use either polynomial long division or the Remainder Theorem to find the remainder.

Method 1: Using the Remainder Theorem

The Remainder Theorem states that if a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, then the remainder is equal to $p(a)$.

In this case, the dividend is $p(x) = x^2 - 4$ and the divisor is $g(x) = x - 2$. This divisor is in the form $(x - a)$, where $a = 2$.

According to the Remainder Theorem, the remainder is $p(2)$.

Substitute $x = 2$ into $p(x) = x^2 - 4$:

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

$p(2) = 4 - 4$

$p(2) = 0$

The remainder when $x^2 - 4$ is divided by $x - 2$ is 0.

Method 2: Using Polynomial Long Division

We can perform long division to divide $x^2 - 4$ by $x - 2$. Write the dividend as $x^2 + 0x - 4$ to include the $x$ term with a zero coefficient.

Divide the leading term of the dividend ($x^2$) by the leading term of the divisor ($x$): $\frac{x^2}{x} = x$. This is the first term of the quotient.

Multiply the divisor ($x - 2$) by the first term of the quotient ($x$): $x(x - 2) = x^2 - 2x$.

Subtract this result from the dividend: $(x^2 + 0x - 4) - (x^2 - 2x) = x^2 + 0x - 4 - x^2 + 2x = 2x - 4$.

Now, divide the leading term of the new dividend ($2x$) by the leading term of the divisor ($x$): $\frac{2x}{x} = 2$. This is the second term of the quotient.

Multiply the divisor ($x - 2$) by the second term of the quotient ($2$): $2(x - 2) = 2x - 4$.

Subtract this result from the current dividend: $(2x - 4) - (2x - 4) = 2x - 4 - 2x + 4 = 0$.

The long division process is shown below:

$\begin{array}{r} x+2\phantom{)} \\ x-2{\overline{\smash{\big)}\,x^2+0x-4\phantom{)}}} \\ \underline{-~\phantom{(}(x^2-2x)\phantom{-b)}} \\ 0+2x-4\phantom{)} \\ \underline{-~\phantom{()}(2x-4)} \\ 0+0\phantom{)} \end{array}$

The remainder is 0.

Both methods show that the remainder is 0.

The correct option is (A) 0.

The graphical representation of a polynomial $y = p(x)$ is the set of all points $(x, y)$ in the coordinate plane such that $y$ is equal to the value of the polynomial $p(x)$ at $x$. For a quadratic polynomial, we are looking at the graph of the equation $y = ax^2 + bx + c$, where $a \neq 0$.

The graph of any equation of the form $y = ax^2 + bx + c$ where $a \neq 0$ is a parabola. A parabola is a symmetric, U-shaped curve.

If $a > 0$, the parabola opens upwards (like a U).

If $a < 0$, the parabola opens downwards (like an inverted U).

Let's consider why the other options are incorrect:

(A) A straight line is the graph of a linear polynomial (degree 1), $y = ax + b$ with $a \neq 0$.

(C) A circle is represented by an equation like $(x-h)^2 + (y-k)^2 = r^2$, which is not a polynomial function $y = p(x)$.

(D) A hyperbola is represented by equations like $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ or $xy = k$, which are also not polynomial functions of the form $y = p(x)$.

Therefore, the graphical representation of a quadratic polynomial is always a parabola.

The correct option is (B) Parabola.

A zero of a polynomial $p(x)$ is a value of $x$ for which the polynomial evaluates to zero, i.e., $p(x) = 0$. To find the zeros of the polynomial $p(x) = (x-1)(x-2)$, we set the polynomial equal to zero.

$p(x) = 0$

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

For the product of two factors to be zero, at least one of the factors must be equal to zero. So, we have two possible cases:

Case 1: The first factor is zero.

$x - 1 = 0$

Adding 1 to both sides, we get:

$x = 1$

Case 2: The second factor is zero.

$x - 2 = 0$

Adding 2 to both sides, we get:

$x = 2$

The values of $x$ for which $p(x) = 0$ are $x = 1$ and $x = 2$. These are the zeros of the polynomial.

The polynomial $p(x) = (x-1)(x-2)$ has two distinct zeros, which are 1 and 2.

Note: If we expand the polynomial, we get $p(x) = x^2 - 2x - x + 2 = x^2 - 3x + 2$. This is a quadratic polynomial (degree 2), and a polynomial of degree $n$ has at most $n$ real zeros.

Therefore, the polynomial $p(x) = (x-1)(x-2)$ has exactly two zeros.

The correct option is (C) 2.

A zero of a polynomial $p(x)$ is a value of the variable $x$ for which the value of the polynomial is equal to 0. In other words, if 'a' is a zero of $p(x)$, then $p(a) = 0$.

We are given the quadratic polynomial $p(x) = kx^2 + 3x + k$, and we are told that one of its zeros is 2. This means that when we substitute $x = 2$ into the polynomial, the result must be 0.

Substitute $x = 2$ into the polynomial $p(x)$:

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

Simplify the expression:

$p(2) = k(4) + 6 + k$

$p(2) = 4k + 6 + k$

$p(2) = 5k + 6$

Since 2 is a zero of the polynomial, $p(2)$ must be equal to 0:

$5k + 6 = 0$

Now, solve this linear equation for $k$. Subtract 6 from both sides of the equation:

$5k = -6$

Divide both sides by 5:

$k = -\frac{6}{5}$

Thus, the value of $k$ is $-\frac{6}{5}$.

The correct option is (B) $-\frac{6}{5}$.

The graph of a quadratic polynomial $y = ax^2 + bx + c$, where $a \neq 0$, is a parabola. Let's analyze each statement provided:

(A) If $a > 0$, the parabola opens upwards. This statement is TRUE. The coefficient 'a' determines the concavity of the parabola. A positive 'a' value results in a parabola that opens upwards, having a minimum point.

(B) If $a < 0$, the parabola opens downwards. This statement is also TRUE. A negative 'a' value results in a parabola that opens downwards, having a maximum point.

(C) The zeros of the polynomial are the y-coordinates where the graph intersects the x-axis. A zero of the polynomial $p(x)$ is a value of $x$ such that $p(x) = 0$. On the graph of $y = p(x)$, these are the points where the curve crosses or touches the x-axis. The points on the x-axis have a y-coordinate of 0. The zeros of the polynomial are the x-coordinates of these intersection points, not the y-coordinates. The y-coordinate at these points is always 0. Therefore, this statement is FALSE.

(D) The number of zeros is at most 2. A quadratic polynomial has a degree of 2. According to the Fundamental Theorem of Algebra, a polynomial of degree $n$ has exactly $n$ complex roots (counting multiplicity). For real polynomials, the number of real zeros (which correspond to the x-intercepts) can be 0, 1 (if the parabola is tangent to the x-axis), or 2 (if the parabola intersects the x-axis at two distinct points). Thus, the maximum number of real zeros for a quadratic polynomial is 2. This statement is TRUE.

Based on the analysis, the statement that is FALSE is (C).

The correct option is (C) The zeros of the polynomial are the y-coordinates where the graph intersects the x-axis.

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): A polynomial of degree 3 can have at most three zeros.

The degree of a polynomial is the highest power of the variable in the polynomial. A polynomial of degree $n$ can have at most $n$ real zeros. A polynomial of degree 3 can have 1, 2, or 3 real zeros (counting multiplicity). It is also true that a polynomial of degree 3 has exactly 3 roots in the complex number system (Fundamental Theorem of Algebra), counting multiplicity. However, when discussing zeros in the context of graphs intersecting the x-axis (which are real zeros), a polynomial of degree 3 can intersect the x-axis at most 3 times.

Therefore, Assertion (A) is True.

Reason (R): The maximum number of real roots a polynomial can have is equal to its degree.

For a polynomial of degree $n$, it is a fundamental property that it can have no more than $n$ distinct real roots. It is possible for a polynomial of degree $n$ to have exactly $n$ real roots (e.g., a polynomial that can be factored into $n$ distinct linear factors like $(x-r_1)(x-r_2)...(x-r_n)$). It is also possible for a polynomial of degree $n$ to have fewer than $n$ real roots (e.g., $x^2+1$ has degree 2 but zero real roots). The statement says the *maximum* number is equal to its degree, which is a correct statement.

Therefore, Reason (R) is True.

Now let's consider if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states a specific case for degree 3. Reason (R) states a general principle that applies to any polynomial degree, including degree 3. Since the maximum number of real roots for a polynomial of degree 3 is limited by its degree, Reason (R) provides the underlying principle that explains why Assertion (A) is true.

Both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Let's evaluate the truthfulness of the Assertion (A) and the Reason (R).

Assertion (A): If the sum of the zeros of a quadratic polynomial $ax^2+bx+c$ is 0, then $b=0$.

Let the quadratic polynomial be $p(x) = ax^2 + bx + c$. A quadratic polynomial has a degree of 2, which means $a \neq 0$. Let the zeros of the polynomial be $\alpha$ and $\beta$. The assertion states that if $\alpha + \beta = 0$, then $b = 0$. We know the relationship between the sum of zeros and the coefficients of a quadratic polynomial.

This statement depends on the formula for the sum of zeros. Let's evaluate the reason first.

Reason (R): The sum of the zeros is given by $-\frac{b}{a}$.

For a quadratic polynomial $ax^2 + bx + c$, the sum of the zeros ($\alpha + \beta$) is indeed given by the formula $-\frac{b}{a}$. This is a standard result derived from Vieta's formulas relating the roots and coefficients of a polynomial.

Therefore, Reason (R) is True.

Now, let's re-evaluate Assertion (A) using the fact that Reason (R) is true.

We are given that the sum of the zeros is 0. So, $\alpha + \beta = 0$.

From Reason (R), we know that $\alpha + \beta = -\frac{b}{a}$.

Equating these two expressions for the sum of zeros:

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

Since $ax^2 + bx + c$ is a quadratic polynomial, the coefficient of $x^2$, $a$, cannot be zero ($a \neq 0$).

For the fraction $-\frac{b}{a}$ to be equal to 0, and knowing that the denominator $a$ is not zero, the numerator $b$ must be equal to 0.

$b = 0$

Thus, if the sum of the zeros of a quadratic polynomial $ax^2+bx+c$ is 0, then $b=0$. This confirms that Assertion (A) is True.

Finally, let's check if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) concludes that $b=0$ if the sum of zeros is 0. This conclusion is directly derived by using the formula for the sum of zeros provided in Reason (R) and setting it equal to 0, along with the condition $a \neq 0$. Therefore, Reason (R) provides the necessary formula to explain why Assertion (A) is true.

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

We need to match each type of polynomial based on its degree with its general form.

(i) Linear Polynomial: A polynomial of degree 1. Its general form is $ax+b$, where $a$ is the coefficient of $x$ and $a \neq 0$ (otherwise, it would be a constant polynomial). The term with the highest power of $x$ is $ax^1$. This matches option (c) $ax+b, a \neq 0$.

