In this chapter, we provide KSEEB Solutions for Class 10 Maths Chapter 8 Polynomials for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB Solutions for Class 10 Maths Chapter 8 Polynomials pdf, free KSEEB solutions for Class 10 Maths Chapter 8 Polynomials book pdf download. Now you will get step by step solution to each question.

**KSEEB SSLC Solutions for Class 10 Maths – Polynomials (English Medium)**

**KSEEB SSLC Solutions for Class 10 Maths Chapter 8** **Exercise 8.1**

**Question 1:**

Find the degree of the following polynomials:

i. x^{2} – 9x + 20

ii. 2x + 4 + 6x^{2}iii. x^{3} + 2x^{2} – 5x – 6

iv. x^{3} + 17x – 21 – x^{2}

**Solution :**

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

i. The degree of the polynomial x^{2} – 9x + 20 is 2.

ii. The degree of the polynomial 2x + 4 + 6x^{2} is 2.

iii. The degree of the polynomial x^{3} + 2x^{2} – 5x – 6 is 3.

iv. The degree of the polynomial x^{3} + 17x – 21 – x^{2} is 3.

**Question 2:**

If f(x) = 2x^{3} + 3x^{2} – 11x + 6, find

i. f(0) f(1)iii. f(-1) iv. f(2) v. f(-3)

**Solution :**

f(x) = 2x^{3} + 3x^{2} – 11x + 6

i. f(0) = 2(0)^{3} + 3(0)^{2} – 11(0) + 6

= 0 + 0 – 0 + 6

= 6

ii. f(1) = 2(1)^{3} + 3(1)^{2} – 11(1) + 6

= 2 + 3 – 11 + 6

= 0

iii. f(-1) = 2(-1)^{3} + 3(-1)^{2} – 11(-1) + 6

= -2 + 3 + 11 + 6

= 18

iv. f(2) = 2(2)^{3} + 3(2)^{2} – 11(2) + 6

= 16 + 12 – 22 + 6

= 12

v. f(-3) = 2(-3)^{3} + 3(-3)^{2} – 11(-3) + 6

= -54 + 27 + 33 + 6

= 12

**Question 3:**

Find the values of the following polynomials

i. g(x) = 7x^{2} + 2x + 14, when x = 1

ii. p(x) = -x^{3} + x^{2} – 6x + 5, when x = 2

iv. p(x) = 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2, when x = -2

**Solution :**

i. g(x) = 7x^{2} + 2x + 14

When x = 1,

g(1) = 7(1)^{2} + 2(1) + 14

= 7 + 2 + 14

= 23

ii. p(x) = -x^{3} + x^{2} – 6x + 5

When x = 2

p(2) = -(2)^{3} + (2)^{2} – 6(2) + 5

= -8 + 4 – 12 + 5

= -11

iv. p(x) = 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2, when x = -2

p(-2) = 2(-2)^{4} – 3(-2)^{3} – 3(-2)^{2} + 6(-2) – 2

= 2 × 16 – 3 × (-8) – 3 × 4 – 12 – 2

= 32 + 24 – 12 – 12 – 2

= 30

**Question 4(i):**

Verify whether the indicated numbers are zeroes of the polynomials in each of the following cases. (For i to vi)

**Solution :**

**Question 4(ii):**

p(x) = x^{2} – 4, x = 2, x = -2

**Solution :**

p(x) = x^{2} – 4

When x = 2

p(2) = (2)^{2} – 4 = 4 – 4 = 0

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

When x = -2

p(-2) = (-2)^{2} – 4 = 4 – 4 = 0

Since p(-2) = 0, it is a zero of the given polynomial.

**Question 4(iii):**

p(x) = x^{3} – 6x^{2} + 11x – 6, x = 1, x = 2 and x = 3

**Solution :**

p(x) = x^{3} – 6x^{2} + 11x – 6

When x = 1

p(1) = (1)^{3} – 6(1)^{2} + 11(1) – 6 = 1 – 6 + 11 – 6 = 0

Since p(1) = 0, it is a zero of the given polynomial.

When x = 2

p(2) = (2)^{3} – 6(2)^{2} + 11(2) – 6 = 8 – 24 + 22 – 6 = 0

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

When x = 3

p(3) = (3)^{3} – 6(3)^{2} + 11(3) – 6 = 27 – 54 + 33 – 6 = 0

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

**Question 4(iv):**

**Solution :**

p(x) = 3x^{3} – 5x^{2} – 11x – 3

When x = 3

p(3) = 3(3)^{3} – 5(3)^{2} – 11(3) – 3

= 81 – 45 – 33 – 3

= 0

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

When x = -1

p(-1) = 3(-1)^{3} – 5(-1)^{2} – 11(-1) – 3

= 3(-1) – 5(1) + 11 – 3

= -3 – 5 + 11 – 3

= 0

Since p(-1) = 0, it is a zero of the given polynomial.

**Question 4(v):**

p(x) = x^{3} – 6x^{2} + 11x – 6, x = 1, x = 2 and x = 3

**Solution :**

p(x) = x^{3} – 6x^{2} + 11x – 6

When x = 1

p(1) = (1)^{3} – 6(1)^{2} + 11(1) – 6 = 1 – 6 + 11 – 6 = 0

Since p(1) = 0, it is a zero of the given polynomial.

