Chapter 2 Polynomials (Class 10 - Maths Old NCERT Textbook Solutions)

Concept:

The zeroes of a polynomial $p(x)$ are the x-coordinates of the points where the graph of the polynomial, $y = p(x)$, intersects the x-axis. To find the number of zeroes for each polynomial, we need to count the number of times its graph cuts or touches the x-axis.

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(i)

In the first graph, the curve intersects the x-axis at exactly one point.

Therefore, the number of zeroes of the polynomial $p(x)$ is 1.

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(ii)

In the second graph, the curve intersects the x-axis at two distinct points.

Therefore, the number of zeroes of the polynomial $p(x)$ is 2.

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(iii)

In the third graph, the curve intersects the x-axis at three distinct points.

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

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(iv)

In the fourth graph, the straight line intersects the x-axis at exactly one point.

Therefore, the number of zeroes of the polynomial $p(x)$ is 1.

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(v)

In the fifth graph, the curve touches the x-axis at exactly one point.

Therefore, the number of zeroes of the polynomial $p(x)$ is 1.

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(vi)

In the sixth graph, the curve intersects the x-axis at four distinct points.

Therefore, the number of zeroes of the polynomial $p(x)$ is 4.

Concept:

The number of zeroes of a polynomial p(x) corresponds to the number of times the graph of the polynomial, y = p(x), intersects or touches the x-axis. Each distinct point of intersection or contact with the x-axis represents a zero of the polynomial.

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(i)

The graph is a horizontal line that is parallel to the x-axis and does not intersect it at any point.

Therefore, the number of zeroes is 0.

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(ii)

The graph intersects the x-axis at exactly one point.

Therefore, the number of zeroes is 1.

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(iii)

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

Therefore, the number of zeroes is 3.

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(iv)

The graph is a parabola that intersects the x-axis at two distinct points.

Therefore, the number of zeroes is 2.

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(v)

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

Therefore, the number of zeroes is 4.

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(vi)

The graph intersects the x-axis at one point and touches the x-axis at two other points. In total, the graph meets the x-axis at three distinct points.

Therefore, the number of zeroes is 3.

Given:

The quadratic polynomial is $p(x) = x^2 + 7x + 10$.

Comparing this with the standard form $ax^2 + bx + c$, we have the coefficients: $a = 1$, $b = 7$, and $c = 10$.

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To Find:

1. The zeroes of the polynomial.

2. To verify the relationship between the zeroes and the coefficients.

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Solution:

1. Finding the Zeroes

To find the zeroes, we set the polynomial equal to zero:

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

We solve this equation by splitting the middle term. We need two numbers whose product is 10 and whose sum is 7. The numbers are 5 and 2.

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

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

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

This gives two possible solutions:

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

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

Therefore, the zeroes of the polynomial are -5 and -2.

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2. Verification of the Relationship

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

Sum of the zeroes:

Calculated sum: $\alpha + \beta = (-5) + (-2) = -7$.

Sum from coefficients: $-\frac{b}{a} = -\frac{7}{1} = -7$.

Product of the zeroes:

Calculated product: $\alpha \beta = (-5) \times (-2) = 10$.

Product from coefficients: $\frac{c}{a} = \frac{10}{1} = 10$.

The relationship between the zeroes and the coefficients is verified.

Given:

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

Comparing this with the standard form $ax^2 + bx + c$, we can write it as $x^2 + 0x - 3$. The coefficients are: $a = 1$, $b = 0$, and $c = -3$.

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To Find:

1. The zeroes of the polynomial.

2. To verify the relationship between the zeroes and the coefficients.

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Solution:

1. Finding the Zeroes

To find the zeroes, we set the polynomial equal to zero:

$x^2 - 3 = 0$

$x^2 = 3$

Taking the square root of both sides:

$x = \pm \sqrt{3}$

Therefore, the zeroes of the polynomial are $\sqrt{3}$ and $-\sqrt{3}$.

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2. Verification of the Relationship

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

Sum of the zeroes:

Calculated sum: $\alpha + \beta = \sqrt{3} + (-\sqrt{3}) = 0$.

Sum from coefficients: $-\frac{b}{a} = -\frac{0}{1} = 0$.

Product of the zeroes:

Calculated product: $\alpha \beta = (\sqrt{3}) \times (-\sqrt{3}) = -3$.

Product from coefficients: $\frac{c}{a} = \frac{-3}{1} = -3$.

The relationship between the zeroes and the coefficients is verified.

Given:

Sum of the zeroes, $\alpha + \beta = -3$.

Product of the zeroes, $\alpha\beta = 2$.

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To Find:

A quadratic polynomial with the given properties.