(ii) Quadratic Polynomial: A polynomial of degree 2. Its general form is $ax^2+bx+c$, where $a$ is the coefficient of $x^2$ and $a \neq 0$. The term with the highest power of $x$ is $ax^2$. This matches option (d) $ax^2+bx+c, a \neq 0$.

(iii) Cubic Polynomial: A polynomial of degree 3. Its general form is $ax^3+bx^2+cx+d$, where $a$ is the coefficient of $x^3$ and $a \neq 0$. The term with the highest power of $x$ is $ax^3$. This matches option (a) $ax^3+bx^2+cx+d, a \neq 0$.

(iv) Constant Polynomial: A polynomial of degree 0 (for a non-zero constant). It is simply a constant value, say $k$. This can be written as $k \cdot x^0$. This matches option (b) $k$ (where $k$ is a constant).

Summarizing the matches:

(i) Linear Polynomial $\to$ (c) $ax+b, a \neq 0$

(ii) Quadratic Polynomial $\to$ (d) $ax^2+bx+c, a \neq 0$

(iii) Cubic Polynomial $\to$ (a) $ax^3+bx^2+cx+d, a \neq 0$

(iv) Constant Polynomial $\to$ (b) $k$ (where $k$ is a constant)

Comparing these matches with the given options, we find that option (A) provides the correct pairing.

The correct option is (A) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b).

The problem provides a case study where the company's profit $P(x)$ from selling $x$ units is given by the quadratic polynomial $P(x) = -x^2 + 10x - 21$. We are told that the zeros of this polynomial represent the break-even points. The break-even points are the values of $x$ where the profit $P(x)$ is zero.

To find the break-even points, we need to find the values of $x$ for which $P(x) = 0$. So, we set the polynomial equal to zero and solve for $x$:

$-x^2 + 10x - 21 = 0$

To make the coefficient of $x^2$ positive, we can multiply the entire equation by -1:

$(-1)(-x^2 + 10x - 21) = (-1)(0)$

$x^2 - 10x + 21 = 0$

Now we need to solve this quadratic equation. We can factor the quadratic expression $x^2 - 10x + 21$. We look for two numbers that multiply to 21 and add up to -10. These numbers are -3 and -7 ($(-3) \times (-7) = 21$ and $(-3) + (-7) = -10$).

So, the factored form of the equation is:

$(x - 3)(x - 7) = 0$

For the product of two factors to be zero, at least one of the factors must be zero. Thus, we have two possible cases:

Case 1:

$x - 3 = 0$

Adding 3 to both sides:

$x = 3$

Case 2:

$x - 7 = 0$

Adding 7 to both sides:

$x = 7$

The zeros of the polynomial $P(x) = -x^2 + 10x - 21$ are $x = 3$ and $x = 7$. These values of $x$ represent the number of units sold at which the profit is zero.

Therefore, the break-even points for the company are 3 and 7.

The correct option is (A) 3 and 7.

A quadratic polynomial can be constructed if the sum and product of its zeros are known. If $\alpha$ and $\beta$ are the zeros of a quadratic polynomial, then the polynomial can be expressed in the form:

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

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

We are given that:

Sum of zeros $= -\frac{1}{4}$

Product of zeros $= \frac{1}{4}$

Substitute these given values into the formula:

$p(x) = k\left(x^2 - \left(-\frac{1}{4}\right)x + \frac{1}{4}\right)$

$p(x) = k\left(x^2 + \frac{1}{4}x + \frac{1}{4}\right)$

The options provided have integer coefficients. To obtain integer coefficients from the expression $x^2 + \frac{1}{4}x + \frac{1}{4}$, we can choose a suitable value for the constant $k$ that clears the denominators. The least common multiple of the denominators (which are 4) is 4. So, let's choose $k = 4$.

Substitute $k = 4$ into the polynomial expression:

$p(x) = 4\left(x^2 + \frac{1}{4}x + \frac{1}{4}\right)$

Distribute the 4 to each term inside the parenthesis:

$p(x) = 4 \cdot x^2 + 4 \cdot \frac{1}{4}x + 4 \cdot \frac{1}{4}$

$p(x) = 4x^2 + 1x + 1$

$p(x) = 4x^2 + x + 1$

This resulting polynomial is $4x^2 + x + 1$. Now, let's compare this with the given options:

(A) $4x^2 + x + 1$

(B) $4x^2 - x + 1$

(C) $4x^2 + x - 1$

(D) $4x^2 - x - 1$

The polynomial we derived matches option (A).

The correct option is (A) $4x^2 + x + 1$.

We are asked to find the remainder when the polynomial $p(x) = x^3 - 2x^2 + 3x - 4$ is divided by the linear polynomial $(x-2)$. We can use the Remainder Theorem to find the remainder efficiently.

The Remainder Theorem states that if a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, then the remainder is equal to $p(a)$.

In this problem, the dividend polynomial is $p(x) = x^3 - 2x^2 + 3x - 4$.

The divisor linear polynomial is $(x - 2)$.

Comparing the divisor $(x - 2)$ with the general form $(x - a)$, we can see that $a = 2$.

According to the Remainder Theorem, the remainder when $p(x)$ is divided by $(x-2)$ is $p(2)$.

Now, we need to substitute $x = 2$ into the expression for $p(x)$ and calculate the value:

$p(2) = (2)^3 - 2(2)^2 + 3(2) - 4$

Let's evaluate each term:

$(2)^3 = 2 \times 2 \times 2 = 8$

$(2)^2 = 2 \times 2 = 4$. So, $2(2)^2 = 2(4) = 8$.

$3(2) = 3 \times 2 = 6$

The constant term is $-4$.

Substitute these values back into the expression for $p(2)$:

$p(2) = 8 - 8 + 6 - 4$

$p(2) = 0 + 6 - 4$

$p(2) = 2$

The value of $p(2)$ is 2.

Therefore, the remainder when $p(x) = x^3 - 2x^2 + 3x - 4$ is divided by $(x-2)$ is 2.

The correct option is (B) 2.

The zeros of a polynomial $p(x)$ are the values of $x$ for which $p(x) = 0$. Graphically, the zeros of a polynomial are the x-coordinates of the points where the graph of $y = p(x)$ intersects or touches the x-axis. The number of zeros of the polynomial is equal to the number of times its graph intersects or touches the x-axis.

Looking at the provided graph (as described in the image-container alt text), the graph of the polynomial $p(x)$ intersects the x-axis at three distinct points.

Since the graph intersects the x-axis at 3 points, the polynomial $p(x)$ has 3 zeros.

The correct option is (C) 3.

We are given the polynomial $p(x) = x^2 - 5x + 6$ and asked to identify its zeros from the given options. A zero of a polynomial $p(x)$ is a value of $x$ for which $p(x) = 0$. To check if a value is a zero, we substitute the value into the polynomial and see if the result is 0.

Let's test each option:

Option (A) $x=1$

Substitute $x=1$ into $p(x)$:

$p(1) = (1)^2 - 5(1) + 6$

$p(1) = 1 - 5 + 6$

$p(1) = -4 + 6$

$p(1) = 2$

Since $p(1) \neq 0$, 1 is not a zero of the polynomial.

Option (B) $x=2$

Substitute $x=2$ into $p(x)$:

$p(2) = (2)^2 - 5(2) + 6$

$p(2) = 4 - 10 + 6$

$p(2) = -6 + 6$

$p(2) = 0$

Since $p(2) = 0$, 2 is a zero of the polynomial.

Option (C) $x=3$

Substitute $x=3$ into $p(x)$:

$p(3) = (3)^2 - 5(3) + 6$

$p(3) = 9 - 15 + 6$

$p(3) = -6 + 6$

$p(3) = 0$

Since $p(3) = 0$, 3 is a zero of the polynomial.

Option (D) $x=0$

Substitute $x=0$ into $p(x)$:

$p(0) = (0)^2 - 5(0) + 6$

$p(0) = 0 - 0 + 6$

$p(0) = 6$

Since $p(0) \neq 0$, 0 is not a zero of the polynomial.

Alternatively, we can find the zeros by factoring the quadratic polynomial $x^2 - 5x + 6$. We look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3 (since $(-2) \times (-3) = 6$ and $(-2) + (-3) = -5$).

So, $p(x)$ can be factored as:

$p(x) = (x - 2)(x - 3)$

To find the zeros, we set $p(x) = 0$:

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

This gives us two possible solutions:

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

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

The zeros of the polynomial are 2 and 3.

Comparing these zeros with the given options, the values that are zeros of the polynomial are 2 and 3.

The correct options are (B) 2 and (C) 3.

We are given the quadratic polynomial $f(x) = x^2 + x + 1$. We are told that $\alpha$ and $\beta$ are the zeros of this polynomial. For a quadratic polynomial in the standard form $ax^2 + bx + c$, the relationships between the zeros and the coefficients are:

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

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

Comparing the given polynomial $f(x) = x^2 + x + 1$ with the standard form $ax^2 + bx + c$, we identify the coefficients:

$a = 1$ (coefficient of $x^2$)

$b = 1$ (coefficient of $x$)

$c = 1$ (constant term)

Now, we can find the sum and product of the zeros:

Sum of zeros: $\alpha + \beta = -\frac{b}{a} = -\frac{1}{1} = -1$

Product of zeros: $\alpha \beta = \frac{c}{a} = \frac{1}{1} = 1$

We need to find the value of the expression $\frac{1}{\alpha} + \frac{1}{\beta}$. To simplify this expression, we can find a common denominator, which is $\alpha \beta$:

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta}{\alpha \beta} + \frac{\alpha}{\alpha \beta} = \frac{\beta + \alpha}{\alpha \beta}$

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

Now, substitute the values we found for the sum of zeros ($\alpha + \beta = -1$) and the product of zeros ($\alpha \beta = 1$) into this expression:

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

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

The value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is -1.

The correct option is (B) -1.

The zeros of a polynomial $p(x)$ are the values of $x$ for which $p(x) = 0$. Graphically, these are the x-coordinates of the points where the graph of $y = p(x)$ intersects or touches the x-axis. The number of real zeros corresponds to the number of such points.

The graph of a quadratic polynomial $y = ax^2 + bx + c$ (where $a \neq 0$) is a parabola. A parabola can interact with the x-axis in one of three ways:

1. It can intersect the x-axis at two distinct points. In this case, the quadratic polynomial has two distinct real zeros.

2. It can not intersect the x-axis at all. In this case, the quadratic polynomial has no real zeros (it has two complex conjugate zeros).

3. It can touch the x-axis at exactly one point. In this case, the vertex of the parabola lies on the x-axis. This happens when the quadratic equation $ax^2 + bx + c = 0$ has exactly one real root, which is a root with multiplicity 2 (a repeated root). While it is a single point of contact, the zero itself is considered to have a multiplicity of 2.

The question states that the graph touches the x-axis at exactly one point. This corresponds to the third case described above.

This means the quadratic polynomial has exactly one real value of $x$ for which $p(x) = 0$, and this zero is repeated.