When x = 2

p(2) = (2)^{3} – 6(2)^{2} + 11(2) – 6 = 8 – 24 + 22 – 6 = 0

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

When x = 3

p(3) = (3)^{3} – 6(3)^{2} + 11(3) – 6 = 27 – 54 + 33 – 6 = 0

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

**Question 4(vi):**

**Solution :**

p(x) = 3x^{3} – 5x^{2} – 11x – 3

When x = 3

p(3) = 3(3)^{3} – 5(3)^{2} – 11(3) – 3

= 81 – 45 – 33 – 3

= 0

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

When x = -1

p(-1) = 3(-1)^{3} – 5(-1)^{2} – 11(-1) – 3

= 3(-1) – 5(1) + 11 – 3

= -3 – 5 + 11 – 3

= 0

Since p(-1) = 0, it is a zero of the given polynomial.

**Question 5(i):**

Find the zeroes of the following quadratic polynomials and verify.

x^{2} + 4x + 4

**Solution :**

Let p(x) = x^{2} + 4x + 4

By factorization,

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

The value of x^{2} + 4x + 4 is zero when (x + 2)^{2} = 0

i.e. when x + 2 = 0

i.e. when x = -2

∴ The zero of x^{2} + 4x + 4 is (-2).

Verification:

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

p(-2) = (-2)^{2} + 4(-2) + 4 = 4 – 8 + 4 = 0

**Question 5(ii):**

x^{2} – 2x – 15

**Solution :**

Let p(x) = x^{2} – 2x – 15

By factorization,

x^{2} – 2x – 15

x^{2} – 5x + 3x – 15

= x(x – 5) + 3(x – 5)

= (x – 5)(x + 3)

The value of x^{2} – 2x – 15 is zero when x – 5 = 0 or x + 3 = 0

i.e., when x = 5 or x = -3

∴ The zeroes of x^{2} – 2x – 15 are 5 and -3.

Verification:

p(x) = x^{2} – 2x – 15

P(5) = (5)^{2} – 2(5) – 15 = 25 – 10 – 15 = 0

P(-3) = (-3)^{2} – 2(-3) – 15 = 9 + 6 – 15 = 0

**Question 5(iii):**

**x ^{2} + 9x – 36 **

**Solution :**

Let p(x) = x^{2} + 9x – 36

By factorization,

x^{2} + 9x – 36

= x^{2} + 12x – 3x – 36

= x(x + 12) – 3(x + 12)

= (x + 12)(x – 3)

The value of x^{2} + 9x – 36 is zero when x + 12 = 0 or x – 3 = 0

i.e., when x = -12 or x = 3

∴ The zeroes of x^{2} + 9x – 36 are -12 and 3.

Verification:

p(x) = x^{2} + 9x – 36

P(-12) = (-12)^{2} + 9(-12) – 36

= 144 – 108 – 36

= 0

p(3) = (3)^{2} + 9(3) – 36 = 9 + 27 – 36 = 0

**Question 5(iv):**

x^{2} + 5x – 14

**Solution :**

Let p(x) = x^{2} + 5x – 14

By factorization,

x^{2} + 5x – 14

= x^{2} + 7x – 2x – 14

= x(x + 7) – 2(x + 7)

= (x + 7)(x – 2)

The value of x^{2} + 5x – 14 is zero when x + 7 = 0 or x – 2 = 0

i.e., when x = -7 or x = 2

∴ The zeroes of the polynomial x^{2} + 5x – 14 are -7 and 2.

Verification:

p(x) = x^{2} + 5x – 14

p(-7) = (-7)^{2} + 5(-7) – 14 = 49 – 35 – 14 = 0

p(2) = (2)^{2} + 5(2) – 14 = 4 + 10 – 14 = 0

**Question 5(v):**

4y^{2} + 8y

**Solution :**

Let p(y) = 4y^{2} + 8y

By factorization,

4y^{2} + 8y = 4y(y + 2)

The value of 4y^{2} + 8y is zero when 4y = 0 or y + 2 = 0

i.e., when y = 0 or y = -2

∴ The zeroes of the polynomial 4y^{2} + 8y are 0 and -2.

Verification:

p(y) = 4y^{2} + 8y

p(0) = 4(0)^{2} + 8(0) = 0

p(-2) = 4(-2)^{2} + 8(-2) = 16 – 16 = 0

**Question 5(vi):**

4a^{2} – 49

**Solution :**

Let p(a) = 4a^{2} – 49

By factorization,

4a^{2} – 49 = (2a)^{2} – (7)^{2} = (2a – 7)(2a + 7)

The value of 4a^{2} – 49 is zero when 2a – 7 = 0 or 2a + 7 = 0

i.e., when 2a = 7 or 2a = -7

**Question 5(vii):**

**Solution :**

**Question 5(viii):**

3x^{2} – x – 4

**Solution :**

Let p(x) = 3x^{2} – x – 4

By factorization,

3x^{2} – x – 4

= 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1(3x – 4)

= (3x – 4)(x + 1)

**Question 6:**

If x = 1 is a zero of the polynomial f(x) = x^{3} – 2x^{2} + 4x + k, find the value of k.

**Solution :**

f(x) = x^{3} – 2x^{2} + 4x + k

Since x = 1 is a zero of given polynomial, f(1) = 0

Thus, we have

(1)^{3} – 2(1)^{2} + 4(1) + k = 0

∴ 1 – 2 + 4 + k = 0

∴ 3 + k = 0

∴ k = -3

**Question 7:**

For what value of k, -4 is a zero of the polynomial x^{2} – x – (2k + 2)?