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Solution:

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

$p(x) = k \cdot [x^2 - (\alpha + \beta)x + \alpha\beta]$, where $k$ is any non-zero real number.

Substituting the given sum and product of the zeroes into this formula:

$p(x) = k \cdot [x^2 - (-3)x + 2]$

$p(x) = k \cdot [x^2 + 3x + 2]$

To find 'a' quadratic polynomial, we can choose the simplest value for $k$, which is $k=1$.

For $k=1$, the required polynomial is:

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

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Final Answer: A required quadratic polynomial is $x^2 + 3x + 2$.

Given:

Cubic polynomial: $p(x) = 3x^3 - 5x^2 - 11x - 3$.

Proposed zeroes: $3, -1, -\frac{1}{3}$.

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Part 1: Verification of Zeroes

To verify if a number is a zero, we substitute it into the polynomial. If the result is 0, it is a zero.

For $x = 3$:

$p(3) = 3(3)^3 - 5(3)^2 - 11(3) - 3 = 3(27) - 5(9) - 33 - 3 \ $$ = 81 - 45 - 33 - 3 = 81 - 81 = 0$.

So, 3 is a zero.

For $x = -1$:

$p(-1) = 3(-1)^3 - 5(-1)^2 - 11(-1) - 3 = 3(-1) - 5(1) + 11 - 3 \ $$ = -3 - 5 + 11 - 3 = -11 + 11 = 0$.

So, -1 is a zero.

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

$p(-\frac{1}{3}) = 3(-\frac{1}{3})^3 - 5(-\frac{1}{3})^2 - 11(-\frac{1}{3}) - 3$

$= 3(-\frac{1}{27}) - 5(\frac{1}{9}) + \frac{11}{3} - 3$

$= -\frac{1}{9} - \frac{5}{9} + \frac{33}{9} - \frac{27}{9} = \frac{-1 - 5 + 33 - 27}{9} = \frac{0}{9} = 0$.

So, $-\frac{1}{3}$ is a zero.

All three given values are verified as zeroes of the polynomial.

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Part 2: Verification of Relationship between Zeroes and Coefficients

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

The polynomial is $p(x) = 3x^3 - 5x^2 - 11x - 3$. Comparing with $ax^3+bx^2+cx+d$, we have:

$a=3, b=-5, c=-11, d=-3$.

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

LHS: $\alpha + \beta + \gamma = 3 + (-1) + (-\frac{1}{3}) = 2 - \frac{1}{3} = \frac{5}{3}$.

RHS: $-\frac{b}{a} = -\frac{-5}{3} = \frac{5}{3}$.

Since LHS = RHS, this is verified.

2. Sum of products of zeroes taken two at a time ($\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$):

LHS: $\alpha\beta + \beta\gamma + \gamma\alpha = (3)(-1) + (-1)(-\frac{1}{3}) + (-\frac{1}{3})(3) = -3 + \frac{1}{3} - 1 \ $$ = -4 + \frac{1}{3} = -\frac{11}{3}$.

RHS: $\frac{c}{a} = \frac{-11}{3}$.

Since LHS = RHS, this is verified.

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

LHS: $\alpha\beta\gamma = (3)(-1)(-\frac{1}{3}) = 1$.

RHS: $-\frac{d}{a} = -\frac{-3}{3} = 1$.

Since LHS = RHS, this is verified.

All relationships are verified.

We will find the zeroes for each quadratic polynomial and verify the relationship between the zeroes and the coefficients using the formulas: Sum of zeroes $= -\frac{b}{a}$ and Product of zeroes $= \frac{c}{a}$.

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(i) $x^2 – 2x – 8$

The polynomial is $p(x) = x^2 - 2x - 8$. Comparing with $ax^2 + bx + c$, we have $a=1$, $b=-2$, $c=-8$.

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

$$x^2 - 2x - 8 = 0$$

Factor the quadratic expression. We need two numbers that multiply to -8 and add to -2. These numbers are 2 and -4.

$$x^2 + 2x - 4x - 8 = 0$$

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

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

Setting each factor to zero gives the zeroes:

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

$x - 4 = 0 \implies x = 4$

The zeroes are $\alpha = -2$ and $\beta = 4$.

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Verification:

Sum of zeroes: $\alpha + \beta = -2 + 4 = 2$

From coefficients: $-\frac{b}{a} = -\frac{(-2)}{1} = 2$

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Product of zeroes: $\alpha \times \beta = (-2) \times 4 = -8$

From coefficients: $\frac{c}{a} = \frac{-8}{1} = -8$

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(ii) $4s^2 – 4s + 1$

The polynomial is $p(s) = 4s^2 - 4s + 1$. Comparing with $as^2 + bs + c$, we have $a=4$, $b=-4$, $c=1$.

