Contents

## GSEB Solutions for Class 10 mathematics – Polynomials (English Medium)

### Exercise-2.1

**Question 1:**

Identify the type of the following polynomials : (on base of power)

- p(x) = x
^{2}– 5x + 6 - p(x) = x
^{2}– x^{3}+ x + q - p(x) = 5x
^{2}+ 8x + 3 - p(x) = x
^{3}

**Solution 1:**

- Here p(x) = x
^{2}– 5x + 6

Degree of this polynomial is 2. So it is a quadratic polynomial in x.

- Here p(x) = x
^{2}– x^{3}+ x + 1

∴ p(x) = -x^{3}+ x^{2}+ x + 1

Degree of this polynomial is 3. So it is a cubic polynomial in x.

- Here p(x) = 5x
^{2}+ 8x + 3

Degree of this polynomial is 2. So it is a quadratic polynomial in x.

- Here p(x) = x
^{3}

Degree of this polynomial is 3. So it is a cubic polynomial in x.

**Question 2:**

Obtain the degree of the following polynomials :

- p(x) = 3x – x
^{4}+ x^{2}+2x^{3}+ 7 - p(x) = x
^{3}– 3x – x^{2}+ 6 - p(x) = 3x – 9
- p(x) = 2x
^{2}– x + 1

**Solution :**

- Here p(x) = 3x – x
^{4 }+ x^{2 }+ 2x^{3 }+ 7

_{∴}p(x) =^{ }-x^{4}+ 2x^{3 }+ x^{2 }+ 3x + 7

The term containing the highest power in this polynomial is -x^{4}. So the degree of the polynomial is**4**. - 2. p(x) = x
^{3 }– 3x – x^{2 }+ 6

_{∴}p(x) = x^{3 }– x^{2 }– 3x + 6

The term containing the highest power in this polynomial is x^{3}. So the degree of the polynomial is**3**. - 3. Here, p(x) = 3x – 9

The term containing the highest power in this polynomial is 3x. So the degree of the polynomial is**1**. - p(x) = 2x
^{2 }– x + 1

The term containing the highest power in this polynomial 2x^{2}. So the degree of the polynomial is**2**.

**Question 3:**

Find the coefficients of the underlined terms :

- p(x) = 10x
^{3}+ 7x^{2}– 3x + 5 - p(x) = 7 – 5x
^{5}+ 3x^{4}+ x^{2}– x - p(x) = 25 – 125x
- p(x) = x
^{3}– x^{2}+ x +7

**Solution :**

- Here, p(x) = 10x
^{3 }+ 7x^{2 }– 3x + 5

Coefficient of the underlined term x^{3}is**10**. - Here, p(x) = 7- 5x
^{5 }+ 3x^{4 }+ x^{2 }– x

Coefficient of the underlined term x^{5}is**-5**. - p(x) = 25 – 125x

Coefficient of the underlined term x is**-125**. - p(x) = x
^{3 }– x^{2}+ x + 7

Coefficient of the underlined term x^{2}is**-1**.

**Question 4:**

Obtain the value of the following polynomials at the given values of x :

- p(x) = 2x
^{3 }+ 3x^{2}+ 7x + 9 ; at x = 0, 1 - p(x) = 3x
^{2}+ 10x + 7 ; at x = -3, 1 - p(x) = x
^{2}– 2x + 5 ; at x = -1, 5 - p(x) = 2x
^{4}– 3x^{3}+ 7x + 5 ; at = -2, 2

**Solution :**

- Here, p(x) = 2x
^{3 }+ 3x^{2 }+ 7x + 9

∴ p(0) = 2(0)^{3 }+ 3(0)^{2 }+ 7(0) + 9

= 0 + 0 + 0 + 9

= 9

Next, p(1) = 2(1)^{3 }+ 3(1)^{2 }+ 7(1) + 9

= 2 + 3 + 7 + 9

= 21 - Here, p(x) = 3x
^{2 }+ 10x + 7

∴ p(-3) = 3(-3)^{2 }+ 10(-3) + 7

= 3(9) + 10(-3) + 7

= 27- 30 + 7

∴ p(-3) = 4

Next, p(1) = 3(1)^{2 }+ 10(1) + 7

= 3 + 10 + 7

∴ p(1) = 20 - Here, p(x) = x
^{2 }– 2x + 5

∴p(-1) = (-1)^{2 }– 2(-1) + 5

= 1 + 2 + 5

∴ p(-1) = 8

Next, p(5) = (5)^{2 }– 2(5) + 5

= 25 – 10 + 5

∴ p(5) = 20 - Here, p(x) = 2x
^{4 }– 3x^{3 }+ 7x + 5

_{∴}p(-2) = 2(-2)^{4 }– 3(-2)^{3 }+ 7(- 2) + 5

= 2(16)- 3(-8) + 7(-2) + 5

= 32 + 24 – 14 + 5

∴p(-2) = 47

Next, p(2) = 2(2)^{4 }– 3(2)^{3 }+ 7(2) + 5

= 2(16) – 3(8) + 7(2) + 5

= 32- 24 + 14 + 5

∴ p(2)= 27

**Question 5:**

Examine the validity of the following statements :