**Solution :**

Let f(x) = x^{2} – x – (2k + 2)

Now, f(-4) = 0

∴ (-4)^{2} – (-4) – 2k – 2 = 0

∴ 16 + 4 – 2k – 2 = 0

∴18 – 2k = 0

∴ 2k = 18

∴ k = 9

**KSEEB SSLC Solutions for Class 10 Maths Chapter 8** **Exercise 8.2**

**Question 1(i):**

Divide p(x) by g(x) in each of the following cases and verify division algorithm. [For (i) to (x)]

p(x) = x^{2} + 4x + 4; g(x) = x + 2

**Solution :**

**Question 1(ii):**

p(x) = 2x^{2} – 9x + 9; g(x) = x – 3

**Solution :**

**Question 1(iii):**

p(x) = x^{3} + 4x^{2} – 5x + 6; g(x) = x + 1

**Solution :**

**Question 1(iv):**

p(x) = 4x^{3} – 6x^{2} + x + 14; g(x) = 2x – 3

**Solution :**

**Question 1(v):**

p(x) = x^{4} – 3x^{2} – 4; g(x) = x + 2

**Solution :**

**Question 1(vi):**

p(x) = x^{3} – 1; g(x) = x – 1

**Solution :**

**Question 1(vii):**

p(x) = x^{6} – 64; g(x) = x – 2

**Solution :**

**Question 1(viii):**

p(x) = x^{7 }– a^{7}; g(x) = x – a

**Solution :**

**Question 1(ix):**

p(x) =x^{4 }– 4x^{2} + 12x + 9; g(x) = x^{2} + 2x – 3

**Solution :**

**Question 1(x):**

p(x) = a^{4} + 4b^{4}; g(x) = a^{2} + 2ab + b^{2}

**Solution :**

**Question 2:**

Find the divisor g(x), when the polynomial p(x) = 4x^{3} + 2x^{2} – 10x + 2 is divided by g(x) and the quotient and remainder obtained are (2x^{2} + 4x + 1) and 5 respectively.

**Solution :**

**Question 3:**

On dividing the polynomial p(x) = x^{3} – 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were

(x – 2) and (-2x + 4), respectively. Find g(x).

**Solution :**

**Question 4:**

A polynomial p(x) is divided by g(x), the obtained quotient q(x) and the remainder r(x) are given in the table. Find p(x) in each case.

Sl. | p(x) | g(x) | q(x) | r(x) |

i ii iii iv v | ? ? ? ? ? | x – 2 x + 3 2x + 1 x – 1 x^{2 }+ 2x + 1 | x^{2 }– x + 1 2x^{2 }+ x + 5 x^{3 }+ 3x^{2 }-x + 1 x^{3 }– x^{2} – x – 1 x^{4} – 2x^{2 }+ 5x -7 | 4 3x + 1 0 2x – 4 4x + 12 |

**Solution :**

- i. p(x) = [g(x) × q(x)] + r(x)

= [(x – 2)(x^{2}– x + 1 )] + 4

= x^{3}– x^{2}+ x – 2x^{2}+ 2x – 2 + 4

= x^{3}– 3x^{2}+ 3x + 2 - p(x) = [g(x) × q(x)] + r(x)

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

= 2x^{3}+ x^{2}+ 5x + 6x^{2}+ 3x + 15 + 3x + 1

= 2x^{3}+ 7x^{2}+ 11x + 16 - p(x) = [g(x) × q(x)] + r(x)

= [(2x + 1)(x^{3}+ 3x^{2}– x + 1)] + 0

= 2x^{4}+ 6x^{3}– 2x^{2}+ 2x + x^{3}+ 3x^{2}– x + 1 +0

= 2x^{4}+ 7x^{3}+ x^{2}+ x + 1 - p(x) = [g(x) × q(x)] + r(x)

= [(x – 1)(x^{3}– x^{2}– x – 1)] + 2x – 4

= x^{4}– x^{3}– x^{2}– x – x^{3}+ x^{2}+ x + 1 + 2x – 4

= x^{4}– 2x^{3}+ 2x – 3 - p(x) = [g(x) × q(x)] + r(x)

= [(x^{2}+ 2x + 1)(x^{4}– 2x^{2}+ 5x – 7)] + 4x +12

= x^{6}– 2x^{4}+ 5x^{3}– 7x^{2}+ 2x^{5}– 4x^{3}+ 10x^{2}– 14x + x^{4}– 2x^{2}+ 5x – 7 + 4x + 12

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

**Question 5(i):**

Find the quotient and remainder on dividing p(x) by g(x)in each of the following cases, without actual division. [For (i) to (iv)]

p(x) = x^{2} + 7x + 10; g(x) = x – 2

**Solution :**

p(x) = x^{2} + 7x + 10 ∴ degree of p(x) is 2.

g(x) = x – 2 ∴ degree of g(x) is 1.

∴ Degree of quotient q(x) = 2 – 1 = 1 and degree of remainder r(x) is zero.