Therefore, if the graph of a quadratic polynomial touches the x-axis at exactly one point, it represents a quadratic polynomial with exactly one real zero (which is a repeated zero).

The correct option is (B) Exactly one (repeated).

A zero of a polynomial $p(x)$ is a value of the variable $x$ for which the value of the polynomial is 0. To find the zeros of a polynomial, we set $p(x) = 0$ and solve for $x$.

We are given the constant polynomial $p(x) = 5$. This polynomial has a constant value of 5, regardless of the value of $x$.

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

$5 = 0$

The equation $5 = 0$ is a false statement. There is no value of $x$ that can make the constant polynomial $p(x) = 5$ equal to zero. Therefore, the polynomial $p(x) = 5$ has no zeros.

Note that the zero polynomial, $p(x) = 0$, is a special case. For the zero polynomial, $p(x)=0$ for every value of $x$, so every real number is a zero of the zero polynomial. However, the given polynomial is $p(x)=5$, which is a non-zero constant polynomial.

Thus, the constant polynomial $p(x) = 5$ has no zero.

The correct option is (D) No zero.

The problem describes a parabolic archway whose height $y$ at a horizontal distance $x$ from the left base is given by the polynomial $y = -x^2 + 6x$. We are told that the zeros of this polynomial represent the points where the archway meets the ground.

The points where the archway meets the ground are those where the height $y$ is zero. To find these points, we need to find the zeros of the polynomial $y = -x^2 + 6x$. We set the polynomial equal to zero and solve for $x$:

$-x^2 + 6x = 0$

We can factor out a common term from the expression on the left side. Both terms, $-x^2$ and $6x$, have $x$ as a common factor. We can factor out $x$:

$x(-x + 6) = 0$

For the product of two factors to be zero, at least one of the factors must be zero. So, we have two possible cases:

Case 1: The first factor is zero.

$x = 0$

Case 2: The second factor is zero.

$-x + 6 = 0$

Add $x$ to both sides:

$6 = x$

The zeros of the polynomial are $x = 0$ and $x = 6$. These are the x-coordinates where the archway meets the ground. Let's assume the archway starts at $x=0$ (the left base) and ends at $x=6$ (the right base) on the x-axis.

The width of the archway at the base is the horizontal distance between these two points. The distance between the points on the x-axis with coordinates $x_1$ and $x_2$ is $|x_2 - x_1|$.

Width $= |6 - 0|$

Width $= 6$ metres

Thus, the width of the archway at the base is 6 metres.

The correct option is (B) 6 metres.

We are given a cubic polynomial in the standard form $ax^3 + bx^2 + cx + d$, where $a$, $b$, $c$, and $d$ are coefficients and $a \neq 0$. We are told that $\alpha$, $\beta$, and $\gamma$ are the zeros (roots) of this polynomial.

There are specific relationships between the zeros of a polynomial and its coefficients, known as Vieta's formulas. For a cubic polynomial $ax^3 + bx^2 + cx + d$, these relationships are:

1. Sum of the zeros: $\alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -\frac{b}{a}$

2. Sum of the products of the zeros taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} = \frac{c}{a}$

3. Product of the zeros: $\alpha\beta\gamma = -\frac{\text{constant term}}{\text{coefficient of } x^3} = -\frac{d}{a}$

The question asks for the product of the zeros, which is $\alpha\beta\gamma$. According to the relationships above, the product of the zeros of a cubic polynomial $ax^3 + bx^2 + cx + d$ is given by $-\frac{d}{a}$.

Comparing this result with the given options, we find that option (D) matches our result.

The correct option is (D) $-\frac{d}{a}$.

We are given a quadratic polynomial $p(x) = ax^2 + bx + c$, and we are told that $\alpha$ and $\beta$ are its zeros. We need to find the value of $\alpha^2 + \beta^2$ in terms of the coefficients $a$, $b$, and $c$.

From the properties of zeros of a quadratic polynomial, we know the sum and product of the zeros in terms of the coefficients:

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

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

We want to find $\alpha^2 + \beta^2$. We can use the algebraic identity for the square of a sum:

$(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$

Rearranging this identity to solve for $\alpha^2 + \beta^2$, we get:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

Now, we substitute the expressions for $(\alpha + \beta)$ and $\alpha \beta$ in terms of $a, b, c$ into this equation:

$\alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right)$

Next, we simplify the expression:

$\left(-\frac{b}{a}\right)^2 = \frac{(-b)^2}{a^2} = \frac{b^2}{a^2}$

$2\left(\frac{c}{a}\right) = \frac{2c}{a}$

So, the expression for $\alpha^2 + \beta^2$ becomes:

$\alpha^2 + \beta^2 = \frac{b^2}{a^2} - \frac{2c}{a}$

To combine the terms on the right-hand side, we find a common denominator, which is $a^2$. We can rewrite $\frac{2c}{a}$ as $\frac{2c \cdot a}{a \cdot a} = \frac{2ac}{a^2}$.

$\alpha^2 + \beta^2 = \frac{b^2}{a^2} - \frac{2ac}{a^2}$

$\alpha^2 + \beta^2 = \frac{b^2 - 2ac}{a^2}$

Comparing this result with the given options, we see that option (A) shows the step-by-step derivation that leads to this result.

The correct option is (A) $(\alpha + \beta)^2 - 2\alpha\beta = (-\frac{b}{a})^2 - 2(\frac{c}{a}) = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2 - 2ac}{a^2}$.

We are asked to find the quotient when the polynomial $p(x) = x^3 + 1$ is divided by the polynomial $g(x) = x + 1$. We can solve this using either polynomial long division or by recognizing the polynomial as a sum of cubes.

Method 1: Using Factoring

The polynomial $p(x) = x^3 + 1$ is in the form of a sum of cubes, $a^3 + b^3$, where $a = x$ and $b = 1$. The factorization formula for a sum of cubes is $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.

Applying this formula to $x^3 + 1^3$:

$x^3 + 1 = (x + 1)(x^2 - x \cdot 1 + 1^2)$

$x^3 + 1 = (x + 1)(x^2 - x + 1)$

The division $p(x) \div g(x)$ is $\frac{x^3 + 1}{x + 1}$. Substituting the factored form:

$\frac{(x + 1)(x^2 - x + 1)}{x + 1}$

Assuming $x + 1 \neq 0$, we can cancel the common factor $(x + 1)$ from the numerator and the denominator:

$\frac{\cancel{(x + 1)}(x^2 - x + 1)}{\cancel{(x + 1)}} = x^2 - x + 1$

Thus, the quotient is $x^2 - x + 1$ and the remainder is 0.

Method 2: Using Polynomial Long Division

We can divide $x^3 + 1$ by $x + 1$ using long division. We can write the dividend as $x^3 + 0x^2 + 0x + 1$ for clarity in the long division process.

$\begin{array}{r} x^2 - x + 1 \\ x+1{\overline{\smash{\big)}\,x^3+0x^2+0x+1}} \\ \underline{-~\phantom{(}(x^3+x^2)\phantom{+0x+1)}} \\ 0-x^2+0x+1 \\ \underline{-~\phantom{()}(-x^2-x)\phantom{+1)}} \\ 0+x+1\phantom{)} \\ \underline{-~\phantom{()}(x+1)} \\ 0+0\phantom{)} \end{array}$

From the long division, the quotient is $x^2 - x + 1$ and the remainder is 0.

Both methods yield the same quotient.

The quotient when $p(x) = x^3 + 1$ is divided by $g(x) = x + 1$ is $x^2 - x + 1$.

Comparing this result with the given options, we find that option (A) matches our quotient.

The correct option is (A) $x^2 - x + 1$.

The points where the graph of a polynomial $y = p(x)$ intersects or touches the x-axis represent the real zeros (or roots) of the polynomial. The number of times the graph intersects or touches the x-axis is equal to the number of real zeros (counting multiplicity in some contexts, but the question usually refers to distinct points of intersection unless specified).

Let's consider the properties of graphs of different types of polynomials:

(A) Linear polynomial: A linear polynomial has degree 1 ($ax+b$, $a \neq 0$). Its graph is a straight line. A non-horizontal straight line intersects the x-axis at exactly one point. Therefore, a linear polynomial cannot have its graph intersect the x-axis at exactly two points.

(B) Quadratic polynomial: A quadratic polynomial has degree 2 ($ax^2+bx+c$, $a \neq 0$). Its graph is a parabola. A parabola can intersect the x-axis at zero points (no real zeros), one point (exactly one real zero with multiplicity 2, where the parabola touches the x-axis), or two distinct points (two distinct real zeros, where the parabola crosses the x-axis). Since a quadratic polynomial can intersect the x-axis at exactly two points, this option applies.

(C) Cubic polynomial: A cubic polynomial has degree 3 ($ax^3+bx^2+cx+d$, $a \neq 0$). Its graph is a curve that goes from $-\infty$ to $+\infty$ (or vice versa) as $x$ goes from $-\infty$ to $+\infty$. A cubic polynomial always has at least one real zero. It can have one real zero, or three distinct real zeros, or one real zero with multiplicity 3, or one real zero with multiplicity 1 and another with multiplicity 2. If a cubic polynomial has one real zero with multiplicity 1 (crossing) and another real zero with multiplicity 2 (touching), its graph intersects the x-axis at exactly two distinct points. For example, the polynomial $p(x) = (x-1)(x-2)^2$ has zeros at $x=1$ (multiplicity 1) and $x=2$ (multiplicity 2). Its graph intersects the x-axis at $x=1$ and touches the x-axis at $x=2$, giving exactly two distinct points of intersection with the x-axis. Therefore, a cubic polynomial can have its graph intersect the x-axis at exactly two points.

(D) Polynomial of degree 4: A polynomial of degree 4 has at most 4 real zeros. It is possible for a polynomial of degree 4 to have exactly two distinct real zeros. For example, the polynomial $p(x) = (x-1)(x-2)(x^2+1)$ has real zeros at $x=1$ and $x=2$ (since $x^2+1=0$ gives complex zeros). The graph of this polynomial intersects the x-axis at exactly these two points. Another example is $p(x) = (x-1)^2(x-2)^2$, which touches the x-axis at $x=1$ and $x=2$, giving exactly two distinct points of intersection. Therefore, a polynomial of degree 4 can have its graph intersect the x-axis at exactly two points.

Based on the analysis, the graphs of quadratic, cubic, and polynomial of degree 4 can intersect the x-axis at exactly two points.

The correct options are (B) Quadratic, (C) Cubic, and (D) Polynomial of degree 4.

We are asked to find a quadratic polynomial whose zeros are given as 3 and -5.

Let the zeros of the quadratic polynomial be $\alpha = 3$ and $\beta = -5$.