- (x + 1) is a factor of p(x) = 3x
^{3}+ 2x^{2}+ 7x + 8 - (x + 2) is a factor of p(x) = x
^{3}+ x^{2}+ x + 2 - (x – 1) is a factor of p(x) = x
^{4}– 2x^{3}+ 3x – 2 - (x – 3) is a factor of p(x) = x
^{2}– 2x – 3

**Question 5(1):**

**Solution :**

If (x + 1) is factor of p(x), then p(-1) = 0.

Now, p(-1) = 3(-1)^{3 }+ 2(-1)^{2 }+ 7(-1) + 8

= 3(-1) + 2(1) + 7(-1) + 8

= -3 + 2 – 7 + 8

= 0

Thus, (x + 1) is a factor of the given polynomial.

∴ The given statement is valid.

**Question 5(2):**

**Solution :**

If(x + 2) is a factor of p(x), then p(-2) = 0.

Now, p(-2) = (-2)^{3 }+ (-2)^{2 }+ (-2) + 2

= (-8) + (4) + (-2) + 2

=-8 + 4 – 2 + 2

= -4 ≠ 0

Thus, (x + 2) is not a factor of the given polynomial.

∴ The given statement is invalid.

**Question 5(3):**

**Solution :**

If(x – 1) is a factor of p(x), then p(1) = 0.

Now, p(1) = (1)^{4 }– 2(1)^{3 }+ 3(1) – 2

= 1 – 2 + 3 – 2

= 0

Thus, (x – 1) is a factor of the given polynomial.

∴ The given statement is valid.

**Question 5(4):**

**Solution :**

If(x – 3) is a factor of p(x) then p(3) = 0.

Now, p(3) = (3)^{2 }– 2(3) – 3

= 9 – 6 – 3

= 0

Thus, (x -3) is a factor of the given polynomial.

∴ The given statement is valid.

**Question 6:**

Factorize the following polynomials :

- p(x) = x
^{3}– x^{2}– x + 1 - p(x) = 5x
^{2}+ 11x + 6 - p(x) = x
^{3}– 3x^{2}+ 9x – 27 - p(x) = x
^{3}+ 2x^{2}+ 3x + 2

**Question 6(1):**

**Solution :**

x^{3} – x^{2 }– x + 1 = x^{2}(x – 1) – 1(x – 1)

= (x – 1)(x^{2 }– 1)

= (x – 1)(x – 1)(x + 1)

= (x – 1)^{2}(x + 1)

**Question 6(2):**

**Solution :**

5x^{2 }+ 11x + 6 = 5x^{2 }+ 5x + 6x + 6

= 5x(x + 1) + 6(x + 1)

= (x + 1)(5x + 6)

**Question 6(3):**

**Solution :**

x^{3 }– 3x^{2 }+ 9x – 27 = x^{2}(x – 3) + 9(x – 3)

= (x – 3)(x^{2 }+ 9)

**Question 6(4):**

**Solution :**

x^{3 }+ 2x^{2 }+ 3x + 2 = __x ^{3 }+ x^{2 }__+

__x__+

^{2 }+ x__2x + 2__

= x

^{2}(x + 1) + x(x + 1) + 2(x + 1)

= (x + 1)(x

^{2 }+ x + 2)

**Question 7:**

Prove that x – 2 is a factor of p(x) = x^{3} – 2x^{2}

**Solution :**

If (x – 2) is a factor of p(x), then p(2)=0.

Now, p(2)= (2)^{3 }– 2(2)^{2
}= 8 – 2(4)

= 8 – 8

= 0

Thus, (x – 2) is a factor of p(x).

### Exercise-2.2

**Question 1:**

Find the number of zeros of the following polynomials .

- p(x) = x
^{2}– x - p(x) = x – x
^{2}– 1 - P(x) = 3x – 2
- p(x) = x
^{3}– x

**Question 1(1):**

**Solution :**

Here p(x)= x^{2 }– x

To find the zeros of p(x), let p(x) = 0

∴ x^{2 }– x = 0

∴ x(x – 1) = 0

∴ x = 0 or x = 1

∴ 0 and 1 are zeroes of p(x).

Hence, the number of zeroes of the given polynomial is 2.