Let q(x) = ax + c and r(x) = k

By using division algorithm, we have

p(x) = [g(x) × q(x)] + r(x)

∴ x^{2} + 7x + 10 = (x – 2)(ax + c) + k

∴ x^{2} + 7x + 10 = ax^{2} + cx – 2ax – 2c + k

∴ x^{2} + 7x + 10 = ax^{2} + (c – 2a)x – 2c + k

∴ a = 1, c – 2a = 7, -2c + k = 10

By solving, we get

a = 1

c – 2a = 7

∴ c – 2(1) = 7

∴ c – 2 = 7

∴ c = 9

-2c + k = 10

∴ -2(9) + k = 10

∴ -18 + k = 10

∴ k = 28

∴ q(x) = ax + c = 1x + 9 = x + 9

r(x) = k = 28

**Question 5(ii):**

Find the quotient and remainder on dividing p(x) in each of the following cases, without actual division. p(x) = 2x^{2} + 14x – 5; g(x) = x + 1

**Solution :**

p(x) = 2x^{2} + 14x – 5 ∴ degree of p(x) is 2.

g(x) = x + 1 ∴ degree of g(x) is 1.

∴ Degree of quotient q(x) = 2 – 1 = 1 and degree of remainder r(x) is zero.

Let q(x) = ax + c and r(x) = k

By using division algorithm, we have

p(x) = [g(x) × q(x)] + r(x)

∴ 2x^{2} + 14x – 5 = (x + 1)(ax + c) + k

∴ 2x^{2} + 14x – 5 = ax^{2} + cx + ax + c + k

∴ 2x^{2} + 14x – 5 = ax^{2} + (c + a)x + c + k

∴ a = 2, c + a = 14, c + k = -5

By solving, we get

a = 2

c + a = 14

∴ c + 2 = 14

∴ c = 12

c + k = -5

∴ 12 + k = -5

∴ k = -17

∴ q(x) = ax + c = 2x + 12

r(x) = k = -17

**Question 5(iii):**

Find the quotient and remainder on dividing p(x) in each of the following cases, without actual division. p(x) = x^{3} + 4x^{2} – 6x + 2; g(x) = x – 3

**Solution :**

p(x) = x^{3} + 4x^{2} – 6x + 2 ∴ degree of p(x) is 3.

g(x) = x – 3 ∴ degree of g(x) is 1.

∴ Degree of quotient q(x) = 3 – 1 = 2 and degree of remainder r(x) is zero.

Let q(x) = ax^{2} + bx + c and r(x) = k

By using division algorithm, we have

p(x) = [g(x) × q(x)] + r(x)

∴ x^{3} + 4x^{2} – 6x + 2 = (x – 3)(ax^{2} + bx + c) + k

∴ x^{3} + 4x^{2} – 6x + 2 = ax^{3} + bx^{2} + cx – 3ax^{2} – 3bx – 3c + k

∴ x^{3} + 4x^{2} – 6x + 2 = ax^{3} + (b – 3a)x^{2} + (c – 3b)x – 3c + k

∴ a = 1, b – 3a = 4, c – 3b = -6, -3c + k = 2

By solving, we get

a = 1

b – 3a = 4

∴ b – 3(1) = 4

∴ b – 3 = 4

∴ b = 7

c – 3b = -6

∴ c – 3(7) = -6

∴ c – 21 = -6

∴ c = 15

-3c + k = 2

∴ -3(15) + k = 2

∴ -45 + k = 2

∴ k = 47

∴ q(x) = ax^{2} + bx + c = 1x^{2} + 7x + 15 = x^{2} + 7x + 15

r(x) = k = 47

**Question 5(iv):**

Find the quotient and remainder on dividing p(x) in each of the following cases, without actual division. p(x) = 4x^{3} + 11x^{2} – 11x + 8; g(x) = x^{2} – x + 1

**Solution :**

p(x) = 4x^{3} + 11x^{2} – 11x + 8 ∴ degree of p(x) is 3.

g(x) = x^{2} – x + 1 ∴ degree of g(x) is 2.

∴ Degree of quotient q(x) = 3 – 2 = 1 and degree of remainder r(x) is zero.

Let q(x) = ax + c and r(x) = k

By using division algorithm, we have

p(x) = [g(x) × q(x)] + r(x)

∴ 4x^{3} + 11x^{2} – 11x + 8 = (x^{2} – x + 1)(ax + c) + k

∴ 4x^{3} + 11x^{2} – 11x + 8 = ax^{3} + cx^{2} – ax^{2} – cx + ax + c + k

∴ 4x^{3} + 11x^{2} – 11x + 8 = ax^{3} + (c – a)x^{2} + (-c + a)x + c + k

∴ a = 4, c – a = 11, -c + a = -11, c + k = 8

By solving, we get

a = 4

c – a = 11

∴ c – 4 = 11

∴ c = 15

c + k = 8

∴ 15 + k = 8

∴ k = -7

∴ q(x) = ax + c = 4x + 15

r(x) = k = -7

**Question 6:**

What must be subtracted from (x^{3} + 5x^{2} + 5x + 8) so that the resulting polynomial is exactly divisible by (x^{2} + 3x – 2)?

**Solution :**

**Question 7:**

What should be added to the polynomial (7x^{3} + 4x^{2} – x – 10) So that the resulting polynomial is exactly divisible by (2x^{2 }+ 3x – 2)?

**Solution :**

**Question 8:**

What should be added to (x^{4} – 1) so that it is exactly divisible by (x^{2} + 2x +1)?

**Solution :**

**Question 9:**

What should be subtracted from (x^{6} – a^{6}), so that it is exactly divisible by (x – a)?