A quadratic polynomial with zeros $\alpha$ and $\beta$ can be written in the form:

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

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

First, calculate the sum of the zeros:

Sum of zeros $= \alpha + \beta = 3 + (-5) = 3 - 5 = -2$

Next, calculate the product of the zeros:

Product of zeros $= \alpha \beta = 3 \times (-5) = -15$

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

$p(x) = k(x^2 - (-2)x + (-15))$

$p(x) = k(x^2 + 2x - 15)$

Since the question asks for *a* quadratic polynomial, we can choose the simplest non-zero value for $k$. Let's take $k = 1$.

For $k = 1$, the polynomial is:

$p(x) = 1 \cdot (x^2 + 2x - 15)$

$p(x) = x^2 + 2x - 15$

This polynomial $x^2 + 2x - 15$ has zeros 3 and -5.

Comparing this result with the given options, we find that option (A) matches our polynomial.

The correct option is (A) $x^2 + 2x - 15$.

We are given a quadratic polynomial in the standard form $ax^2 + bx + c$. The zeros (or roots) of this polynomial are the values of $x$ that satisfy the equation $ax^2 + bx + c = 0$.

The nature of the zeros of a quadratic equation is determined by the value of the discriminant. The discriminant is given by the expression $b^2 - 4ac$, where $a$, $b$, and $c$ are the coefficients of the quadratic polynomial.

Let the discriminant be denoted by $\Delta$. So, $\Delta = b^2 - 4ac$.

The relationship between the value of the discriminant and the nature of the roots of a quadratic equation $ax^2 + bx + c = 0$ is as follows:

1. If $\Delta > 0$ ($b^2 - 4ac > 0$), the equation has two distinct real roots (zeros).

2. If $\Delta = 0$ ($b^2 - 4ac = 0$), the equation has two real and equal roots (zeros). In this case, the quadratic polynomial is a perfect square, and its graph touches the x-axis at exactly one point.

3. If $\Delta < 0$ ($b^2 - 4ac < 0$), the equation has no real roots (zeros); it has two complex conjugate roots.

The question states that the zeros of the quadratic polynomial are equal. According to the relationship described above, the zeros of a quadratic polynomial are equal if and only if the discriminant is equal to 0.

Therefore, if the zeros of a quadratic polynomial are equal, the discriminant $b^2 - 4ac$ is equal to 0.

The correct option is (C) Equal to 0.

A polynomial is an expression that can be written in the form $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $a_i$ are coefficients (real numbers) and $n$ is a non-negative integer. The operations involved must be limited to addition, subtraction, and multiplication, with variables raised only to non-negative integer powers.

Let's examine each expression given in the options:

(A) $x^{-2} + 3x - 1$

This expression contains the term $x^{-2}$. The exponent of the variable $x$ in this term is -2, which is a negative integer. According to the definition of a polynomial, the exponents of the variables must be non-negative integers.

Therefore, this expression is NOT a polynomial.

(B) $\sqrt{5}x^3 - \frac{2}{7}x$

This expression contains the terms $\sqrt{5}x^3$ and $-\frac{2}{7}x$. The coefficients are $\sqrt{5}$ and $-\frac{2}{7}$, which are real numbers. The exponents of the variable $x$ are 3 and 1. Both 3 and 1 are non-negative integers. The operations involved are subtraction and multiplication. This fits the definition of a polynomial.

Therefore, this expression IS a polynomial.

(C) $y^5 - |y| + 2$

This expression contains the term $|y|$, which represents the absolute value of $y$. The absolute value function $|y|$ cannot be expressed as a term $ky^n$ where $n$ is a non-negative integer. Polynomials do not include operations like absolute value applied directly to the variable in this form.

Therefore, this expression is NOT a polynomial.

Based on our analysis, both expression (A) and expression (C) are not polynomials.

Option (D) states that both (A) and (C) are not polynomials.

The correct option is (D) Both (A) and (C).

We are given that the graph of a polynomial is a straight line parallel to the x-axis. We need to identify the type of polynomial this graph represents.

A straight line parallel to the x-axis has a constant y-coordinate for all values of x. The equation of such a line can be written in the form:

$y = k$

where $k$ is a constant real number.

The graph of a polynomial $p(x)$ is represented by the equation $y = p(x)$. Therefore, if the graph is a straight line parallel to the x-axis, it means that the polynomial $p(x)$ must be equal to a constant value $k$ for all values of $x$.

$p(x) = k$

By definition, a polynomial of the form $p(x) = k$, where $k$ is a real number, is called a constant polynomial. If $k \neq 0$, the degree of the constant polynomial is 0. If $k = 0$, the polynomial is the zero polynomial, and its graph is the x-axis itself, which is also a straight line parallel to itself.

Let's consider the other options:

(A) A linear polynomial ($ax+b$, with $a \neq 0$) has a graph that is a straight line, but it is not parallel to the x-axis (unless $a=0$, which makes it a constant polynomial). It intersects the x-axis at one point (unless it is the x-axis itself, which is a constant polynomial).

(B) A quadratic polynomial ($ax^2+bx+c$, with $a \neq 0$) has a graph that is a parabola, not a straight line.

(C) A cubic polynomial ($ax^3+bx^2+cx+d$, with $a \neq 0$) has a graph that is a cubic curve, not a straight line.

Thus, a straight line parallel to the x-axis represents a constant polynomial.

The correct option is (D) Constant.

We are given the quadratic polynomial $2x^2 - 4x + 5$. Let the zeros of this polynomial be $\alpha$ and $\beta$. For a standard quadratic polynomial $ax^2 + bx + c$, the relationships between its zeros and coefficients are given by Vieta's formulas.

Comparing the given polynomial $2x^2 - 4x + 5$ with the standard form $ax^2 + bx + c$, we identify the coefficients:

$a = 2$

$b = -4$

$c = 5$

The formulas for the sum and product of the zeros are:

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

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

Substitute the values of $a$, $b$, and $c$ into these formulas:

Sum of zeros: $\alpha + \beta = -\frac{-4}{2} = \frac{4}{2} = 2$

Product of zeros: $\alpha \beta = \frac{5}{2}$

We need to find the value of the expression $\frac{1}{\alpha} + \frac{1}{\beta}$. To simplify this expression, we combine the fractions by finding a common denominator, which is $\alpha \beta$:

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta}{\alpha \beta} + \frac{\alpha}{\alpha \beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{\alpha + \beta}{\alpha \beta}$

Now, substitute the values we calculated for the sum of zeros ($\alpha + \beta = 2$) and the product of zeros ($\alpha \beta = \frac{5}{2}$) into the simplified expression:

$\frac{\alpha + \beta}{\alpha \beta} = \frac{2}{\frac{5}{2}}$

To divide by a fraction, we multiply by its reciprocal:

$\frac{2}{\frac{5}{2}} = 2 \times \frac{2}{5} = \frac{4}{5}$

So, the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is $\frac{4}{5}$.

The correct option is (A) $\frac{4}{5}$.

We are asked to find the quotient when the polynomial $p(x) = x^3 - 3x^2 + 5x - 3$ is divided by the polynomial $g(x) = x^2 - x + 1$. We will use polynomial long division.

The division is set up as follows:

Divide the leading term of the dividend ($x^3$) by the leading term of the divisor ($x^2$):

$\frac{x^3}{x^2} = x$

This is the first term of the quotient.

Multiply the divisor ($x^2 - x + 1$) by the first term of the quotient ($x$):

$x(x^2 - x + 1) = x^3 - x^2 + x$

Subtract this result from the dividend:

$(x^3 - 3x^2 + 5x - 3) - (x^3 - x^2 + x)$

$= x^3 - 3x^2 + 5x - 3 - x^3 + x^2 - x$

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

Now, divide the leading term of the new polynomial ($-2x^2$) by the leading term of the divisor ($x^2$):

$\frac{-2x^2}{x^2} = -2$

This is the second term of the quotient.

Multiply the divisor ($x^2 - x + 1$) by the second term of the quotient ($-2$):

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

Subtract this result from the current polynomial ($-2x^2 + 4x - 3$):

$(-2x^2 + 4x - 3) - (-2x^2 + 2x - 2)$

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

$= 2x - 1$

The degree of the remainder ($2x - 1$, degree 1) is less than the degree of the divisor ($x^2 - x + 1$, degree 2), so the division process is complete.

The polynomial long division can be written as:

$\begin{array}{r} x - 2\phantom{)} \\ x^2-x+1{\overline{\smash{\big)}\,x^3-3x^2+5x-3}} \\ \underline{-~\phantom{()}(x^3-x^2+x)\phantom{-3)}} \\ 0-2x^2+4x-3\phantom{)} \\ \underline{-~\phantom{()}(-2x^2+2x-2)} \\ 0+2x-1\phantom{)} \end{array}$

The quotient is the expression at the top of the division, which is $x - 2$. The remainder is $2x - 1$.

The question asks for the quotient.

The quotient when $x^3 - 3x^2 + 5x - 3$ is divided by $x^2 - x + 1$ is $x - 2$.

The correct option is (A) $x - 2$.

A zero of a polynomial $p(x)$ is a value of the variable $x$ for which the value of the polynomial is equal to 0. If 'a' is a zero of $p(x)$, then $p(a) = 0$.

We are given the cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$, and we are told that one of its zeros is 0. This means that when we substitute $x = 0$ into the polynomial, the result must be 0.

Substitute $x = 0$ into the polynomial $p(x)$:

$p(0) = a(0)^3 + b(0)^2 + c(0) + d$

Now, simplify the expression:

$a(0)^3 = a \times 0 = 0$

$b(0)^2 = b \times 0 = 0$

$c(0) = c \times 0 = 0$

So, the expression for $p(0)$ becomes:

$p(0) = 0 + 0 + 0 + d$

$p(0) = d$

Since 0 is a zero of the polynomial, $p(0)$ must be equal to 0:

$d = 0$

Thus, if one zero of the cubic polynomial $ax^3 + bx^2 + cx + d$ is 0, the value of the constant term $d$ must be 0.

The correct option is (A) 0.

A polynomial is an algebraic expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.

The degree of a polynomial is the highest power of the variable in a polynomial with non-zero coefficients. For a polynomial in one variable, the degree is the highest exponent of the variable present in the expression.

An example of a polynomial of degree 3 is: $P(x) = 4x^3 - 2x^2 + 5x - 1$

The given polynomial is $p(x) = 7x^4 - 3x^3 + \sqrt{5}x^2 - \frac{1}{2}x + 9$.

The coefficient of $x^3$ is the numerical factor multiplying $x^3$. In this polynomial, it is $-3$.

The coefficient of $x^2$ is the numerical factor multiplying $x^2$. In this polynomial, it is $\sqrt{5}$.

The coefficient of $x$ is the numerical factor multiplying $x$. In this polynomial, it is $-\frac{1}{2}$.

The constant term is the term that does not contain the variable $x$. In this polynomial, it is $9$.

We need to check if the given expressions satisfy the definition of a polynomial, which requires variables to have only non-negative integer exponents.