**Question 1(2):**

**Solution :**

Here p(x) = x – x^{2 }– 1

To find the zeroes of p(x), let p(x) = 0

∴ x – x^{2 }– 1 = 0

∴ x^{2 }– x + 1= 0

(∵ The square of a real number cannot be negative.)

Hence, the number of zeroes of the given polynomial is 0.

**Question 1(3):**

**Solution :**

**Question 1(4):**

**Solution :**

Here p(x) = x^{3 }– x

To find the zeroes of p(x), let p(x) = 0

∴ x^{3 }– x = 0

∴ x(x^{2 }– 1) = 0

∴ x(x + 1)(x – 1) = 0

∴ x = 0 or x = -1 or x = 1

∴ 0, -1 and 1 are the zeros of p(x).

Thus, the number of zeroes of the given polynomial is 3.

**Question 2:**

Find the number of zeros and real zeros of p(x) = x^{3} + 1. Show them by a graph.

**Solution :**

For p(x) = x^{3 }+ 1, to find the zeroes of p(x), let p(x) = 0.

∴ x^{3 }+ 1 = 0

∴ (x + 1)(x^{2 }– x + 1) = 0

∴ x + 1 = 0 or x^{2 }– x + 1 = 0

∴ x = -1 and real zeroes of x^{2 }– x + 1 are not possible.

∴ -1 is the only real zero.

Thus, number of zeros is for the given equation is 1 and that real zero is 0.

For p(x) = x^{3 }+ 1, taking x = -2, -1, 0 and 2.

We can obtain the following table:

x | -2 | -1 | 0 | 2 |

p(x) = x^{3 }+ 1 |
-7 | 0 | 1 | 9 |

Plotting these points on a graph paper as shown in the figure, we can see that the graph intersects the x-axis at (-1, 0) so, -1 is the zero of p(x).

**Question 3:**

Draw the graph of p(x) = x^{2} + 1 and find the real zeros of this polynomial.

**Solution :**

Taking values as x = -3, -2, -1, 0, 1, 2, 3 in p(x), we obtain the following table:

X | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

p(x) = x^{2 }+ 1 |
10 | 5 | 2 | 1 | 2 | 5 | 10 |

Plotting all these points on a graph paper and joining them as shown in the figure.

We see that the graph does not intersect the x-axis at any point. So p(x) has no real zero. The graph lies in the upper half-plane of the x-axis

**Question 4:**

From the figure 2.10 find the number of zeros of y = p(x).

**Solution :**

- The graph of y = p(x) intersects the x-axis at one point. So, the number of real zeroes of p(x) is 1.
- The graph of y = p(x) does not intersect the x-axis at any point. So, the number of real zeroes of p(x) is zero or p(x) had no real 0.
- The graph of y = p(x) intersects the x-axis at three distinct points. So the number of real zeroes of p(x) is 3.
- The graph of y = p(x) intersects the x-axis at two distinct points. So, the number of real zeroes of p(x) is 2.
- The graph y = p(x) intersect the x-axis at four distinct points. So, the number of real zeroes of p(x) is 4.
- The graph of y = p(x) intersects the x-axis at three distinct points. So, number of real zeroes of p(x) is 3.

**Question 5:**

Find the number of zeros and zeros of p(x) = x2 – 4. Represent them graphically.

**Solution :
**Here, p(x) = x

^{2 }– 4

To find the zeroes of p(x), let p(x) = 0

∴ x

^{2 }– 4 = 0

(x – 2)(x + 2) = 0

x = 2 or x = -2

The real zeroes are 2 and -2, which are 2 in all.

To draw the graph of this polynomial, we take some different values of x and prepare following table:

X | -3 | -2 | 0 | 2 | 3 |

p(x) = x^{2 }– 4 |
5 | 0 | -4 | 0 | 5 |

Plot these points on a graph paper as shown in the figure. Here we join all these point (-3, 5), (-2, 0), (0, -4), (2, 0) and (3, 5) we get the shape of the graph as a ∪ as shown, i.e. opening upwards.

We can see that this graph intersects the x-axis at two distinct points (-2, 0) and (2, 0). The x-coordinates of these point are considered as zeroes of this polynomial.

Thus, -2 and 2 are the zeroes of p(x).

### Exercise-2.3

**Question 1:**

Prove that 4 and 1 are the zeros of the quadratic polynomial p(x) = x2 – 5x + 4. Also verifr the

relationship between the zeros and the coefficients of p(x).

**Solution :**

Here p(x) = x^{2 }– 5x + 4

∴ p(4) = (4)^{2 }– 5(4) + 4

= 16 – 20 + 4

= 0

Also p(1) = (1)^{2 }– 5(1)+ 4

= 1- 5 + 4

= 0

Hence, 4 and 1 are the zeroes of the quadratic polynomial p(x) = x^{2 }– 5x + 4.