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

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

Factor the quadratic expression. This is a perfect square trinomial: $(2s - 1)^2$.

$$(2s - 1)^2 = 0$$

Setting the factor to zero gives the zeroes:

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

The polynomial has a repeated zero at $s = \frac{1}{2}$. So, the zeroes are $\alpha = \frac{1}{2}$ and $\beta = \frac{1}{2}$.

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Verification:

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

From coefficients: $-\frac{b}{a} = -\frac{(-4)}{4} = \frac{4}{4} = 1$

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Product of zeroes: $\alpha \times \beta = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

From coefficients: $\frac{c}{a} = \frac{1}{4}$

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(iii) $6x^2 – 3 – 7x$

Rearrange the polynomial in standard form: $p(x) = 6x^2 - 7x - 3$. Comparing with $ax^2 + bx + c$, we have $a=6$, $b=-7$, $c=-3$.

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

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

Factor the quadratic expression. We look for two numbers that multiply to $a \times c = 6 \times (-3) = -18$ and add to $b = -7$. These numbers are 2 and -9.

Rewrite the middle term $-7x$ as $2x - 9x$:

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

Group the terms and factor:

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

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

Setting each factor to zero gives the zeroes:

$3x + 1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3}$

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

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

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Verification:

Sum of zeroes: $\alpha + \beta = -\frac{1}{3} + \frac{3}{2} = \frac{-1 \times 2 + 3 \times 3}{6} = \frac{-2 + 9}{6} = \frac{7}{6}$

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

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

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

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(iv) $4u^2 + 8u$

The polynomial is $p(u) = 4u^2 + 8u$. Comparing with $au^2 + bu + c$, we have $a=4$, $b=8$, $c=0$.

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To find the zeroes, set $p(u) = 0$:

$$4u^2 + 8u = 0$$

Factor out the common term $4u$:

$$4u(u + 2) = 0$$

Setting each factor to zero gives the zeroes:

$4u = 0 \implies u = 0$

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

The zeroes are $\alpha = 0$ and $\beta = -2$.

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Verification:

Sum of zeroes: $\alpha + \beta = 0 + (-2) = -2$

From coefficients: $-\frac{b}{a} = -\frac{8}{4} = -2$

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Product of zeroes: $\alpha \times \beta = 0 \times (-2) = 0$

From coefficients: $\frac{c}{a} = \frac{0}{4} = 0$

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(v) $t^2 – 15$

The polynomial is $p(t) = t^2 - 15$. Comparing with $at^2 + bt + c$, we have $a=1$, $b=0$, $c=-15$.

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

$$t^2 - 15 = 0$$

$$t^2 = 15$$

$$t = \pm \sqrt{15}$$

The zeroes are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.

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Verification:

Sum of zeroes: $\alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0$

From coefficients: $-\frac{b}{a} = -\frac{0}{1} = 0$

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

From coefficients: $\frac{c}{a} = \frac{-15}{1} = -15$

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(vi) $3x^2 – x – 4$

The polynomial is $p(x) = 3x^2 - x - 4$. Comparing with $ax^2 + bx + c$, we have $a=3$, $b=-1$, $c=-4$.

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

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

Factor the quadratic expression. We look for two numbers that multiply to $a \times c = 3 \times (-4) = -12$ and add to $b = -1$. These numbers are 3 and -4.

Rewrite the middle term $-x$ as $3x - 4x$:

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

Group the terms and factor:

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

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

Setting each factor to zero gives the zeroes:

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

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

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

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Verification:

Sum of zeroes: $\alpha + \beta = -1 + \frac{4}{3} = \frac{-3}{3} + \frac{4}{3} = \frac{-3+4}{3} = \frac{1}{3}$

From coefficients: $-\frac{b}{a} = -\frac{(-1)}{3} = \frac{1}{3}$

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

From coefficients: $\frac{c}{a} = \frac{-4}{3} = -\frac{4}{3}$

Let the sum of the zeroes of a quadratic polynomial be $S$ and the product of the zeroes be $P$. A quadratic polynomial with sum of zeroes $S$ and product of zeroes $P$ can be written as:

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

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

We will use this formula for each part, taking $k=1$ for simplicity to find 'a' quadratic polynomial.

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(i) Sum = $\frac{1}{4}$, Product = $-1$

Here, $S = \frac{1}{4}$ and $P = -1$.

Using the formula with $k=1$:

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

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

To get integer coefficients, we can take $k=4$.

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

A quadratic polynomial is $4x^2 - x - 4$.