**Solution :**

**Question 10(i):**

Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and (i) degree p(x) = degree q(x) (ii) degree q(x) = degree r(x) (iii) degree r(x) = 0

**Solution :**

**Question 10(ii):**

Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and degree q(x) = degree r(x)

**Solution :**

**Question 10(iii):**

Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and (i) degree p(x) = degree q(x) (ii) degree q(x) = degree r(x) (iii) degree r(x) = 0

**Solution :**

**KSEEB SSLC Solutions for Class 10 Maths Chapter 8** **Exercise 8.3**

**Question 1(i):**

In each of the following cases, use the remainder theorem and find the remainder when p(x) is divided by g(x). Verify the result by actual division. [For (i) to (vi)]

p(x) = x^{3} + 3x^{2} – 5x + 8, g(x) = x – 3

**Solution :**

By remainder theorem, the required remainder is equal to p(3).

p(x) = x^{3} + 3x^{2} – 5x + 8

∴ p(3) = (3)^{3} + 3(3)^{2} – 5(3) + 8

= 27 + 27 – 15 + 8

= 47

∴ Required remainder = p(3) = 47

Verification by actual division:

p(x) = x^{3} + 3x^{2} – 5x + 8, g(x) = x – 3

**Question 1(ii):**

p(x) = 4x^{3} – 10x^{2} + 12x – 3, g(x) = x + 1

**Solution :**

By remainder theorem, the required remainder is equal to p(-1).

p(x) = 4x^{3} – 10x^{2} + 12x – 3

∴ p(-1) = 4(-1)^{3} – 10(-1)^{2} + 12(-1) – 3

= -4 – 10 – 12 – 3

= -29

∴ Required remainder = p(-1) = -29

Verification by actual division:

p(x) = 4x^{3} – 10x^{2} + 12x – 3, g(x) = x + 1

**Question 1(iii):**

p(x) = 2x^{4} – 5x^{2} + 15x – 6, g(x) = x – 2

**Solution :**

By remainder theorem, the required remainder is equal to p(2).

p(x) = 2x^{4} – 5x^{2} + 15x – 6, g(x) = x – 2

∴ p(2) = 2(2)^{4} – 5(2)^{2} + 15(2) – 6

= 32 – 20 + 30 – 6

= 36

∴ Required remainder = p(2) = 36

Verification by actual division:

p(x) = 2x^{4} – 5x^{2} + 15x – 6, g(x) = x – 2

**Question 1(iv):**

p(x) = 4x^{3} – 12x^{2} + 14x – 3, g(x) = 2x – 1

**Solution :**

**Question 1(v):**

p(x) = 7x^{3} – x^{2} + 2x – 1, g(x) = 1 – 2x

**Solution :**

**Question 1(vi):**

**Solution :**

**Question 2(i):**

Find the remainder using reminder theorem when

(2x^{2} + 3x^{2} + x + 1) is divided by

x – 1

**Solution :**

Let p(x) = 2x^{3} + 3x^{2} + x + 1

x – 1 = 0

∴ x = 1

∴ p(1) = 2(1)^{3} + 3(1)^{2} + 1 + 1

= 2 + 3 + 1 + 1

= 7

∴ Remainder = r(x) = 7

**Question 2(ii):**

Find the remainder using reminder theorem when

(2x^{2} + 3x^{2} + x + 1) is divided by

**Solution :**

**Question 2(iii):**

Find the remainder using reminder theorem when

(2x^{2} + 3x^{2} + x + 1) is divided by

3 + 2x

**Solution :**

**Question 2(iv):**

Find the remainder using reminder theorem when

(2x^{2} + 3x^{2} + x + 1) is divided by x

**Solution :**

Let p(x) = 2x^{3} + 3x^{2} + x + 1

x = 0

∴ p(0) = 2(0)^{3} + 3(0)^{2} + 0 + 1 = 1

∴ Remainder = r(x) = 1

**Question 2(v):**

Find the remainder using reminder theorem when

(2x^{2} + 3x^{2} + x + 1) is divided by

x+1/3

**Solution :**

Let p(x) = 2x^{3} + 3x^{2} + x + 1

**Question 3:**

If the polynomials (2x^{3 }+ ax^{2} + 3x – 5) and (x^{3} + x^{2} – 4x – a) leave the same remainder when divided by

(x – 1), find the value of a.

**Solution :**

Let p(x) = 2x^{3} + ax^{2} + 3x – 5 and g(x) = x^{3} + x^{2} – 4x – a

By remainder theorem, the two remainders are p(1) and g(1).

By the given condition, p(1) = g(1)

∴ p(1) = 2(1)^{3} + a(1)^{2} + 3(1) – 5

= 2 + a + 3 – 5

= a

And, g(1) = (1)^{3} + (1)^{2} – 4(1) – a

= 1 + 1 – 4 – a

= -2 – a

Since p(1) = g(1), we get

a = -2 – a

∴ a + a = -2

∴ 2a = -2

**Question 4:**

If the polynomials (ax^{3} + 4x^{2} – 3x + 4) and (x^{3} + x^{2} – 4x + a) leave the same remainder when divided by (x – 3),

find the value of a.

**Solution :**

Let p(x) = ax^{3} + 4x^{2} – 3x + 4 and g(x) = x^{3} + x^{2} – 4x + a

By remainder theorem, the two remainders are p(3) and g(3).

By the given condition, p(3) = g(3)

∴ p(3) = a(3)^{3} + 4(3)^{2} – 3(3) + 4

= 27a + 36 – 9 + 4

= 27a + 31

And, g(3) = (3)^{3} + (3)^{2} – 4(3) + a

= 27 + 9 – 12 + a

= 24 + a

Since p(3) = g(3), we get

27a + 31 = 24 + a

∴ 27a – a = 24 – 31

∴ 26a = -7

**Question 5:**

If p(x) = x^{4} – 2x^{3} + 3x^{2} – ax + b is a polynomial such that when it is divided by (x + 1) and (x – 1), the remainders are respectively 19 and 5. Determine the remainder when p(x) is divided by (x – 2).