In p(x) = x^{2 }– 5x + 4

a = 1, b = -5, c = 4

Now, sum of the zeroes

= 4 + 1

= 5

**Question 2:**

Find the zeros of the following quadratic polynomials

- p(x) = x2 + 4x – 21
- p(x) = 6×2 – 11x + 5
- p(x) = 4×2 + 9x + 5
- p(x) = 3×2 + 5x – 8
- p(x) = x2 – 81
- p(x) = x2 – x – 6

**Question 2(1):**

**Solution :**

Here p(x) = x^{2 }+ 4x – 21, then to find its zeroes let p(x) = 0,

x^{2 }+ 4x – 21 = 0

∴ x^{2 }+ 7x – 3x – 21 = 0

∴ x(x + 7) – 3(x + 7) = 0

∴ (x + 7)(x – 3) = 0

∴ x = -7 or x = 3-7 and 3 are the zeroes of the quadratic polynomial p(x).

**Question 2(2):**

**Solution :**

To find the zeroes of p(x), let p(x) = 0,

6x^{2 }– 11x + 5 = 0

∴ 6x^{2 }– 6x – 5x + 5 = 0

∴ 6x(x – 1) – 5(x – 5) = 0

∴ (x – 1)(6x – 5) = 0

**Question 2(3):**

**Solution :**

To find the zeroes of p(x), let p(x) = 0,

4x^{2 }+ 9x + 5 = 0

∴4x^{2 }+ 4x + 5x + 5 = 0

∴4x(x + 1) + 5(x + 1) = 0

∴ (x + 1)(4x + 5) = 0

**Question 2(4):**

**Solution :**

To find the zeroes of p(x), let p(x) = 0,

3x^{2 }+ 5x – 8 = 0

∴ 3x^{2 }+ 8x – 3x – 8 = 0

∴ x(3x + 8) – 1(3x + 8) = 0

∴ (3x + 8)(x – 1) = 0

**Question 2(5):**

**Solution :**

To find the zeroes of p(x), let p(x) = 0,

x^{2 }– 81= 0

∴ (x)^{2 }– (9) ^{2 }= 0

∴ (x – 9)(x + 9) = 0

∴ x = 9 or x = -9

-9 and 9 are the zeroes of the quadratic polynomial p(x).

**Question 2(6):**

**Solution :**

To find the zeroes of p(x), let p(x) = 0,

x^{2 }– x – 6 = 0

∴ x^{2 }– 3x + 2x – 6 = 0

∴ x(x – 3) + 2 (x – 3) = 0

∴ (x – 3)(x + 2) = 0

∴ x = 3 or x = -2

3 and -2 are the zeroes of the quadratic polynomial p(x).

**Question 3:**

Find the zeros, the sum of the zeros and the product of the zeros of the quadratic polynomial p(x) = 3x^{2} – x – 4

**Solution :**

Here, p(x) = 3x^{2 }– x – 4

= 3x^{2 }– 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

=(3x – 4)(x + 1)

To find the zeroes of p(x), let p(x) = 0,

(3x – 4)(x + 1) = 0

**Question 4:**

Obtain a quadratic polynomial with the following conditions :

- The sum of zeros = 2; the product of zeros = -3
- The sum of zeros = -3; the product of zeros = -4
- The sum of zeros = \(\frac{1}{3}\); the product of zeros = \(\frac{1}{2}\)

**Question 4(1):**

**Solution :**

The sum of zeroes = 2; the product of zeroes = -3

Suppose a and b are the zeroes of the quadratic polynomial p(x).

∴c = -3, a = -3(k) = -3k

Now, p(x) = ax^{2 }+ bx + c

= kx^{2 }– 2kx – 3k (∵ Substituting values of a, b andc)

p(x) = k(x^{2 }– 2x – 3), k ≠ 0

Hence p(x) is the required quadratic polynomial.

**Question 4(2):**

**Solution :**

The sum of zeroes = -3; product of zeroes = -4

Suppose a and b are zeroes of the quadratic polynomial p(x).

∴ The sum of zeroes = a + b

∴c = -4a = -4k

Now, p(x) = ax^{2 }+ bx + c

= kx^{2 }+ 3kx – 4k (∵ Substituting values of a, b and c)

= k(x^{2 }– 3x – 4), k ≠ 0

Hence p(x) is the required quadratic polynomial.