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(ii) Sum = $\sqrt{2}$, Product = $\frac{1}{3}$

Here, $S = \sqrt{2}$ and $P = \frac{1}{3}$.

Using the formula with $k=1$:

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

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

To clear the fraction, we can take $k=3$.

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

A quadratic polynomial is $3x^2 - 3\sqrt{2}x + 1$.

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(iii) Sum = 0, Product = $\sqrt{5}$

Here, $S = 0$ and $P = \sqrt{5}$.

Using the formula with $k=1$:

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

$$p(x) = x^2 + \sqrt{5}$$

A quadratic polynomial is $x^2 + \sqrt{5}$.

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(iv) Sum = 1, Product = 1

Here, $S = 1$ and $P = 1$.

Using the formula with $k=1$:

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

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

A quadratic polynomial is $x^2 - x + 1$.

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(v) Sum = $-\frac{1}{4}$, Product = $\frac{1}{4}$

Here, $S = -\frac{1}{4}$ and $P = \frac{1}{4}$.

Using the formula with $k=1$:

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

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

To clear the fractions, we can take $k=4$.

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

A quadratic polynomial is $4x^2 + x + 1$.

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(vi) Sum = 4, Product = 1

Here, $S = 4$ and $P = 1$.

Using the formula with $k=1$:

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

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

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

To Find:

The quotient and remainder when the polynomial $p(x) = 2x^2 + 3x + 1$ is divided by the polynomial $g(x) = x + 2$.

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Solution using Polynomial Long Division:

We set up the long division as follows:

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

Step-by-step Explanation:

1. Divide the first term of the dividend ($2x^2$) by the first term of the divisor ($x$). This gives $2x$. Write $2x$ as the first term of the quotient.
2. Multiply the divisor $(x+2)$ by $2x$ to get $2x(x+2) = 2x^2 + 4x$.
3. Subtract this result from the dividend: $(2x^2+3x) - (2x^2+4x) = -x$. Bring down the next term (+1).
4. The new dividend is $-x+1$. Divide the first term of this new dividend ($-x$) by the first term of the divisor ($x$). This gives $-1$. Write $-1$ as the second term of the quotient.
5. Multiply the divisor $(x+2)$ by $-1$ to get $-1(x+2) = -x-2$.
6. Subtract this result from the new dividend: $(-x+1) - (-x-2) = -x+1+x+2 = 3$.

The process stops because the degree of the remainder (3, which has degree 0) is less than the degree of the divisor ($x+2$, which has degree 1).

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Result:

The quotient is $q(x) = 2x - 1$.

The remainder is $r(x) = 3$.

To Find:

The quotient and remainder when $p(x) = 3x^3 + x^2 + 2x + 5$ is divided by $g(x) = 1 + 2x + x^2$.

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Solution using Polynomial Long Division:

First, we write the dividend and the divisor in standard form (descending order of their degrees).

Dividend: $p(x) = 3x^3 + x^2 + 2x + 5$

Divisor: $g(x) = x^2 + 2x + 1$

Now, we perform the long division:

$\begin{array}{r} 3x - 5\phantom{)} \\ x^2+2x+1{\overline{\smash{\big)}\,3x^3+x^2+2x+5\phantom{)}}} \\ \underline{-~\phantom{()}(3x^3+6x^2+3x)\phantom{)}} \\ -5x^2-x+5\phantom{)} \\ \underline{-~\phantom{()}(-5x^2-10x-5)} \\ 9x+10\phantom{)} \end{array} $

Step-by-step Explanation:

1. Divide $3x^3$ by $x^2$ to get $3x$. This is the first term of the quotient.
2. Multiply $3x$ by the divisor $(x^2+2x+1)$ to get $3x^3+6x^2+3x$.
3. Subtract this from the dividend: $(3x^3+x^2+2x) - (3x^3+6x^2+3x) = -5x^2-x$. Bring down the next term (+5).
4. The new dividend is $-5x^2-x+5$. Divide $-5x^2$ by $x^2$ to get $-5$. This is the second term of the quotient.
5. Multiply $-5$ by the divisor $(x^2+2x+1)$ to get $-5x^2-10x-5$.
6. Subtract this from the new dividend: $(-5x^2-x+5) - (-5x^2-10x-5) = 9x+10$.

The degree of the remainder ($9x+10$, degree 1) is less than the degree of the divisor ($x^2+2x+1$, degree 2), so the division stops.

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Result:

The quotient is $q(x) = 3x - 5$.

The remainder is $r(x) = 9x + 10$.

Given:

Dividend: $p(x) = 3x^2 - x^3 - 3x + 5$

Divisor: $g(x) = x - 1 - x^2$

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Part 1: Polynomial Long Division

First, we write both polynomials in standard form (descending order of degrees).