(a) $4x^2 - 3x + 7$

In this expression, the exponents of the variable $x$ are $2$, $1$ (in $-3x = -3x^1$), and $0$ (in $7 = 7x^0$). All these exponents ($2, 1, 0$) are non-negative integers. The operations involved are subtraction, multiplication, and addition, with constant coefficients. Therefore, this expression is a polynomial.

(b) $y + \frac{2}{y^2}$

This expression can be rewritten as $y^1 + 2y^{-2}$. The term $2y^{-2}$ has an exponent of $-2$ for the variable $y$. Since $-2$ is a negative integer, which is not a non-negative integer, this expression is not a polynomial.

(c) $\sqrt{z} + z^2$

This expression can be rewritten as $z^{1/2} + z^2$. The term $z^{1/2}$ has an exponent of $1/2$ for the variable $z$. Since $1/2$ is not an integer (it's a rational number), this expression is not a polynomial.

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

We need to find the value of $p(x)$ at $x = -1$. Substitute $x = -1$ into the polynomial:

$p(-1) = (-1)^2 - 3(-1) - 4$

$p(-1) = 1 - (-3) - 4$

$p(-1) = 1 + 3 - 4$

$p(-1) = 4 - 4$

$p(-1) = 0$

We need to find the value of $p(x)$ at $x = 4$. Substitute $x = 4$ into the polynomial:

$p(4) = (4)^2 - 3(4) - 4$

$p(4) = 16 - 12 - 4$

$p(4) = 4 - 4$

$p(4) = 0$

Thus, the value of the polynomial $p(x)$ at $x = -1$ is $0$ and at $x = 4$ is $0$.

To find the zero of the linear polynomial $p(x) = 3x - 6$, we need to find the value of $x$ for which $p(x) = 0$.

Set the polynomial equal to zero:

Substitute the expression for $p(x)$ into equation (i):

Now, solve for $x$ from equation (ii):

$3x = 6$

$x = \frac{6}{3}$

$x = 2$

Thus, the zero of the linear polynomial $p(x) = 3x - 6$ is $x = 2$.

We can verify this by substituting $x=2$ back into $p(x)$: $p(2) = 3(2) - 6 = 6 - 6 = 0$.

To check if $x = -2$ is a zero of the polynomial $p(x) = x^3 + 4x^2 + 5x + 2$, we need to evaluate the polynomial at $x = -2$. If $p(-2) = 0$, then $x = -2$ is a zero.

Substitute $x = -2$ into the polynomial $p(x)$:

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

Calculate each term:

$(-2)^3 = -8$

$(-2)^2 = 4$, so $4(-2)^2 = 4(4) = 16$

$5(-2) = -10$

Substitute these values back into the expression for $p(-2)$:

$p(-2) = -8 + 16 - 10 + 2$

Perform the addition and subtraction:

$p(-2) = (-8 + 16) + (-10 + 2)$

$p(-2) = 8 + (-8)$

$p(-2) = 0$

Since $p(-2) = 0$, we conclude that $x = -2$ is a zero of the polynomial $p(x) = x^3 + 4x^2 + 5x + 2$.

Consider the linear polynomial $p(x) = ax + b$, where $a \neq 0$.

The zero of this polynomial is the value of $x$ for which $p(x) = 0$. Setting the polynomial equal to zero:

Solving for $x$ from equation (i):

$ax = -b$

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

Geometrically, the expression $p(x) = ax + b$ represents a linear function of $x$. The graph of this linear function, $y = ax + b$, is a straight line in the Cartesian coordinate system.

The zero of the polynomial, $x = -\frac{b}{a}$, is the value of $x$ where the function's value $y$ is equal to zero. On the graph $y = ax + b$, the points where $y = 0$ are the points where the graph intersects the x-axis.

Therefore, the geometrical meaning of the zero of a linear polynomial $ax + b$ is the x-coordinate of the point where the graph of the linear function $y = ax + b$ intersects the x-axis.

The zeros of a polynomial $p(x)$ are the values of $x$ for which $p(x) = 0$.

When considering the graph of a polynomial $y = p(x)$, the zeros of the polynomial are the x-coordinates of the points where the graph intersects the x-axis, because at these points the y-coordinate is $0$, meaning $p(x) = 0$.

The given graph is that of a quadratic polynomial $y = ax^2 + bx + c$, which is a parabola.

According to the description, the parabola intersects the x-axis at two points.

Since there are two points of intersection with the x-axis, there are two distinct x-values for which $y=0$.

Therefore, based on the given graph, the quadratic polynomial has two zeros.

For a quadratic polynomial $ax^2 + bx + c$, where $a \neq 0$, let the zeros be $\alpha$ and $\beta$.

The relationship between the zeros and the coefficients is given by:

1. Sum of the zeros: The sum of the zeros ($\alpha + \beta$) is equal to the negative of the coefficient of $x$ divided by the coefficient of $x^2$.

$\alpha + \beta = -\frac{b}{a}$

2. Product of the zeros: The product of the zeros ($\alpha \beta$) is equal to the constant term divided by the coefficient of $x^2$.

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

The given quadratic polynomial is $4x^2 - 5x + 3$.

Comparing this polynomial with the standard form of a quadratic polynomial $ax^2 + bx + c$, we can identify the coefficients:

$a = 4$

$b = -5$

$c = 3$

The sum of the zeros ($\alpha + \beta$) of a quadratic polynomial is given by the formula $-\frac{b}{a}$.

Sum of zeros $= -\frac{(-5)}{4}$

Sum of zeros $= \frac{5}{4}$

The product of the zeros ($\alpha \beta$) of a quadratic polynomial is given by the formula $\frac{c}{a}$.

Product of zeros $= \frac{3}{4}$

Thus, the sum of the zeros is $\frac{5}{4}$ and the product of the zeros is $\frac{3}{4}$.

Let the quadratic polynomial be $p(x) = ax^2 + bx + c$, and let its zeros be $\alpha$ and $\beta$.

We are given the sum of the zeros:

$\alpha + \beta = -5$

We are given the product of the zeros:

$\alpha \beta = 6$

We know that for a quadratic polynomial $ax^2 + bx + c$, the sum of the zeros is $\alpha + \beta = -\frac{b}{a}$ and the product of the zeros is $\alpha \beta = \frac{c}{a}$.

A quadratic polynomial with sum of zeros $S$ and product of zeros $P$ can be written in the form $k(x^2 - Sx + P)$, where $k$ is any non-zero real number.

In this case, the sum of zeros $S = -5$ and the product of zeros $P = 6$.

Substitute these values into the general form:

$p(x) = k(x^2 - (-5)x + 6)$

$p(x) = k(x^2 + 5x + 6)$

We can choose any non-zero value for $k$. The simplest choice is $k=1$.

For $k=1$, the polynomial is:

$p(x) = 1(x^2 + 5x + 6)$

$p(x) = x^2 + 5x + 6$

Thus, a quadratic polynomial whose sum of zeros is $-5$ and product of zeros is $6$ is $x^2 + 5x + 6$. Other polynomials are scalar multiples of this, like $2x^2 + 10x + 12$ (for $k=2$), etc.

We perform the polynomial long division of $6x^2 - 7x - 3$ by $2x - 3$.

$\begin{array}{r} 3x+1 \phantom{)} \\ 2x-3{\overline{\smash{\big)}\,6x^2-7x-3}} \\ \underline{-\phantom{(}(6x^2-9x)} \\ \phantom{0x^2+}2x-3 \\ \underline{-\phantom{()}(2x-3)} \\ \phantom{0x^2-0x+}0 \end{array}$

The quotient is $3x + 1$ and the remainder is $0$.

The Division Algorithm for Polynomials states that:

Given any polynomial $P(x)$ and any non-zero polynomial $G(x)$, there exist unique polynomials $Q(x)$ and $R(x)$ such that:

$P(x) = G(x) \cdot Q(x) + R(x)$

where $R(x) = 0$ or $\text{deg}(R(x)) < \text{deg}(G(x))$.

Here, $P(x)$ is the dividend, $G(x)$ is the divisor, $Q(x)$ is the quotient, and $R(x)$ is the remainder.

To find the zeros of the polynomial $p(x)$, we set $p(x)$ equal to zero.

We have $p(x) = (x - 5)(x + 2)$.

Setting $p(x) = 0$, we get:

$(x - 5)(x + 2) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

So, either $x - 5 = 0$ or $x + 2 = 0$.

Solving the first equation:

$x - 5 = 0$

$x = 5$

Solving the second equation:

$x + 2 = 0$

$x = -2$

Thus, the zeros of the polynomial $p(x) = (x - 5)(x + 2)$ are $x = 5$ and $x = -2$.

Given that $x = 2$ is a zero of the polynomial $p(x) = 2x^2 + ax - 6$.

By the definition of a zero, if $x=2$ is a zero, then $p(2)$ must be equal to $0$.

Substitute $x = 2$ into the polynomial $p(x)$:

$p(2) = 2(2)^2 + a(2) - 6$

$p(2) = 2(4) + 2a - 6$

$p(2) = 8 + 2a - 6$

$p(2) = 2a + 2$

Since $p(2) = 0$, we have:

Now, solve for $a$:

$2a = -2$

$a = \frac{-2}{2}$

$a = -1$

Therefore, the value of $a$ is -1.

To find the remainder when the polynomial $p(x) = x^3 - 2x^2 + x - 1$ is divided by $(x - 1)$, we can use the Remainder Theorem.

The Remainder Theorem states that if a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, then the remainder is $p(a)$.

In this case, the polynomial is $p(x) = x^3 - 2x^2 + x - 1$, and the divisor is $(x - 1)$. Comparing $(x - 1)$ with $(x - a)$, we have $a = 1$.

According to the Remainder Theorem, the remainder is $p(1)$.

Now, we evaluate $p(x)$ at $x = 1$:

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

$p(1) = 1 - 2(1) + 1 - 1$

$p(1) = 1 - 2 + 1 - 1$

$p(1) = -1$

Thus, the remainder when $p(x) = x^3 - 2x^2 + x - 1$ is divided by $(x - 1)$ is -1.

Geometrically, the points where the graph of a polynomial intersects or touches the x-axis represent the zeros of the polynomial.

If the graph of a quadratic polynomial touches the x-axis at exactly one point, it means that the polynomial has exactly one zero.

For a quadratic polynomial, having exactly one zero implies that the two roots of the corresponding quadratic equation are equal or repeated.

Furthermore, if the graph of a quadratic polynomial (a parabola) touches the x-axis at a single point, that point is the vertex of the parabola, and the vertex lies directly on the x-axis.

Given the quadratic polynomial $p(x) = x^2 - \sqrt{2}x - 12$.

The standard form of a quadratic polynomial is $ax^2 + bx + c$.

Comparing the given polynomial with the standard form, we have:

$a = 1$

$b = -\sqrt{2}$

$c = -12$

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

The sum of the zeros of a quadratic polynomial is given by the formula $\alpha + \beta = -\frac{b}{a}$.