**Question 4(3):**

**Solution :**

**Question 5:**

Obtain the quadratic or the cubic polynomial as the case may be in the standard form with the following coefficients :

- a = 6, b = 17, c = 11
- a = 1, b = -1, c = -1, d = 1
- a = 5, b = 7, c = 2,
- a = 1, b = -3, c = -1, d = 3
- a = 3, b = -5, c = -11, d = -3

**Solution :**

- The quadratic polynomial is,

p(x) = ax^{2 }+ bx + c; a ≠ 0, a, b, c ∊ R.

∴ Substituting a = 6, b = 17 and c = 11 we get the required quadratic polynomial

p(x) = 6x^{2 }+ 17x + 11.

- The cubic polynomial is,

p(x) = ax^{3 }+ bx^{2 }+ cx + d; a ≠ 0, a, b, c, d ∊ R.

∴ Substituting a = 1, b = 1 and c = 1 we get the required cubic polynomial

p(x) = x^{3 }– x^{2 }– x + 1.

- The quadratic polynomial is,

p(x) = ax^{2 }+ bx + c; a ≠ 0, a, b, c ∊ R.

∴ Substituting a = 5, b = 7 and c = 2 we get the required quadratic polynomial

p(x) = 5x^{2 }+ 7x + 2.

- The cubic polynomial is,

p(x) = ax^{3 }+ bx^{2 }+ cx + d; a ≠ 0, a, b, c, d ∊ R.

∴ Substituting a = 1, b = -3, c = -1 and d = 3, we get the required cubic polynomial

p(x) = x^{3 }- 3x^{2 }– x + 3.

- The cubic polynomial is,

p(x) = ax^{3 }+ bx^{2 }+ cx + d; a ≠ 0, a, b, c, d ∊ R.

∴ Substituting a = 3, b = -5, c = -11 and d = -3, we get the required cubic polynomial

p(x) = 3x^{3 }- 5x^{2 }– 11x – 3.

### Exercise-2.4

**Question 1:**

Divide the following polynomial p(x) by the polynomial s(x).

- p(x) = 2x
^{3}– 13x^{2}+ 23x – 12, s(x) = 2x – 3 - p(x) = \(\frac{2}{3}\) x
^{2}+ 5x + 6, s(x) = x + 6 - p(x) = 40x
^{2}+ 11x – 63, s(x) = 8x – 9 - p(x) = 2x
^{3}+ 9x^{2}+ 13x + 6, s(x) = 2x^{2}+ 5x + 3 - p(x) = x
^{4}+ 4x^{3}+ 5x^{2}– 7x – 3, s(x) = x2 – 1

**Question 1(1):**

**Solution :**

Here,

Dividend = p(x) = 2x^{3 }– 13x^{2 }+ 13x + 23x – 12 and

Divisor = s(x) = 2x – 3

Thus, the quotient polynomial q(x) = x^{2 }– 5x + 4 and the reminder polynomial r(x) = 0.

**Question 1(2):**

**Solution :**

**Question 1(3):**

**Solution :**

Here,

Dividend = p(x) = 40x^{2 }+ 121x – 63 and

Divisor = s(x) = 8x – 9

Thus, the quotient polynomial q(x) = 5x + 7 and

the reminder polynomial r(x) = 0.

**Question 1(4):**

**Solution :**

Here,

Dividend = p(x)= 2x^{3 }+ 9x^{2 }+ 13x + 6 and

Divisor = s(x)= 2x^{2 }+ 5x + 3

Thus, the quotient polynomial q(x) = x + 2

and the reminder polynomial r(x) = 0

**Question 1(5):**

**Solution :**

Here,

Dividend polynomial = p(x) = x^{4 }+ 4x^{3 }+ 5x^{2 }– 7x – 3 Divisor polynomial = s(x) = x^{2 }– 1.

Thus, the quotient polynomial q(x) = x^{2 }+ 4x + 6 and the remainder polynomial r(x) = 3 – 3x.

**Question 2:**

Find the remainder polynomial when the cubic polynomial x^{3} – 3x^{2} + 4x + 5 is divided by x – 2.

**Solution :**

Here we do the calculation by the synthetic division method.

Dividend polynomial p(x) = x^{3 }– 3x^{2 }+ 4x + 5 and the divisor polynomial s(x) = x – 2

Coefficients of x^{3}, x^{2}, x and x^{0} in the dividend polynomial are 1, -3, 4 and 5 respectively.

Taking x – 2 = 0 we get x = 2.

Thus, the reminder obtained when the cubic polynomial x^{3 }– 3x^{2 }+ 4x + 5 is divided by x – 2 is r(x) = 9.

**Question 3:**

3 is a zero of p(x) = 3x^{3} – x^{2} – ax – 45. Find ‘a’.