Dividend: $p(x) = -x^3 + 3x^2 - 3x + 5$

Divisor: $g(x) = -x^2 + x - 1$

Now, we perform the long division:

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

From the division, we get:

Quotient, $q(x) = x - 2$

Remainder, $r(x) = 3$

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Part 2: Verification of the Division Algorithm

The division algorithm for polynomials states that:

Dividend = Divisor × Quotient + Remainder

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

Let's verify this by calculating the right-hand side (RHS).

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

First, multiply the divisor and the quotient:

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

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

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

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

Now, add the remainder:

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

The result, $-x^3 + 3x^2 - 3x + 5$, is equal to the original dividend $p(x)$.

Since LHS = RHS, the division algorithm is verified.

Given:

The polynomial is $p(x) = 2x^4 - 3x^3 - 3x^2 + 6x - 2$.

Two of its zeroes are $\sqrt{2}$ and $-\sqrt{2}$.

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To Find:

All the zeroes of the polynomial $p(x)$.

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Solution:

Since $\sqrt{2}$ and $-\sqrt{2}$ are zeroes of the polynomial, $(x - \sqrt{2})$ and $(x + \sqrt{2})$ must be factors of $p(x)$.

The product of these factors will also be a factor of $p(x)$.

Product = $(x - \sqrt{2})(x + \sqrt{2}) = x^2 - (\sqrt{2})^2 = x^2 - 2$.

So, $g(x) = x^2 - 2$ is a factor of $p(x)$. To find the other factors, we divide $p(x)$ by $g(x)$.

$\begin{array}{r} 2x^2 -3x + 1\phantom{)} \\ x^2+0x-2{\overline{\smash{\big)}\,2x^4-3x^3-3x^2+6x-2\phantom{)}}} \\ \underline{-~\phantom{()}(2x^4+0x^3-4x^2)\phantom{-b)}} \\ -3x^3+\phantom{0}x^2+6x-2\phantom{)} \\ \underline{-~\phantom{()}(-3x^3+0x^2+6x)} \\ x^2+0x-2\phantom{)} \\ \underline{-~\phantom{()}(x^2+0x-2)} \\ 0\phantom{)} \end{array} $

The quotient is $q(x) = 2x^2 - 3x + 1$. The remainder is 0, as expected.

The remaining zeroes of $p(x)$ are the zeroes of the quotient $q(x)$. To find them, we set $q(x) = 0$.

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

We solve this quadratic equation by splitting the middle term. We need two numbers that multiply to $2 \times 1 = 2$ and add to $-3$. The numbers are $-2$ and $-1$.

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

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

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

This gives two more zeroes:

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

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

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Final Answer:

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

(i) $p(x) = x^3 – 3x^2 + 5x – 3$, $g(x) = x^2 – 2$

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

Quotient = $x - 3$

Remainder = $7x - 9$

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(ii) $p(x) = x^4 – 3x^2 + 4x + 5$, $g(x) = x^2 + 1 – x$

First, write the divisor in standard form: $g(x) = x^2 - x + 1$.

$\begin{array}{r} x^2+x-3\phantom{)} \\ x^2-x+1{\overline{\smash{\big)}\,x^4+0x^3-3x^2+4x+5\phantom{)}}} \\ \underline{-~\phantom{()}(x^4-x^3+\phantom{0}x^2)\phantom{-b)}} \\ x^3-4x^2+4x+5\phantom{)} \\ \underline{-~\phantom{()}(x^3-\phantom{0}x^2+\phantom{0}x)\phantom{-b)}} \\ -3x^2+3x+5\phantom{)} \\ \underline{-~\phantom{()}(-3x^2+3x-3)} \\ 8\phantom{)} \end{array} $

Quotient = $x^2 + x - 3$

Remainder = $8$

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(iii) $p(x) = x^4 – 5x + 6$, $g(x) = 2 – x^2$

First, write the divisor in standard form: $g(x) = -x^2 + 2$.

$\begin{array}{r} -x^2 - 2\phantom{)} \\ -x^2+0x+2{\overline{\smash{\big)}\,x^4+0x^3+0x^2-5x+6\phantom{)}}} \\ \underline{-~\phantom{()}(x^4+0x^3-2x^2)\phantom{-b)}} \\ 2x^2-5x+6\phantom{)} \\ \underline{-~\phantom{()}(2x^2+0x-4)} \\ -5x+10\phantom{)} \end{array} $

Quotient = $-x^2 - 2$

Remainder = $-5x + 10$

The first polynomial is a factor of the second if the remainder after division is zero.