Sum of zeros $= -\frac{(-\sqrt{2})}{1} = \sqrt{2}$

The product of the zeros of a quadratic polynomial is given by the formula $\alpha \beta = \frac{c}{a}$.

Product of zeros $= \frac{-12}{1} = -12$

Therefore, the sum of the zeros is $\sqrt{2}$ and the product of the zeros is $-12$.

Let the sum of the zeros of the quadratic polynomial be $S$ and the product of the zeros be $P$.

We are given:

Sum of zeros, $S = 4$

Product of zeros, $P = 1$

A quadratic polynomial can be written in terms of the sum and product of its zeros as:

$k(x^2 - (\text{Sum of zeros})x + \text{Product of zeros})$

or

$k(x^2 - Sx + P)$, where $k$ is a non-zero constant.

Substitute the given values of $S$ and $P$ into the formula:

$k(x^2 - (4)x + 1)$

$k(x^2 - 4x + 1)$

We can take $k = 1$ for the simplest quadratic polynomial.

So, the polynomial is $x^2 - 4x + 1$.

Therefore, the quadratic polynomial is $x^2 - 4x + 1$.

Let the given quadratic polynomial be $p(x) = x^2 - 3x - 10$.

To find the zeros of the polynomial by factorisation, we set $p(x) = 0$ and factor the expression.

$x^2 - 3x - 10 = 0$

We need to find two numbers that multiply to give the constant term ($-10$) and add up to give the coefficient of the middle term ($-3$).

The numbers are $2$ and $-5$, because $2 \times (-5) = -10$ and $2 + (-5) = -3$.

Now, we split the middle term ($-3x$) using these two numbers:

$x^2 + 2x - 5x - 10 = 0$

Group the terms and factor by grouping:

$(x^2 + 2x) + (-5x - 10) = 0$

Factor out the common factor from each group:

$x(x + 2) - 5(x + 2) = 0$

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

$(x + 2)(x - 5) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

So, we set each factor equal to zero and solve for $x$:

Case 1: $x + 2 = 0$

$x = -2$

Case 2: $x - 5 = 0$

$x = 5$

The zeros of the quadratic polynomial $x^2 - 3x - 10$ are -2 and 5.

We classify each polynomial based on its degree and the number of terms it contains.

(a) $5x^3 + 2x - 1$

Degree: The highest power of the variable $x$ is 3. Thus, the degree is 3.

Number of terms: There are three terms ($5x^3$, $2x$, and $-1$). Thus, the number of terms is 3.

Classification:

Based on degree 3: Cubic polynomial.

Based on 3 terms: Trinomial.

Combined classification: Cubic trinomial.

(b) $y^2$

Degree: The highest power of the variable $y$ is 2. Thus, the degree is 2.

Number of terms: There is one term ($y^2$). Thus, the number of terms is 1.

Classification:

Based on degree 2: Quadratic polynomial.

Based on 1 term: Monomial.

Combined classification: Quadratic monomial.

(c) $z + 9$

Degree: The highest power of the variable $z$ is 1 (since $z = z^1$). The degree of the constant term 9 is 0. The highest degree is 1. Thus, the degree is 1.

Number of terms: There are two terms ($z$ and $9$). Thus, the number of terms is 2.

Classification:

Based on degree 1: Linear polynomial.

Based on 2 terms: Binomial.

Combined classification: Linear binomial.

Given the quadratic polynomial $p(x) = x^2 - x - 2$.

The standard form of a quadratic polynomial is $ax^2 + bx + c$.

Comparing $x^2 - x - 2$ with $ax^2 + bx + c$, we have:

$a = 1$

$b = -1$

$c = -2$

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

According to the properties of quadratic polynomials:

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

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

Using the values of $a$, $b$, and $c$ for the given polynomial:

Sum of zeros: $\alpha + \beta = -\frac{(-1)}{1} = 1$

Product of zeros: $\alpha \beta = \frac{-2}{1} = -2$

We need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.

First, combine the fractions by finding a common denominator, which is $\alpha \beta$:

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta}{\alpha \beta} + \frac{\alpha}{\alpha \beta} = \frac{\alpha + \beta}{\alpha \beta}$

Now, substitute the calculated values of $(\alpha + \beta)$ and $(\alpha \beta)$ into the expression:

$\frac{\alpha + \beta}{\alpha \beta} = \frac{1}{-2}$

$\frac{1}{\alpha} + \frac{1}{\beta} = -\frac{1}{2}$

Thus, the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is $-\frac{1}{2}$.

The degree of a polynomial is the highest power of the variable in the polynomial.

A cubic polynomial is a polynomial with a degree of 3.

According to the Fundamental Theorem of Algebra, a polynomial of degree $n$ has exactly $n$ complex roots (counting multiplicity). The real zeros are a subset of these complex roots.

A polynomial of degree $n$ can have at most $n$ distinct real zeros.

Since a cubic polynomial has a degree of 3, the maximum number of distinct real zeros it can have is 3.

Here is a rough graph of a cubic polynomial with 3 distinct zeros. The points where the graph crosses the x-axis are the zeros of the polynomial.

In the graph above, the curve crosses the x-axis at three distinct points, indicating three distinct real zeros.

An example of a quadratic polynomial whose graph does not intersect the x-axis is one that has no real zeros.

Consider the quadratic polynomial $p(x) = x^2 + 1$.

To find the zeros, we set $p(x) = 0$:

$x^2 + 1 = 0$

$x^2 = -1$

The equation $x^2 = -1$ has no real solutions, because the square of any real number is non-negative ($\geq 0$). The solutions are imaginary numbers ($x = \pm i$).

Alternatively, for a quadratic polynomial $ax^2 + bx + c$, the graph does not intersect the x-axis if the discriminant $\Delta = b^2 - 4ac$ is negative ($\Delta < 0$).

For $p(x) = x^2 + 1$, we have $a = 1$, $b = 0$, and $c = 1$.

The discriminant is $\Delta = (0)^2 - 4(1)(1) = 0 - 4 = -4$.

Since $\Delta = -4 < 0$, the polynomial has no real zeros.

Geometrically, the graph of $y = x^2 + 1$ is a parabola that opens upwards (since $a=1 > 0$) and its vertex is at $(0, 1)$. Since the vertex is above the x-axis and the parabola opens upwards, the graph never intersects the x-axis.

Therefore, a quadratic polynomial whose graph does not intersect the x-axis is $x^2 + 1$.

Given the polynomial $p(x, y) = x^2 + xy + y^2$.

We need to find the value of the polynomial when $x = 1$ and $y = -1$.

Substitute the values of $x$ and $y$ into the expression:

$p(1, -1) = (1)^2 + (1)(-1) + (-1)^2$

Calculate each term:

$(1)^2 = 1$

$(1)(-1) = -1$

$(-1)^2 = 1$

Now, substitute these values back into the expression:

$p(1, -1) = 1 + (-1) + 1$

$p(1, -1) = 1 - 1 + 1$

$p(1, -1) = 0 + 1$

$p(1, -1) = 1$

Thus, the value of the polynomial $p(x, y) = x^2 + xy + y^2$ when $x = 1$ and $y = -1$ is 1.

For a quadratic polynomial $ax^2 + bx + c$, the zeros are the roots of the corresponding quadratic equation $ax^2 + bx + c = 0$.

The nature of the roots of a quadratic equation is determined by its discriminant, which is given by $\Delta = b^2 - 4ac$.

The roots are:

1. Real and distinct if $\Delta > 0$.

2. Real and equal if $\Delta = 0$.

3. Complex (non-real) if $\Delta < 0$.

Given that the zeros of the quadratic polynomial are equal, this means the corresponding quadratic equation has real and equal roots.

According to the discriminant criterion, the roots are equal if and only if the discriminant is zero.

So, we must have:

Substituting the expression for the discriminant:

This equation represents the relationship between the coefficients $a$, $b$, and $c$ when the zeros of the quadratic polynomial are equal.

The relationship can also be written as:

Thus, if the zeros of the quadratic polynomial $ax^2 + bx + c$ are equal, the relationship between its coefficients is $b^2 - 4ac = 0$ or $b^2 = 4ac$.

The zeros of a polynomial are the values of $x$ for which $p(x) = 0$.

Geometrically, the real zeros of a polynomial correspond to the points where the graph of the polynomial intersects or touches the x-axis (the x-intercepts).

The provided image description states that the "Graph of a cubic polynomial intersecting x-axis at three points".

By observing the graph (as described), we can count the number of points where the curve crosses or touches the x-axis.

The graph intersects the x-axis at three distinct points.

Therefore, the number of real zeros of the polynomial $p(x)$ is 3.

The zeros of a polynomial $p(x)$ are the values of $x$ for which $p(x) = 0$.

Geometrical Meaning: The real zeros of a polynomial are the x-coordinates of the points where the graph of $y = p(x)$ intersects or touches the x-axis.

Illustrations with Graphs:

The number of real zeros is the number of times the graph crosses or touches the x-axis.

1. Linear Polynomial (e.g., $p(x) = ax + b$, $a \neq 0$):

Always has 1 real zero. The graph is a line intersecting the x-axis at one point.

2. Quadratic Polynomial (e.g., $p(x) = ax^2 + bx + c$, $a \neq 0$):

The graph is a parabola. Can have 0, 1, or 2 real zeros.

Case A: 2 distinct real zeros.

Graph intersects the x-axis at two distinct points.

Case B: 1 real zero (repeated).

Graph touches the x-axis at exactly one point (the vertex).

Case C: No real zeros.

Graph does not intersect the x-axis.

Let the given quadratic polynomial be $p(x) = x^2 + 7x + 10$.

Finding the Zeros by Factorisation:

To find the zeros, we set $p(x) = 0$:

$x^2 + 7x + 10 = 0$

We factor the quadratic expression. We look for two numbers that multiply to 10 and add up to 7. These numbers are 2 and 5 ($2 \times 5 = 10$ and $2 + 5 = 7$).

Split the middle term:

$x^2 + 2x + 5x + 10 = 0$

Group the terms and factor:

$(x^2 + 2x) + (5x + 10) = 0$

$x(x + 2) + 5(x + 2) = 0$

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

$(x + 2)(x + 5) = 0$

Set each factor equal to zero to find the zeros:

Case 1: $x + 2 = 0 \implies x = -2$

Case 2: $x + 5 = 0 \implies x = -5$

So, the zeros of the polynomial are $\alpha = -2$ and $\beta = -5$.

Verification of the Relationship between Zeros and Coefficients:

The standard form of a quadratic polynomial is $ax^2 + bx + c$.

Comparing $x^2 + 7x + 10$ with $ax^2 + bx + c$, we have the coefficients:

$a = 1$

$b = 7$

$c = 10$

Relationship 1: Sum of zeros = $-\frac{b}{a}$

Sum of zeros: $\alpha + \beta = -2 + (-5) = -7$

$- \frac{b}{a} = - \frac{7}{1} = -7$

Since $-7 = -7$, the relationship for the sum of zeros is verified.