**Solution :**

Given that, 3 is a zero of polynomial p(x)

⇒ p(3) = 0

For p(x) = 3x^{3 }– x^{2 }– ax – 45

∴ 3(3)^{3 }– (3)^{2 }– a(3) – 45 = 0

∴ 3(27) – 9 – 3a – 45 = 0

∴ 81- 54 – 3a = 0

∴ 3a = 27

∴ a = 9

Hence, if 3 is a zero of the polynomial p(x), then a = 9.

**Question 4:**

The product of two polynomials is 6×3 + 29×2 + 44x + 21 and one of the polynomials is 3x + 7. Find the other polynomial.

**Solution :**

**Question 5:**

If polynomial p(x) is divided by x2 + 3x + 5, the quotient polynomial and the remainder polynomials are 2×2 + x + 1 and x – 3 respectively. Find p(x).

**Solution :**

Here, Divisor polynomial = s(x) = x^{2 }+ 3x + 5

Quotient polynomial = q(x) = 2x^{2 }+ x + 1

Remainder polynomial = r(x) = x – 3.

We know that,

Dividend = Divisor × Quotient + Remainder

∴ p(x) = s(x).q(x) + r(x)

= (x^{2 }+ 3x + 5)(2x^{2 }+ x + 1) + (x – 3)

= 2x^{4 }+ x^{3 }+ x^{2 }+ 6x^{3 }+ 3x^{2 }+ 3x + 10x^{2 }+ 5x + 5 + x – 3

=2x^{4 }+ 7x^{3 }+ 14x^{2 }+ 9x + 2

∴ Required polynomial is p(x) = 2x^{4 }+ 7x^{3 }+ 14x^{2 }+ 9x + 2

**Question 6:**

Divide p(x) = x^{3} – 4x^{2} + 5x – 2 by x – 2 Find r(x).

**Solution :**

Here we do the calculation by the synthetic division method.

Here, dividend polynomial = p(x) = x^{3 }– 4x^{2 }+ 5x – 2.

Divisor polynomial = s(x) = x – 2.

Coefficients of x^{3}, x^{2} x and x^{0} in the dividend polynomial are q, -4, 5 and -2 and taking x – 2 = 0 we get x = 2.

Thus, polynomial p(x) = x^{3 }– 4x ^{2 }+ 5x – 2 is divided by x – 2, the remainder polynomial r(x) = 0.

**Question 7:**

There are x^{4} + 57x + 15 pens to be distributed in a class of x^{2} + 4x + 2 students. Each student should get the maximum possible number of pens. Find the number of pens received by each student and the number of pens left undistributed (x ∈ N).

**Solution :**

Here, we divide the number of pens x^{4 }+ 57x + 15 by the number of students x^{2 }+ 4x + 2. The quotient polynomial obtained will be the number of pens each student receives and the remainder polynomial gives the number of pens left undistributed.

Here, p(x) = x^{4 }+ 57x + 15 and

s(x) = x^{2 }+ 4x + 2

Each student will get q(x) = x^{2 }– 4x + 14 (the quotient polynomial) number of pens and the number of pens undistributed are r(x) = 9x – 13 (the remainder polynomial).

**Question 8:**

A trader bought 2×2 – x + 2 TV sets for ₹ 8×4 + 71 – 6. Find the price of one TV set.

**Solution :**

2x^{2 }– x + 2 TV sets can be bought in Rs. 8x^{4}+7x-6.

∴ The price of one TV set is same as the quotient polynomial obtained by dividing 8x^{4 }+ 7x – 6 by 2x^{2 }– x + 2

Thus, the price of one TV set is Rs. 4x^{2 }+ 2x – 3.

**Question 9:**

\(-\sqrt{2}\) and \(\sqrt{2}\) are two of the zeros of p(x) = 2×4 + 7×3 – 8×2 – 14x + 8. Find the remaining zeros of p(x).

**Solution :**

∴ The quotient polynomial is q(x) = 2x^{2 }+ 7x – 4 and the remainder polynomial is r(x) = 0

∴ p(x) = 2x^{4 }+ 7x^{3 }– 8x^{2 }– 14x + 8

= (x^{2 }– 2)(2x^{2 }+ 7x – 4)

∴ It is clear that, the zeroes of the second factor 2x^{2 }+ 7x – 4 are the remaining zeroes of the polynomial p(x).

Now, to find the zeroes of q(x) = 2x^{2 }+ 7x – 4 we take q(x) = 0.