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(i) $t^2 – 3$, $2t^4 + 3t^3 – 2t^2 – 9t – 12$

$\begin{array}{r} 2t^2 + 3t + 4\phantom{)} \\ t^2+0t-3{\overline{\smash{\big)}\,2t^4+3t^3-2t^2-9t-12\phantom{)}}} \\ \underline{-~\phantom{()}(2t^4+0t^3-6t^2)\phantom{-b)}} \\ 3t^3+4t^2-9t-12\phantom{)} \\ \underline{-~\phantom{()}(3t^3+0t^2-9t)} \\ 4t^2+0t-12\phantom{)} \\ \underline{-~\phantom{()}(4t^2+0t-12)} \\ 0\phantom{)} \end{array} $

Since the remainder is 0, yes, the first polynomial is a factor of the second.

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(ii) $x^2 + 3x + 1$, $3x^4 + 5x^3 – 7x^2 + 2x + 2$

$\begin{array}{r} 3x^2 - 4x + 2\phantom{)} \\ x^2+3x+1{\overline{\smash{\big)}\,3x^4+5x^3-7x^2+2x+2\phantom{)}}} \\ \underline{-~\phantom{()}(3x^4+9x^3+3x^2)\phantom{-b)}} \\ -4x^3-10x^2+2x+2\phantom{)} \\ \underline{-~\phantom{()}(-4x^3-12x^2-4x)} \\ 2x^2+6x+2\phantom{)} \\ \underline{-~\phantom{()}(2x^2+6x+2)} \\ 0\phantom{)} \end{array} $

Since the remainder is 0, yes, the first polynomial is a factor of the second.

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(iii) $x^3 – 3x + 1$, $x^5 – 4x^3 + x^2 + 3x + 1$

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

Since the remainder is 2 (not 0), no, the first polynomial is not a factor of the second.

Given:

The polynomial is $p(x) = 3x^4 + 6x^3 - 2x^2 - 10x - 5$.

Two of its zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.

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To Find:

All the other zeroes of the polynomial.

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Solution:

Since $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are zeroes, the corresponding factors are $(x - \sqrt{\frac{5}{3}})$ and $(x + \sqrt{\frac{5}{3}})$.

The product of these factors is also a factor of the polynomial:

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

To avoid fractions, we can say that $3(x^2 - \frac{5}{3}) = 3x^2 - 5$ is a factor of the polynomial.

Now, we divide the given polynomial by $3x^2 - 5$ to find the other factors.

$\begin{array}{r} x^2+2x+1\phantom{)} \\ 3x^2+0x-5{\overline{\smash{\big)}\,3x^4+6x^3-2x^2-10x-5\phantom{)}}} \\ \underline{-~\phantom{()}(3x^4+0x^3-5x^2)\phantom{-b)}} \\ 6x^3+3x^2-10x-5\phantom{)} \\ \underline{-~\phantom{()}(6x^3+0x^2-10x)} \\ 3x^2+0x-5\phantom{)} \\ \underline{-~\phantom{()}(3x^2+0x-5)} \\ 0\phantom{)} \end{array} $

The quotient is $q(x) = x^2 + 2x + 1$. The remaining zeroes are the zeroes of this quotient.

Set the quotient to zero to find the other zeroes:

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

This is a perfect square trinomial:

$(x + 1)^2 = 0$

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

Since the factor is squared, the zero $x = -1$ has a multiplicity of 2.

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Final Answer: The other two zeroes of the polynomial are -1 and -1.

Given:

Dividend, $p(x) = x^3 - 3x^2 + x + 2$

Quotient, $q(x) = x - 2$

Remainder, $r(x) = -2x + 4$

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To Find:

The divisor polynomial, $g(x)$.

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Solution:

According to the division algorithm for polynomials:

Dividend = (Divisor × Quotient) + Remainder

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

We can rearrange this to solve for $g(x)$:

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

$g(x) = \frac{p(x) - r(x)}{q(x)}$

First, calculate $p(x) - r(x)$:

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

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

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

Now, we need to divide this result by the quotient $q(x) = x - 2$.

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

The result of the division is $x^2 - x + 1$.

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Final Answer: The divisor polynomial is $g(x) = x^2 - x + 1$.

The division algorithm is $p(x) = g(x) \cdot q(x) + r(x)$, where deg $r(x)$ < deg $g(x)$ or $r(x) = 0$.

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(i) deg p(x) = deg q(x)

This condition is met when the divisor $g(x)$ is a constant (a polynomial of degree 0).

Example:

Let $p(x) = 2x^2 - 2x + 14$

Let $g(x) = 2$

Dividing $p(x)$ by $g(x)$, we get:

$q(x) = x^2 - x + 7$

$r(x) = 0$

Here, deg $p(x) = 2$ and deg $q(x) = 2$.