Relationship 2: Product of zeros = $\frac{c}{a}$

Product of zeros: $\alpha \beta = (-2) \times (-5) = 10$

$\frac{c}{a} = \frac{10}{1} = 10$

Since $10 = 10$, the relationship for the product of zeros is verified.

The zeros of the polynomial are -2 and -5, and the relationship between the zeros and coefficients is successfully verified.

Let the given quadratic polynomial be $p(x) = 6x^2 - 3 - 7x$.

Rewrite the polynomial in standard form $ax^2 + bx + c$:

$p(x) = 6x^2 - 7x - 3$

Comparing this with $ax^2 + bx + c$, we identify the coefficients:

$a = 6$

$b = -7$

$c = -3$

Finding the Zeros by Factorisation:

To find the zeros, we set $p(x) = 0$:

$6x^2 - 7x - 3 = 0$

We use the method of splitting the middle term. We need two numbers whose product is $a \times c = 6 \times (-3) = -18$ and whose sum is $b = -7$. The numbers are $2$ and $-9$ ($2 \times (-9) = -18$ and $2 + (-9) = -7$).

Split the middle term $-7x$ into $+2x - 9x$:

$6x^2 + 2x - 9x - 3 = 0$

Group the terms and factor by grouping:

$(6x^2 + 2x) + (-9x - 3) = 0$

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

Factor out the common binomial factor $(3x + 1)$:

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

Set each factor equal to zero to find the zeros:

Case 1: $3x + 1 = 0$

$3x = -1$

$x = -\frac{1}{3}$

Case 2: $2x - 3 = 0$

$2x = 3$

$x = \frac{3}{2}$

So, the zeros of the polynomial are $\alpha = -\frac{1}{3}$ and $\beta = \frac{3}{2}$.

Verification of the Relationship between Zeros and Coefficients:

Relationship 1: Sum of zeros = $-\frac{b}{a}$

Sum of zeros: $\alpha + \beta = -\frac{1}{3} + \frac{3}{2}$

$\alpha + \beta = \frac{(-1 \times 2) + (3 \times 3)}{6} = \frac{-2 + 9}{6} = \frac{7}{6}$

From coefficients: $-\frac{b}{a} = -\frac{(-7)}{6} = \frac{7}{6}$

Since $\frac{7}{6} = \frac{7}{6}$, the relationship for the sum of zeros is verified.

Relationship 2: Product of zeros = $\frac{c}{a}$

Product of zeros: $\alpha \beta = (-\frac{1}{3}) \times (\frac{3}{2})$

$\alpha \beta = -\frac{1 \times 3}{3 \times 2} = -\frac{3}{6} = -\frac{1}{2}$

From coefficients: $\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$

Since $-\frac{1}{2} = -\frac{1}{2}$, the relationship for the product of zeros is verified.

The zeros of the polynomial are $-\frac{1}{3}$ and $\frac{3}{2}$, and the relationship between the zeros and coefficients is verified.

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

Step 1: Find the sum of the zeros.

Sum of zeros, $S = \alpha + \beta$

$S = \frac{1}{4} + (-1) = \frac{1}{4} - 1$

$S = \frac{1 - 4}{4} = -\frac{3}{4}$

Step 2: Find the product of the zeros.

Product of zeros, $P = \alpha \beta$

$P = \left(\frac{1}{4}\right) \times (-1)$

$P = -\frac{1}{4}$

Step 3: Construct the quadratic polynomial.

A quadratic polynomial with sum of zeros $S$ and product of zeros $P$ is given by the formula:

$k(x^2 - Sx + P)$, where $k$ is any non-zero constant.

Substitute the values of $S$ and $P$:

$p(x) = k\left(x^2 - \left(-\frac{3}{4}\right)x + \left(-\frac{1}{4}\right)\right)$

$p(x) = k\left(x^2 + \frac{3}{4}x - \frac{1}{4}\right)$

To obtain a polynomial with integer coefficients, we can choose $k$ to be the least common multiple of the denominators, which is 4.

Let $k = 4$.

$p(x) = 4\left(x^2 + \frac{3}{4}x - \frac{1}{4}\right)$

$p(x) = 4x^2 + 4\left(\frac{3}{4}x\right) - 4\left(\frac{1}{4}\right)$

$p(x) = 4x^2 + 3x - 1$

The quadratic polynomial is $4x^2 + 3x - 1$.

Step 4: Verify the relationship between the zeros and coefficients.

For the polynomial $4x^2 + 3x - 1$, the coefficients are:

$a = 4$

$b = 3$

$c = -1$

The zeros we started with are $\alpha = \frac{1}{4}$ and $\beta = -1$.

Verification for Sum of Zeros:

Sum of zeros from the found polynomial should be $-\frac{b}{a}$.

Sum of given zeros: $\alpha + \beta = \frac{1}{4} + (-1) = -\frac{3}{4}$

Sum of zeros from coefficients: $-\frac{b}{a} = -\frac{3}{4}$

Since $-\frac{3}{4} = -\frac{3}{4}$, the relationship is verified for the sum of zeros.

Verification for Product of Zeros:

Product of zeros from the found polynomial should be $\frac{c}{a}$.

Product of given zeros: $\alpha \beta = \left(\frac{1}{4}\right) \times (-1) = -\frac{1}{4}$

Product of zeros from coefficients: $\frac{c}{a} = \frac{-1}{4} = -\frac{1}{4}$

Since $-\frac{1}{4} = -\frac{1}{4}$, the relationship is verified for the product of zeros.

The relationship between the zeros and coefficients is verified for the polynomial $4x^2 + 3x - 1$.

Given polynomials: $p(x) = x^3 - 3x^2 + 5x - 3$ and $g(x) = x^2 - 2$.

We perform polynomial long division to divide $p(x)$ by $g(x)$.

$\begin{array}{r} x \phantom{+0x} - 3 \phantom{)} \\ x^2 - 2{\overline{\smash{\big)}\,x^3 - 3x^2 + 5x - 3}} \\ \underline{-~\phantom{(}(x^3 \phantom{-3x^2} - 2x)}\phantom{-3)} \\ \phantom{x^3} - 3x^2 + 7x - 3 \\ \underline{-~\phantom{()} (-3x^2 \phantom{+7x} + 6)} \\ \phantom{x^3 -3x^2} 7x - 9 \end{array}$

From the long division, we get the quotient $Q(x) = x - 3$ and the remainder $R(x) = 7x - 9$.

The Division Algorithm for Polynomials states that for any polynomial $p(x)$ and a non-zero polynomial $g(x)$, there exist unique polynomials $Q(x)$ (quotient) and $R(x)$ (remainder) such that:

$p(x) = g(x) \cdot Q(x) + R(x)$

where $R(x) = 0$ or $\text{deg}(R(x)) < \text{deg}(g(x))$.

Here, $\text{deg}(R(x)) = \text{deg}(7x - 9) = 1$ and $\text{deg}(g(x)) = \text{deg}(x^2 - 2) = 2$. Since $1 < 2$, the condition $\text{deg}(R(x)) < \text{deg}(g(x))$ is satisfied.

Now, we verify the algorithm by calculating $g(x) \cdot Q(x) + R(x)$:

$g(x) \cdot Q(x) + R(x) = (x^2 - 2)(x - 3) + (7x - 9)$

First, multiply $(x^2 - 2)$ by $(x - 3)$:

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

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

Now, add the remainder $R(x) = 7x - 9$:

$(x^3 - 3x^2 - 2x + 6) + (7x - 9)$

$= x^3 - 3x^2 + (-2x + 7x) + (6 - 9)$

$= x^3 - 3x^2 + 5x - 3$

This result is equal to $p(x)$.

So, $p(x) = g(x) \cdot Q(x) + R(x)$ is verified:

The Division Algorithm is verified.

Let the dividend be $p(x) = x^4 - 6x^3 + 16x^2 - 25x + 10$ and the divisor be $g(x) = x^2 - 2x + k$.

The given remainder is $R(x) = x + a$.

We perform polynomial long division of $p(x)$ by $g(x)$.

$\begin{array}{r} x^2 - 4x + (8-k) \\ x^2 - 2x + k{\overline{\smash{\big)}\,x^4 - 6x^3 + 16x^2 - 25x + 10}} \\ \underline{-~\phantom{(}(x^4 - 2x^3 + kx^2)} \\ \phantom{x^4} - 4x^3 + (16-k)x^2 - 25x \phantom{+10} \\ \underline{-~\phantom{()} (-4x^3 + 8x^2 - 4kx)}\phantom{+10} \\ \phantom{x^4-4x^3} (16-k-8)x^2 + (-25+4k)x + 10 \\ \phantom{x^4-4x^3} (8-k)x^2 + (4k-25)x + 10 \\ \underline{-~\phantom{()}( (8-k)x^2 - 2(8-k)x + k(8-k) )} \\ \phantom{x^4-4x^3+(8-k)x^2} ((4k-25) - (-16+2k))x + (10 - k(8-k)) \\ \phantom{x^4-4x^3+(8-k)x^2} (4k-25+16-2k)x + (10 - 8k + k^2) \\ \phantom{x^4-4x^3+(8-k)x^2} (2k-9)x + (k^2 - 8k + 10) \end{array}$

The remainder obtained from the division is $(2k - 9)x + (k^2 - 8k + 10)$.

We are given that the remainder is $x + a$.

Equating the coefficients of the remainder we obtained with the given remainder $x + a$:

Coefficient of $x$:

Constant term:

Solve the equation for $k$ from the coefficient of $x$:

$2k - 9 = 1$

$2k = 1 + 9$

$2k = 10$

$k = \frac{10}{2}$

$k = 5$

Now substitute the value of $k=5$ into the equation for the constant term to find $a$:

$a = k^2 - 8k + 10$

$a = (5)^2 - 8(5) + 10$

$a = 25 - 40 + 10$

$a = -15 + 10$

$a = -5$

Thus, the values are $k = 5$ and $a = -5$.

Let the given polynomial be $p(x) = x^4 - 6x^3 - 26x^2 + 138x - 35$.

The degree of the polynomial is 4, so it has at most 4 real zeros.

We are given two zeros: $\alpha_1 = 2 + \sqrt{3}$ and $\alpha_2 = 2 - \sqrt{3}$.

Since the coefficients of the polynomial are real, if $(a + \sqrt{b})$ is a zero, then its conjugate $(a - \sqrt{b})$ must also be a zero.

If $x = 2 + \sqrt{3}$ is a zero, then $(x - (2 + \sqrt{3}))$ is a factor.

If $x = 2 - \sqrt{3}$ is a zero, then $(x - (2 - \sqrt{3}))$ is a factor.

The product of these factors is also a factor of $p(x)$. Let this factor be $g(x)$.

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

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

Using the identity $(A - B)(A + B) = A^2 - B^2$, where $A = (x - 2)$ and $B = \sqrt{3}$:

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

$g(x) = (x^2 - 4x + 4) - 3$

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

So, $x^2 - 4x + 1$ is a factor of $p(x)$. To find the other zeros, we divide $p(x)$ by $g(x)$.