### Exercise-2

**Question 1:**

State whether the following statements are true or false :

- \(\frac{7}{5}\) is a zero of the linear polynomial p(x) = 5x + 7.
- p(x) = x
^{2}+ 2x + 1 has two distinct zeros. - The cubic polynomial p(x) = x
^{3}+ x^{2}– x – 1 has two distinct zeros. - The graph of the cubic polynomial p(x) = x
^{3}meets the X—axis at only one point. - Any quadratic polynomial p(x) has at least one zero, x ∈ R

**Solution :**

2. False.

Reason: To find the zeros of p(x), let p(x) = 0

∴ x^{2} + 2x + 1 = 0

∴ (x + 1)^{2 }= 0

∴ x + 1 = 0 or x = -1

∴ x = -1 or x = -1

Here, both the zeroes are equal, i.e. -1, and hence not distinct.

3. True.

Reason: To find the zeroes of p(x), let p(x) = 0

∴x^{3 }+ x^{2 }– x – 1 = 0

∴x^{2}(x + 1) – 1(x + 1) = 0

∴ (x + 1)(x^{2 }– 1) = 0

∴ (x + 1)(x + 1)(x – 1) = 0

∴ x + 1 = 0 or x + 1 = 0 or x – 1 = 0

∴ x = -1 o r x = -1 or x = 1

∴ Two distinct zeroes of p(x) are 1 and -1.

Hence, p(x) has at the most two distinct zeroes.

4. True.

Reason: The graph of p(x) = x^{3 } is the x-axis then p(x) = 0

∴ x^{3 }= 0

∴ x = 0

∴ y = p(0) = 0

Thus, the graph of p(x) = x^{3} meets the x-axis at only one point (0, 0).

5. False

If the graph of the quadratic polynomial p(x) does not intersect the x-axis at any point, then the quadratic polynomial does not have any real zero.

Example: polynomial p(x) = x^{2 }+ 4x + 5, does not have any real zeros.

**Question 2:**

Find the zeros and number of zeros of p(x) = x^{2} + 9x + 18. Show them on a graph.

**Solution :**

Here, p(x) = x^{2 }+ 9x + 18

To find the zeroes of p(x), let p(x) = 0

∴ x^{2 }+ 9x + 18 = 0

∴ x^{2 }+ 6x + 3x + 18 = 0

∴ x(x + 6) +3(x + 6) = 0

∴ (x + 6)(x + 3) = 0

∴ x = -6 or x = -3

Thus, the distinct zeroes -6 and -3, i.e. 2 in all.To draw the graph of this polynomial, we take different values of x and prepare the following table:

x | -7 | -6 | -5 | -4 | -3 | -2 |

p(x)= x^{2 }+ 9x + 18 |
4 | 0 | -2 | -2 | 0 | 4 |

On plotting these points on a graph paper and joining the points (-7, 4), (-6, 0), (-5, -2), (-4, -2), (-3, 0) and (-2, 4). As shown in the figure the shape of the graph is a **U, **as shown i.e., it opens upwards.

Also the graph intersects the x-axis at two distinct points (-6, 0) and (-3, 0). Their x-coordinates are considered as the zeroes of this polynomial.

Thus, -6 and -3 are the zeroes of p(x).

**Question 3:**

Find the zeros, the sum and the product of zeros of p(x) = 4x^{2} + 12x + 5.

**Solution :**

For 4x^{2 }+ 12x + 5

= 4x^{2 }+ 10x + 2x + 5

= 2x(2x + 5) + 1(2x + 5)

= (2x + 5)(2x + 1)

To find the zeroes of p(x) = 0

∴ (2x + 5)(2x + 1) = 0

**Question 4:**

-4 and 9 are the sum and product of the zeros respectively of a quadratic polynomial. Find the quadratic polynomial.

**Solution :**

Suppose, a and b are the zeroes of the quadratic polynomial p(x) = ax^{2 }+ bx + c.

Thus, c = 9a = 9k

Now, p(x) = ax^{2 }+ bx + c

∴ p(x) = kx^{2 }+ 4kx + 9k

= k(x^{2 }+ 4x + 9), k ≠ 0

∴ Required quadratic polynomial is k(x^{2 }+ 4x + 9), k ≠ 0.

**Question 5:**

Find q(x) and r(x), for the quadratic polynomial p(x) = 11x – 21 + 2x^{2} when divided by 1 + 2x

**Solution :**

Here, Dividend = p(x) = 11x – 21 + 2x^{2
}= 2x^{2 }+ 11x – 21 (∵ rearranging the terms)

Divisor = s(x) = 1 + 2x = 2x+1

∴ Quotient polynomial is q(x) = x + 5 and the remainder polynomial is r(x) = -26.

**Question 6:**

Divide 2x^{3} + 3x^{2} – 11x – 6 by x^{2} + x – 6

**Solution :**

Here, Dividend = p(x) = 2x^{3 }+ 3x^{2 }– 11x – 6

Divisor = s(x) = x^{2 }+ x – 6

∴ Quotient polynomial q(x) = 2x + 1 and the

remainder polynomial r(x) = 0.