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(ii) deg q(x) = deg r(x)

We need the degree of the quotient to be equal to the degree of the remainder. Also, the degree of the remainder must be less than the degree of the divisor.

Example:

Let $p(x) = x^3 + x^2 + x + 1$

Let $g(x) = x^2 - 1$

Dividing $p(x)$ by $g(x)$:

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

$q(x) = x + 1$

$r(x) = 2x + 2$

Here, deg $q(x) = 1$ and deg $r(x) = 1$.

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(iii) deg r(x) = 0

This means the remainder $r(x)$ must be a non-zero constant.

Example:

Let $p(x) = x^3 + 1$

Let $g(x) = x^2$

Dividing $p(x)$ by $g(x)$:

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

$q(x) = x$

$r(x) = 1$

Here, deg $r(x) = 0$.

(i) $p(x) = 2x^3 + x^2 – 5x + 2$; Zeroes: $\frac{1}{2}$, 1, – 2

Part 1: Verification of Zeroes

We substitute each given number into the polynomial $p(x)$.

$p(\frac{1}{2}) = 2(\frac{1}{2})^3 + (\frac{1}{2})^2 - 5(\frac{1}{2}) + 2 = 2(\frac{1}{8}) + \frac{1}{4} - \frac{5}{2} + 2 \ $$ = \frac{1}{4} + \frac{1}{4} - \frac{10}{4} + \frac{8}{4} = \frac{1+1-10+8}{4} = \frac{0}{4} = 0$.

$p(1) = 2(1)^3 + (1)^2 - 5(1) + 2 = 2 + 1 - 5 + 2 = 5 - 5 = 0$.

$p(-2) = 2(-2)^3 + (-2)^2 - 5(-2) + 2 = 2(-8) + 4 + 10 + 2 \ $$ = -16 + 16 = 0$.

Since the result is 0 in all cases, the given numbers are indeed the zeroes of the polynomial.

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Part 2: Verification of Relationship between Zeroes and Coefficients

Let the zeroes be $\alpha = \frac{1}{2}$, $\beta = 1$, $\gamma = -2$.

For $p(x) = 2x^3 + x^2 - 5x + 2$, we have coefficients $a=2, b=1, c=-5, d=2$.

Sum of zeroes: $\alpha + \beta + \gamma = \frac{1}{2} + 1 - 2 = \frac{1}{2} - 1 = -\frac{1}{2}$.

From coefficients: $-\frac{b}{a} = -\frac{1}{2}$. (Verified)

Sum of products of zeroes taken two at a time:

$\alpha\beta + \beta\gamma + \gamma\alpha = (\frac{1}{2})(1) + (1)(-2) + (-2)(\frac{1}{2}) = \frac{1}{2} - 2 - 1 \ $$ = \frac{1}{2} - 3 = -\frac{5}{2}$.

From coefficients: $\frac{c}{a} = \frac{-5}{2}$. (Verified)

Product of zeroes: $\alpha\beta\gamma = (\frac{1}{2})(1)(-2) = -1$.

From coefficients: $-\frac{d}{a} = -\frac{2}{2} = -1$. (Verified)

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(ii) $p(x) = x^3 – 4x^2 + 5x – 2$; Zeroes: 2, 1, 1

Part 1: Verification of Zeroes

$p(2) = (2)^3 - 4(2)^2 + 5(2) - 2 = 8 - 4(4) + 10 - 2 = 8 - 16 + 10 - 2 \ $$ = 18 - 18 = 0$.

$p(1) = (1)^3 - 4(1)^2 + 5(1) - 2 = 1 - 4 + 5 - 2 = 6 - 6 = 0$.

Since 1 is listed twice, it is a zero with multiplicity 2. The given numbers are the zeroes.

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Part 2: Verification of Relationship between Zeroes and Coefficients

Let the zeroes be $\alpha = 2$, $\beta = 1$, $\gamma = 1$.

For $p(x) = x^3 - 4x^2 + 5x - 2$, we have coefficients $a=1, b=-4, c=5, d=-2$.

Sum of zeroes: $\alpha + \beta + \gamma = 2 + 1 + 1 = 4$.

From coefficients: $-\frac{b}{a} = -\frac{-4}{1} = 4$. (Verified)

Sum of products of zeroes taken two at a time:

$\alpha\beta + \beta\gamma + \gamma\alpha = (2)(1) + (1)(1) + (1)(2) = 2 + 1 + 2 = 5$.

From coefficients: $\frac{c}{a} = \frac{5}{1} = 5$. (Verified)

Product of zeroes: $\alpha\beta\gamma = (2)(1)(1) = 2$.