We perform polynomial long division:

$\begin{array}{r} x^2 - 2x - 35 \\ x^2 - 4x + 1{\overline{\smash{\big)}\,x^4 - 6x^3 - 26x^2 + 138x - 35}} \\ \underline{-~\phantom{(}(x^4 - 4x^3 + x^2)} \\ \phantom{x^4} - 2x^3 - 27x^2 + 138x \phantom{-35} \\ \underline{-~\phantom{()} (-2x^3 + 8x^2 - 2x)}\phantom{-35} \\ \phantom{x^4-2x^3} - 35x^2 + 140x - 35 \\ \underline{-~\phantom{()}(-35x^2 + 140x - 35)} \\ \phantom{x^4-2x^3-35x^2+140x} 0 \end{array}$

The quotient is $Q(x) = x^2 - 2x - 35$ and the remainder is $0$.

Thus, $p(x) = (x^2 - 4x + 1)(x^2 - 2x - 35)$.

The other zeros are the zeros of the quotient $Q(x) = x^2 - 2x - 35$.

To find the zeros of $x^2 - 2x - 35$, we factor the quadratic expression:

We need two numbers that multiply to $-35$ and add up to $-2$. These numbers are $5$ and $-7$.

$x^2 - 2x - 35 = (x + 5)(x - 7)$

Set the factors to zero:

$x + 5 = 0 \implies x = -5$

$x - 7 = 0 \implies x = 7$

The other two zeros of the polynomial are -5 and 7.

Let the given polynomial be $p(x) = 3x^4 + 6x^3 - 2x^2 - 10x - 5$.

The degree of the polynomial is 4, which means it has at most 4 real zeros.

We are given two zeros: $\alpha_1 = \sqrt{\frac{5}{3}}$ and $\alpha_2 = -\sqrt{\frac{5}{3}}$.

If $x = \sqrt{\frac{5}{3}}$ is a zero, then $(x - \sqrt{\frac{5}{3}})$ is a factor of $p(x)$.

If $x = -\sqrt{\frac{5}{3}}$ is a zero, then $(x - (-\sqrt{\frac{5}{3}})) = (x + \sqrt{\frac{5}{3}})$ is a factor of $p(x)$.

Since $(x - \sqrt{\frac{5}{3}})$ and $(x + \sqrt{\frac{5}{3}})$ are factors, their product is also a factor of $p(x)$.

Let $g(x) = \left(x - \sqrt{\frac{5}{3}}\right)\left(x + \sqrt{\frac{5}{3}}\right)$.

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

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

To find the other zeros, we divide $p(x)$ by $g(x)$. It's easier to divide by a multiple of $g(x)$ with integer coefficients. Let's use $3 \cdot g(x) = 3\left(x^2 - \frac{5}{3}\right) = 3x^2 - 5$. This polynomial $3x^2-5$ has the same zeros as $x^2 - \frac{5}{3}$.

We perform polynomial long division of $3x^4 + 6x^3 - 2x^2 - 10x - 5$ by $3x^2 - 5$:

$\begin{array}{r} x^2 + 2x + 1 \\ 3x^2 - 5{\overline{\smash{\big)}\,3x^4 + 6x^3 - 2x^2 - 10x - 5}} \\ \underline{-~\phantom{(}(3x^4 \phantom{+6x^3} - 5x^2)}\phantom{-10x-5} \\ \phantom{3x^4} 6x^3 + 3x^2 - 10x - 5 \\ \underline{-~\phantom{(}(6x^3 \phantom{+3x^2} - 10x)}\phantom{-5} \\ \phantom{3x^4+6x^3} 3x^2 \phantom{-10x} - 5 \\ \underline{-~\phantom{()} (3x^2 \phantom{+0x}- 5)} \\ \phantom{3x^4+6x^3+3x^2-10x} 0 \end{array}$

The quotient is $Q(x) = x^2 + 2x + 1$ and the remainder is $0$.

Thus, $p(x) = (3x^2 - 5)(x^2 + 2x + 1)$. The zeros of $p(x)$ are the zeros of $3x^2 - 5$ (which we already know) and the zeros of $x^2 + 2x + 1$.

To find the remaining zeros, we find the zeros of the quotient polynomial $Q(x) = x^2 + 2x + 1$.

Set $Q(x) = 0$:

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

Factor the quadratic expression. This is a perfect square trinomial:

$(x + 1)^2 = 0$

Set the factor equal to zero:

$x + 1 = 0$

$x = -1$

Since the factor $(x+1)$ is squared, the zero $x=-1$ has a multiplicity of 2. This means it is a repeated zero.

Therefore, the other two zeros of the polynomial are -1 and -1.

Given:

Dividend, $p(x) = x^3 - 3x^2 + x + 2$

Quotient, $Q(x) = x - 2$

Remainder, $R(x) = -2x + 4$

Divisor, $g(x)$ (to be found)

Solution:

According to the Division Algorithm for polynomials:

$p(x) = g(x) \cdot Q(x) + R(x)$

We want to find $g(x)$. Rearranging the formula, we get:

$g(x) \cdot Q(x) = p(x) - R(x)$

$g(x) = \frac{p(x) - R(x)}{Q(x)}$

First, calculate $p(x) - R(x)$:

$p(x) - R(x) = (x^3 - 3x^2 + x + 2) - (-2x + 4)$

$p(x) - R(x) = x^3 - 3x^2 + x + 2 + 2x - 4$

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

Now, we divide $x^3 - 3x^2 + 3x - 2$ by $Q(x) = x - 2$ to find $g(x)$.

We perform polynomial long division:

$\begin{array}{r} x^2 - x + 1 \\ x - 2{\overline{\smash{\big)}\,x^3 - 3x^2 + 3x - 2}} \\ \underline{-~\phantom{(}(x^3 - 2x^2)} \\ \phantom{x^3} - x^2 + 3x \phantom{-2} \\ \underline{-~\phantom{()} (-x^2 + 2x)}\phantom{-2} \\ \phantom{x^3-x^2} x - 2 \\ \underline{-~\phantom{()}(x - 2)} \\ \phantom{x^3-x^2+x} 0 \end{array}$

The quotient of this division is $x^2 - x + 1$, and the remainder is 0.

Therefore, $g(x) = x^2 - x + 1$.

The polynomial $g(x)$ is $x^2 - x + 1$.

Given the quadratic polynomial $p(x) = x^2 - 2x + (k+1)$.

Comparing this polynomial with the standard form $ax^2 + bx + c$, we identify the coefficients:

$a = 1$

$b = -2$

$c = k+1$

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

According to the relationship between the zeros and coefficients of a quadratic polynomial:

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

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

Substitute the values of $a$, $b$, and $c$ into these formulas:

Sum of zeros: $\alpha + \beta = -\frac{(-2)}{1} = 2$

Product of zeros: $\alpha \beta = \frac{k+1}{1} = k+1$

We are given the condition that the sum of the zeros is equal to the product of the zeros:

Substitute the expressions we found for $\alpha + \beta$ and $\alpha \beta$ into this equation:

Now, solve for $k$:

$k + 1 = 2$

$k = 2 - 1$

$k = 1$

Therefore, the value of $k$ is 1.

Let the dividend be $p(x) = x^4 + x^3 + 8x^2 + ax + b$ and the divisor be $g(x) = x^2 + 1$.

Since $p(x)$ is exactly divisible by $g(x)$, the remainder when $p(x)$ is divided by $g(x)$ must be 0.

We perform polynomial long division of $p(x)$ by $g(x) = x^2 + 0x + 1$.

$\begin{array}{r} x^2 + x + 7 \\ x^2 + 0x + 1{\overline{\smash{\big)}\,x^4 + x^3 + 8x^2 + ax + b}} \\ \underline{-~\phantom{(}(x^4 + 0x^3 + x^2)}\phantom{+ax+b} \\ \phantom{x^4} x^3 + 7x^2 + ax + b \\ \underline{-~\phantom{(}(x^3 + 0x^2 + x)}\phantom{+b} \\ \phantom{x^4+x^3} 7x^2 + (a-1)x + b \\ \underline{-~\phantom{()} (7x^2 + 0x + 7)} \\ \phantom{x^4+x^3+7x^2} (a-1)x + (b-7) \end{array}$

The remainder obtained from the division is $(a-1)x + (b-7)$.

For the polynomial to be exactly divisible by $x^2 + 1$, the remainder must be equal to 0.

So, we set the remainder equal to $0x + 0$:

Equating the coefficients of the corresponding powers of $x$ on both sides:

Equating the coefficients of $x$:

Solving for $a$: $a = 1$

Equating the constant terms:

Solving for $b$: $b = 7$

Therefore, the values of $a$ and $b$ are $a=1$ and $b=7$.

Let the given quadratic polynomial be $f(x) = ax^2 + bx + c$. Since $a \neq 0$ and $c \neq 0$, the polynomial is quadratic, and none of its zeros are zero.

Let $\alpha$ and $\beta$ be the zeros of the polynomial $f(x)$.

According to the relationship between the zeros and coefficients of a quadratic polynomial:

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

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

We need to find a quadratic polynomial whose zeros are the reciprocals of $\alpha$ and $\beta$. Let the new zeros be $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.

Let $S'$ be the sum of the new zeros and $P'$ be the product of the new zeros.

Sum of new zeros: $S' = \frac{1}{\alpha} + \frac{1}{\beta}$

Combine the fractions:

$S' = \frac{\beta + \alpha}{\alpha \beta}$

$S' = \frac{\alpha + \beta}{\alpha \beta}$

Substitute the sum and product of the original zeros:

$S' = \frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{a} \times \frac{a}{c} = -\frac{b}{c}$

Product of new zeros: $P' = \left(\frac{1}{\alpha}\right) \times \left(\frac{1}{\beta}\right)$

$P' = \frac{1}{\alpha \beta}$

Substitute the product of the original zeros:

$P' = \frac{1}{\frac{c}{a}} = \frac{a}{c}$

A quadratic polynomial with sum of zeros $S'$ and product of zeros $P'$ can be written as $k(x^2 - S'x + P')$, where $k$ is any non-zero constant.

Substitute the values of $S'$ and $P'$:

$p'(x) = k\left(x^2 - \left(-\frac{b}{c}\right)x + \frac{a}{c}\right)$

$p'(x) = k\left(x^2 + \frac{b}{c}x + \frac{a}{c}\right)$

For simplicity, we can choose $k=c$ (since $c \neq 0$).

$p'(x) = c\left(x^2 + \frac{b}{c}x + \frac{a}{c}\right)$

$p'(x) = cx^2 + c\left(\frac{b}{c}x\right) + c\left(\frac{a}{c}\right)$

$p'(x) = cx^2 + bx + a$

Thus, a quadratic polynomial whose zeros are the reciprocals of the zeros of $ax^2 + bx + c$ is $cx^2 + bx + a$.