**Solution :**

p(x) = x^{4} – 2x^{3} + 3x^{2} – ax + b

Now, p(-1) = 19 …..(given)

∴ (-1)^{4} – 2(-1)^{3} + 3(-1)^{2} – a(-1) + b = 19

∴ 1 + 2 + 3 + a + b = 19

∴ 6 + a + b = 19

∴ a + b = 13 ……(1)

Also, p(1) = 5 ….(given)

∴ (1)^{4} – 2(1)^{3} + 3(1)^{2} – a(1) + b = 5

∴ 1 – 2 + 3 – a + b = 5

∴ 2 – a + b = 5

∴ -a + b = 3 …….(2)

Adding equations (1) and (2), we get

2b = 16

∴ b = 8

Substituting b = 8 in equation (1), we get

a + 8 = 13

∴ a = 5

Substituting values of a and b in p(x), we have

p(x) = x^{4} – 2x^{3} + 3x^{2} – 5x + 8

∴ p(2) = (2)^{4} – 2(2)^{3} + 3(2)^{2} – 5(2) + 8

= 16 – 16 + 12 – 10 + 8

= 10

∴ Remainder = 10

**Question 6:**

The polynomials (2x^{3} – 5x^{2} + x + a) and (ax^{3} + 2x^{2} – 3) when divided by (x-2) leave the remainder R_{1} and R_{2} respectively. Find the value of ‘a’ in each of the following cases if

i. R_{1} = R_{2}ii. 2R_{1} + R_{2} = 0

iii R_{1}+ R_{2} = 0

iv R_{1} – 2R_{2} = 0

**Solution :**

Let p(x) = 2x^{3} – 5x^{2} + x + a and g(x) = ax^{3} + 2x^{2} – 3

By the remainder theorem, the two remainders are p(2) and g(2).

Now, p(2) = R_{1} ….(given)

∴ 2(2)^{3} – 5(2)^{2} + 2 + a = R_{1}∴ 16 – 20 + 2 + a = R_{1}∴ -2 + a = R_{1}Also, g(2) = R_{2} ….(given)

∴ a(2)^{3} + 2(2)^{2} – 3 = R_{2}∴ 8a + 8 – 3 = R_{2}∴ 8a + 5 = R_{2}

**KSEEB SSLC Solutions for Class 10 Maths Chapter 8** **Exercise 8.4**

**Question 1(i):**

In each of the following cases, use factor theorem to find whether g(x) is a factor of polynomial p(x) or not. [For (i) to (viii)]

p(x) = x^{3} – 3x^{2} + 6x – 20, g(x) = x – 2

**Solution :**

p(x) = x^{3} – 3x^{2} + 6x – 20 and g(x) = x – 2

By factor theorem, g(x) = (x – 2) is a factor of p(x) if p(2) = 0

∴ p(2) = (2)^{3} – 3(2)^{2} + 6(2) – 20

= 8 – 12 + 12 – 20

= -12

Since p(2) ≠ 0, g(x) is not a factor of p(x).

**Question 1(ii):**

p(x) = 2x^{4} + x^{3} + 4x^{2} – x – 7, g(x) = x + 2

**Solution :**

p(x) = 2x^{4} + x^{3} + 4x^{2} – x – 7, g(x) = x + 2

By factor theorem, g(x) = x + 2 is a factor of p(x) if p(-2) = 0

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

= 32 – 8 + 16 + 2 – 7

= 35

Since p(-2) ≠ 0, g(x) is not a factor of p(x).

**Question 1(iii):**

p(x) = 3x^{4} + 17x^{3} + 9x^{2} – 7x – 10, g(x) = x + 5

**Solution :**

p(x) = 3x^{4} + 17x^{3} + 9x^{2} – 7x – 10, g(x) = x + 5

By factor theorem, g(x) = x + 5 is a factor of p(x) if p(-5) = 0

∴ p(-5) = 3(-5)^{4} + 17(-5)^{3} + 9(-5)^{2} – 7(-5) – 10

= 3(625) + 17(-125) + 9(25) + 35 – 10

= 1875 – 2125 + 225 + 35 – 10

= 0

Since p(-5) = 0, g(x) is a factor of p(x).

**Question 1(iv):**

p(x) = x^{3} – 6x^{2} – 19x + 84, g(x) = x – 7

**Solution :**

p(x) = x^{3} – 6x^{2} – 19x + 84, g(x) = x – 7

By factor theorem, g(x) = x – 7 is a factor of p(x) if p(7) = 0

∴ p(7) = (7)^{3} – 6(7)^{2} – 19(7) + 84

= 343 – 6(49) – 133 + 84

= 343 – 294 – 133 + 84

= 0

Since p(7) = 0, g(x) is a factor of p(x).

**Question 1(v):**

**Solution :**

**Question 1(vi):**

p(x) = 3x^{3} + x^{2} – 20x + 12, g(x) = 3x – 2

**Solution :**

**Question 1(vii):**

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

**Solution :**

p(x) = x^{3} – 6x^{2} + 11x – 6,

g(x) = x^{2} – 3x + 2

= x^{2} – 2x – x + 2

= x(x – 2) – 1(x – 2)

= (x – 2)(x – 1)

By factor theorem, g(x) = (x – 2)(x – 1) is a factor of p(x) if p(2) = 0 and p(1) = 0.