**Question 7:**

4 is a zero of the cubic polynomial p(x) = x^{3} – 3x^{2} – 6x + 8. Find the remaining zeros of p(x)

**Solution :**

We will do the calculation by the synthetic division method.

Here 4 is the zero of polynomial p(x).

So (x – 4) is the factor of p(x).

∴ Dividend = p(x) = x^{3 }– 3x^{2 }– 6x + 8 and

Divisor polynomial = s(x) = x – 4.

Coefficients of x^{3}, x^{2}, x and x° are 1, -3, -6 and 8 respectively. Taking x – 4 = 0

∴ x = 4

∴ p(x) = x^{3 }– 3x^{2 }– 6x + 8

= (x – 4) (x^{2 }+ x – 2)

= (x – 4)(x + 2)(x – 1)

To find the remaining zeroes, let p(x)=0

(x – 4)(x + 2)(x – 1) = 0

∴ The remaining zeroes of p(x) are -2 and 1.

**Question 8:**

The product of two polynomials is 3x^{4} + 5x^{3} – 21x^{2} – 53x — 30. If one of them is x^{2} – x – 6, find the other polynomial.

**Solution :**

∴ The quotient polynomial q(x) = 3x^{2 }+ 8x + 5 and the remainder polynomial r(x) = 0.

∴ The polynomial 3x^{2 }+ 8x + 5 is a factor of 3x^{4 }+ 5x^{3 }– 21x^{2 }– 53x – 30.

**Question 9:**

2 + \( \sqrt{3}\) and 2 – \(\sqrt{3}\) are the zeros of p(x) = x^{4} – 6x^{3} – 26x^{2} + 138x – 35. Find the remaining zeros of p(x).

**Solution :**

**Question 10:**

Select a proper option (a), (b), (c) or (d) from given options :

**Question 10(1):**

The linear polynomial p(x) = 7x (1) – 3 has the zero ..

**Solution :**

**Question 10(2):**

The cubic polynomial p(x) = x^{3} – x has….. zeros.

**Solution :**

d. 3

p(x) = x^{3 }– x

= x(x^{2 }– 1) = x(x – 1)(x + 1)

∴ x = 0, x = 1, x = -1

∴ The cubic polynomial has 3 zeroes.

**Question 10(3):**

The graph of p(x) = 3x – 2 – x^{2} intersects the X-axis in ….. points.

**Solution :**

c. 2

The zeroes of p(x) are the intersections points of the equation, p(x) = 3x – 2 – x^{2} with the x-axis.

∴ Whatever the number of distinct zeros of p(x) that many number of distinct intersection points can be found on the x-axis.

To find zeros of p(x)= 3x – 2 – x^{2},

Let p(x) = 0

∴ 3x – 2 – x^{2 }= 0

∴ x^{2 }– 3x + 2 = 0

∴ x^{2 }– 2x – x + 2 = 0

∴ x(x – 2) – 1(x – 2) = 0

∴ (x – 1)(x – 2) = 0

∴x = 1 or x = 2

∴ The graph of p(x) intersect the x-axis in 2 distinct points which are (1, 0) and (2, 0).

**Question 10(4):**

The sum of the zeros of 3x^{2} + 5x – 2 is …..

**Solution :**

**Question 10(5):**

The graph of p(x) = 3x + 5 represents …….

**Solution :**

a. a straight line

p(x) = 3x + 5 is a linear polynomial. So its graph is a straight line.

**Question 10(6):**

quadratic polynomial has no zero. Its graph ……

**Solution :**

d. is in any one half plane of the x-axis

The number of intersection points of a quadratic polynomial is the number of real zeroes of that quadratic polynomial.

Here, the quadratic polynomial has no zero.

∴ It is in any one half plane of the x-axis.

**Question 10(7):**

For the graph in figure 2.11 y = p(x) has … zeros.

**Solution :**

d. 4

Here, the graph intersects the x-axis in four points.

∴ p(x) has 4 zeroes.

**Question 10(8):**

The product of the zeros of x^{2} – 4x + 3 is …..

**Solution :**

**Question 10(9):**

a = 3, b = 5, c = 7, d = 11 in the standard notation gives the cubic polynomial …..

**Solution :**

d. 3x^{3 }+ 5x^{2 }+ 7x + 11

The standard form of a cubic polynomial is

p(x) = ax^{3 }+ bx^{2 }+ cx + d

∴ For a = 3, b = 5, c = 7 and d = 11,

p(x) = 3x^{3 }+ 5x^{2 }+ 7x + 11.