From coefficients: $-\frac{d}{a} = -\frac{-2}{1} = 2$. (Verified)

Given:

Let the zeroes of the cubic polynomial be $\alpha, \beta, \gamma$.

Sum of zeroes: $\alpha + \beta + \gamma = 2$.

Sum of the product of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = -7$.

Product of zeroes: $\alpha\beta\gamma = -14$.

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To Find:

A cubic polynomial with these properties.

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Solution:

A cubic polynomial can be constructed using the following formula:

$p(x) = k \cdot [x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - (\alpha\beta\gamma)]$

where $k$ is any non-zero constant.

Substituting the given values into the formula:

$p(x) = k \cdot [x^3 - (2)x^2 + (-7)x - (-14)]$

$p(x) = k \cdot [x^3 - 2x^2 - 7x + 14]$

To find 'a' cubic polynomial, we can take the simplest case where $k=1$.

$p(x) = x^3 - 2x^2 - 7x + 14$

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Final Answer: A required cubic polynomial is $x^3 - 2x^2 - 7x + 14$.

Given:

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

The zeroes are in the form $a-b, a, a+b$.

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To Find:

The values of $a$ and $b$.

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Solution:

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

For the polynomial $p(x) = x^3 - 3x^2 + x + 1$, the coefficients are $A=1, B=-3, C=1, D=1$.

Using the relationship for the sum of zeroes:

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

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

$3a = 3$

$a = 1$

Now we have the value of $a$. The zeroes are $1-b, 1, 1+b$.

Using the relationship for the product of zeroes:

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

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

Substitute $a=1$ into the equation:

$(1-b)(1)(1+b) = -1$

$1 - b^2 = -1$

$2 = b^2$

$b = \pm\sqrt{2}$

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Final Answer: The values are $a = 1$ and $b = \pm\sqrt{2}$.

Given:

Polynomial $p(x) = x^4 - 6x^3 - 26x^2 + 138x - 35$.

Two zeroes are $2 + \sqrt{3}$ and $2 - \sqrt{3}$.

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To Find:

The other two zeroes of the polynomial.

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Solution:

If $2 + \sqrt{3}$ and $2 - \sqrt{3}$ are zeroes, then $(x - (2 + \sqrt{3}))$ and $(x - (2 - \sqrt{3}))$ are factors.

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

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

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

$g(x) = (x-2)^2 - (\sqrt{3})^2 = (x^2 - 4x + 4) - 3 = x^2 - 4x + 1$.

To find the other zeroes, we divide $p(x)$ by $g(x)$.

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

The quotient is $q(x) = x^2 - 2x - 35$. The other zeroes are the zeroes of this quotient.

Set $q(x) = 0$:

$x^2 - 2x - 35 = 0$

Factor the quadratic equation:

$(x-7)(x+5) = 0$

The other zeroes are $x = 7$ and $x = -5$.

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Final Answer: The other zeroes of the polynomial are 7 and -5.

Given:

Dividend $p(x) = x^4 - 6x^3 + 16x^2 - 25x + 10$.

Divisor $g(x) = x^2 - 2x + k$.

Remainder $r(x) = x + a$.

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To Find:

The values of $k$ and $a$.

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Solution:

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

$\begin{array}{r} x^2 - 4x + (8-k)\phantom{)} \\ x^2-2x+k{\overline{\smash{\big)}\,x^4-6x^3+16x^2-25x+10\phantom{)}}} \\ \underline{-~\phantom{()}(x^4-2x^3+\phantom{0}kx^2)\phantom{-b)}} \\ -4x^3+(16-k)x^2-25x\phantom{)} \\ \underline{-~\phantom{()}(-4x^3+\phantom{0}8x^2-4kx)} \\ (8-k)x^2+(-25+4k)x+10\phantom{)} \\ \underline{-~\phantom{()}((8-k)x^2+(-16+2k)x+k(8-k))} \\ (-25+4k - (-16+2k))x + (10 - k(8-k))\phantom{)}\\ (2k-9)x + (10-8k+k^2)\phantom{)} \end{array} $

The remainder we calculated is $(2k-9)x + (k^2 - 8k + 10)$.

We are given that the remainder is $x+a$. We can equate the coefficients of the calculated remainder and the given remainder.

Comparing the coefficient of $x$:

$2k-9 = 1$

$2k = 10$

$k = 5$

Comparing the constant term:

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

Substitute the value of $k=5$ into this equation:

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

$25 - 40 + 10 = a$

$-15 + 10 = a$

$a = -5$

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Final Answer: The values are $k = 5$ and $a = -5$.