∴ p(2) = (2)^{3} – 6(2)^{2} + 11(2) – 6

= 8 – 24 + 22 – 6

= 0

And, p(1) = (1)^{3} – 6(1)^{2} + 11(1) – 6

= 1 – 6 + 11 – 6

= 0

Since p(2) = 0 and p(1) = 0, g(x) is a factor of p(x).

**Question 1(viii):**

p(x) = 2x^{4} + 3x^{3} – 2x^{2 }– 9x – 12, g(x) = x^{2} – 3

**Solution :**

**Question 2:**

Find the value of a if (x – 5) is a factor of (x^{3} – 3x^{2} + ax – 10).

**Solution :**

Let p(x) = x^{3} – 3x^{2} + ax – 10

Since (x – 5) is a factor of p(x), p(5) = 0

∴ (5)^{3} – 3(5)^{2} + a(5) – 10 = 0

∴ 125 – 3(25) + 5a – 10 = 0

∴ 125 – 75 + 5a – 10 = 0

∴ 40 + 5a = 0

∴ 5a = -40

**Question 3:**

For what value of a, (x + 2) is a factor of (4x^{4 }+ 2x^{3} – 3x^{2} + 8x + 5a).

**Solution :**

Let p(x) = 4x^{4} + 2x^{3} – 3x^{2} + 8x + 5a

By factor theorem, x + 2 is a factor of p(x) if p(-2) = 0

∴ 4(-2)^{4} + 2(-2)^{3} – 3(-2)^{2} + 8(-2) + 5a = 0

∴ 64 – 16 – 12 – 16 + 5a = 0

∴ 20 + 5a = 0

∴ 5a = -20

**Question 4:**

if (x^{3} + ax^{2} – bx + 10) is divisible by x^{2} – 3x + 2, find the values of a and b.

**Solution :**

Let p(x) = x^{3} + ax^{2} – bx + 10 and

g(x) = x^{2} – 3x + 2

= x^{2} – 2x – x + 2

= x(x – 2) – 1(x – 2)

= (x – 2)(x – 1)

Since p(x) is divisible by g(x) = (x – 2)(x – 1),

p(2) = 0 and p(1) = 0

Consider p(2) = 0

∴ (2)^{3} + a(2)^{2} – b(2) + 10 = 0

∴ 8 + 4a – 2b + 10 = 0

∴ 18 + 4a – 2b = 0

∴ 4a – 2b = -18

Dividing throughout by 2, we get

2a – b = -9 ……(1)

Now, consider p(1) = 0

∴ (1)^{3} + a(1)^{2} – b(1) + 10 = 0

∴ 1 + a – b + 10 = 0

∴ a – b + 11 = 0

∴ a – b = -11 ….(2)

Subtracting equation (2) from (1), we get

a = 2

Substituting a = 2 in equation (2), we get

2 – b = -11

-b = -11 – 2

-b = -13

∴ b = 13

∴ a = 2 and b = 13

**Question 5:**

Find the values of a and b so that (x^{4} + ax^{3} + 2x^{2} – 3x + b) is divisible by (x^{2} – 1).

**Solution :**

Let p(x) = x^{4} + ax^{3} + 2x^{2} – 3x + b and

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

Since p(x) is divisible by g(x) = (x – 1)(x + 1),

p(1) = 0 and p(-1) = 0

Consider p(1) = 0

∴ (1)^{4} + a(1)^{3} + 2(1)^{2} – 3(1) + b = 0

∴ 1 + a + 2 – 3 + b = 0

∴ a + b = 0

∴ a = -b

Now, consider p(-1) = 0

∴ (-1)^{4} + a(-1)^{3} + 2(-1)^{2} – 3(-1) + b = 0

∴ 1 – a + 2 + 3 + b = 0

∴ -a + b + 6 = 0

∴ -a + b = -6 .….(1)

Substituting a = -b in equation (1), we get

-(-b) + b = -6

∴ b + b = -6

∴ 2b = -6

∴ b = -3

∴ a = -b = -(-3) = 3

∴ a = 3 and b = -3

**Question 6:**

**Solution :**

Comparing equations (1) and (2), we have

4a + b = a + 4b

∴ 4a – a = 4b – b

∴ 3a = 3b

∴ a = b

**Question 7:**

What must be added to x^{3} – 2x^{2} – 12x – 9 so that it is exactly divisible by x^{2} + x – 6?

**Solution :**

Let p(x) = x^{3} – 2x^{2} – 12x – 9 and

g(x) = x^{2} + x – 6

By division algorithm,

p(x) = [g(x) × q(x)] + r(x)

∴ p(x) + [-r(x)] = g(x) × q(x)

Let us divide p(x) by g(x) to find the remainder r(x).

∴ 3x + 27 must be added to p(x) so that it is exactly divisible by g(x).

**Question 8:**

What must be added to 2x^{3} + 3x^{2} – 22x + 12 so that the result is exactly divisible by (2x^{2} + 5x – 14)?

**Solution :**

Let p(x) = 2x^{3} + 3x^{2} – 22x + 12 and

g(x) = 2x^{2} + 5x – 14

By division algorithm,

p(x) = [g(x) × q(x)] + r(x)

∴ p(x) + [-r(x)] = g(x) × q(x)

Let us divide p(x) by g(x) to find the remainder r(x).

∴ 3x + 2 must be added to p(x) so that it is exactly divisible by g(x).

**Question 9:**

What must be subtracted from x^{3} – 6x^{2} – 5x + 80 so that the result is exactly divisible by x^{2} + x – 12?

**Solution :**

Let p(x) = x^{3} – 6x^{2} – 5x + 8 and g(x) = x^{2} + x – 12

By division algorithm,

p(x) = [g(x) × q(x)] + r(x)

∴ p(x) – r(x) = g(x) × q(x)

Let us divide p(x) by g(x) to find the remainder r(x).

∴ 14x – 4 must be subtracted from p(x) so that it is exactly divisible by x^{2} + x – 12.

**Question 10:**

In each of the following two polynomials, find the value of a, if x + a is a factor.

i. x^{3} + ax^{2} – 2x + a + 4

ii. x^{4} – a^{2}x^{2} + 3x – a

**Solution :**

i. Let, p(x) = x^{3} + ax^{2} – 2x + a + 4

Since (x + a) is a factor of p(x), p(-a) = 0

∴ (-a)^{3} + a(-a)^{2} – 2(-a) + a + 4 = 0

∴ -a^{3} + a^{3} + 2a + a + 4 = 0

∴ 3a + 4 = 0

∴ 3a = -4

∴ a=-4/3

ii. Let, p(x) = x^{4} – a^{2}x^{2} + 3x – a

Since (x + a) is a factor of p(x), p(-a) = 0

∴ (-a)^{4} – a^{2}(-a)^{2} + 3(-a) – a = 0

∴ a^{4} – a^{4} – 3a – a = 0

∴ -4a =0

∴ a = 0

**Exercise 8.5:**

**Question 1(i):**

Find the quotient and remainder using synthetic division. [For (i) to (ix)] (x^{3} + x^{2} – 3x + 5) ÷ (x – 1)

**Solution :**

Dividend: x^{3} + x^{2} – 3x + 5

Divisor: x – 1

Synthetic division:

∴ q(x) = x^{2} + 2x – 1 and r(x) = 4

**Question 1(ii):**

(3x^{3} – 2x^{2} + 7x – 5) ÷ (x + 3)

**Solution :**

Dividend: 3x^{3} – 2x^{2} + 7x – 5

Divisor: x + 3

∴ q(x) = 3x^{2} – 11x + 40 and r(x) = -125

**Question 1(iii):**

( 4x^{3} – 16x^{2} – 9x – 36 ) ÷ (x + 2)

**Solution :**

Dividend: 4x^{3} – 16x^{2} – 9x – 36

Divisor: x + 2

Synthetic division:

∴ q(x) = 4x^{2} – 24x + 39 and r(x) = -114

**Question 1(iv):**

(2x^{3} – x^{2} + 22x – 24) ÷ (x + 2)

**Solution :**

Dividend: 2x^{3} – x^{2} + 22x – 24

Divisor: x + 2

Synthetic division:

∴ q(x) = 2x^{2} – 5x + 32 and r(x) = -88

**Question 1(v):**

(6x^{4} – 29x^{3} + 40x^{2} – 12) ÷ (x – 3)

**Solution :**

Dividend: 6x^{4} – 29x^{3} + 40x^{2} – 12

Divisor: x – 3

Synthetic division:

∴ q(x) = 6x^{3} – 11x^{2} + 7x + 21 and r(x) = 51

**Question 1(vi):**

(8x^{4 }– 27x^{2} + 6x + 9) ÷ (x + 1)

**Solution :**

Dividend: 8x^{4 }– 27x^{2} + 6x + 9

Divisor: x + 1

Synthetic division:

∴ q(x) = 8x^{3} – 8x^{2} – 19x + 25 and r(x) = -16

**Question 1(vii):**

(3x^{3 }– 4x^{2} – 10x + 6) ÷ (3x – 2)

**Solution :**

Dividend: 3x^{3 }– 4x^{2} – 10x + 6

**Question 1(viii):**

(8x^{4} – 2x^{2} + 6x – 5) ÷ (4x + 1)

**Solution :**

Dividend: 8x^{4} – 2x^{2} + 6x – 5

**Question 1(ix):**

(2x^{4} – 7x^{3} – 13x^{2} + 63x – 48) ÷ (2x – 1)

**Solution :**

Dividend: 2x^{4} – 7x^{3} – 13x^{2} + 63x – 48

**Question 2:**

If the quotient obtained on dividing (x^{4} + 10x^{3} + 35x^{2} + 50x + 29) by (x + 4) is (x^{3} – ax^{2} + bx + 6) then find a, b and also the remainder.

**Solution :**

Dividend: p(x) = x^{4} + 10x^{3} + 35x^{2} + 50x + 29

Divisor: g(x) = x + 4

Quotient: q(x) = x^{3} – ax^{2} + bx + 6 and

Synthetic division

∴ q(x) = x^{3} + 6x^{2} + 11x + 6 and r(x) = 5

But q(x) = x^{3} – ax^{2} + bx + 6

On comparing,

a = -6 and b = 11

Remainder, r(x) = 5

**Question 3:**

If the quotient obtained on dividing (8x^{4} – 2x^{2} + 6x – 7) by (2x + 1) is (4x^{3} + px^{2} – qx + 3) then find p, q and also the remainder.

**Solution :**

Dividend: p(x) = 8x^{4} – 2x^{2} + 6x – 7

**All Chapter KSEEB Solutions For Class 10 Maths**

—————————————————————————–**All Subject KSEEB Solutions For Class 10**

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