KSEEB SSLC Solutions for Class 10 Maths Chapter 8 Polynomials

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. x2 – 9x + 20
ii. 2x + 4 + 6x2
iii. x3 + 2x2 – 5x – 6
iv. x3 + 17x – 21 – x2
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-1

Solution :

The degree of a polynomial is the highest power of its variable.
i. The degree of the polynomial x2 – 9x + 20 is 2.
ii. The degree of the polynomial 2x + 4 + 6x2 is 2.
iii. The degree of the polynomial x3 + 2x2 – 5x – 6 is 3.
iv. The degree of the polynomial x3 + 17x – 21 – x2 is 3.
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-1s

Question 2:

If f(x) = 2x3 + 3x2 – 11x + 6, find
i. f(0) f(1)iii. f(-1) iv. f(2) v. f(-3)

Solution :

f(x) = 2x3 + 3x2 – 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) = 7x2 + 2x + 14, when x = 1
ii. p(x) = -x3 + x2 – 6x + 5, when x = 2
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-3
iv.  p(x) = 2x4 – 3x3 – 3x2 + 6x – 2, when x = -2

Solution :

i. g(x) = 7x2 + 2x + 14
When x = 1,
g(1) = 7(1)2 + 2(1) + 14
= 7 + 2 + 14
= 23

ii. p(x) = -x3 + x2 – 6x + 5
When x = 2
p(2) = -(2)3 + (2)2 – 6(2) + 5
= -8 + 4 – 12 + 5
= -11
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-3s
iv. p(x) = 2x4 – 3x3 – 3x2 + 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)
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-4(i)

Solution :
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-4(i)s

Question 4(ii):

p(x) = x2 – 4, x = 2, x = -2

Solution :

p(x) = x2 – 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) = x3 – 6x2 + 11x – 6, x = 1, x = 2 and x = 3
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-4(iii)

Solution :

p(x) = x3 – 6x2 + 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.
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-4(iii)s

Question 4(iv):

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-4(iv)
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-4(iv)

Solution :

p(x) = 3x3 – 5x2 – 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.
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-4(iv)s

Question 4(v):

p(x) = x3 – 6x2 + 11x – 6, x = 1, x = 2 and x = 3

Solution :

p(x) = x3 – 6x2 + 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):

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-4(vi)
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-4(vi)

Solution :

p(x) = 3x3 – 5x2 – 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.
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-4(vi)s

Question 5(i):

Find the zeroes of the following quadratic polynomials and verify.
x2 + 4x + 4

Solution :

Let p(x) = x2 + 4x + 4
By factorization,
x2 + 4x + 4 = x2 + 2(x)(2) + (2)2 = (x + 2)2

The value of x2 + 4x + 4 is zero when (x + 2)2 = 0
i.e. when x + 2 = 0
i.e. when x = -2
∴ The zero of x2 + 4x + 4 is (-2).

Verification:
P(x) = x2 + 4x + 4
p(-2) = (-2)2 + 4(-2) + 4 = 4 – 8 + 4 = 0

Question 5(ii):

x2 – 2x – 15

Solution :

Let p(x) = x2 – 2x – 15
By factorization,
x2 – 2x – 15
x2 – 5x + 3x – 15
= x(x – 5) + 3(x – 5)
= (x – 5)(x + 3)

The value of x2 – 2x – 15 is zero when x – 5 = 0 or x + 3 = 0
i.e., when x = 5 or x = -3
∴ The zeroes of x2 – 2x – 15 are 5 and -3.

Verification:
p(x) = x2 – 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):

x2 + 9x – 36 

Solution :

Let p(x) = x2 + 9x – 36
By factorization,
x2 + 9x – 36
= x2 + 12x – 3x – 36
= x(x + 12) – 3(x + 12)
= (x + 12)(x – 3)

The value of x2 + 9x – 36 is zero when x + 12 = 0 or x – 3 = 0
i.e., when x = -12 or x = 3
∴ The zeroes of x2 + 9x – 36 are -12 and 3.

Verification:
p(x) = x2 + 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):

x2 + 5x – 14

Solution :

Let p(x) = x2 + 5x – 14
By factorization,
x2 + 5x – 14
= x2 + 7x – 2x – 14
= x(x + 7) – 2(x + 7)
= (x + 7)(x – 2)

The value of x2 + 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 x2 + 5x – 14 are -7 and 2.

Verification:
p(x) = x2 + 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):

4y2 + 8y

Solution :

Let p(y) = 4y2 + 8y
By factorization,
4y2 + 8y = 4y(y + 2)

The value of 4y2 + 8y is zero when 4y = 0 or y + 2 = 0
i.e., when y = 0 or y = -2
∴ The zeroes of the polynomial 4y2 + 8y are 0 and -2.

Verification:
p(y) = 4y2 + 8y
p(0) = 4(0)2 + 8(0) = 0
p(-2) = 4(-2)2 + 8(-2) = 16 – 16 = 0

Question 5(vi):

4a2 – 49

Solution :

Let p(a) = 4a2 – 49
By factorization,
4a2 – 49 = (2a)2 – (7)2 = (2a – 7)(2a + 7)

The value of 4a2 – 49 is zero when 2a – 7 = 0 or 2a + 7 = 0
i.e., when 2a = 7 or 2a = -7
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-5(vi)s

Question 5(vii):
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-5(vii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-5(vii)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-5(vii)s

Question 5(viii):

3x2 – x – 4

Solution :

Let p(x) = 3x2 – x – 4
By factorization,
3x2 – x – 4
= 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.1-5(viii)s

Question 6:

If x = 1 is a zero of the polynomial f(x) = x3 – 2x2 + 4x + k, find the value of k.

Solution :

f(x) = x3 – 2x2 + 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 x2 – x – (2k + 2)?

Solution :

Let f(x) = x2 – 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) = x2 + 4x + 4; g(x) = x + 2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(i)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(i)s

Question 1(ii):

p(x) = 2x2 – 9x + 9; g(x) = x – 3

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(ii)s

Question 1(iii):

p(x) = x3 + 4x2 – 5x + 6; g(x) = x + 1

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(iii)s

Question 1(iv):

p(x) = 4x3 – 6x2 + x + 14; g(x) = 2x – 3

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(iv)s

Question 1(v):

p(x) = x4 – 3x2 – 4; g(x) = x + 2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(v)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(v)s

Question 1(vi):

p(x) = x3 – 1; g(x) = x – 1

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(vi)s

Question 1(vii):

p(x) = x6 – 64; g(x) = x – 2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(vii)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(vii)s

Question 1(viii):

p(x) = x7 – a7; g(x) = x – a

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(viii)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(viii)s

Question 1(ix):

p(x) =x4 – 4x2 + 12x + 9; g(x) = x2 + 2x – 3

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(ix)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(ix)s

Question 1(x):

p(x) = a4 + 4b4; g(x) = a2 + 2ab + b2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(x)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-1(x)s

Question 2:

Find the divisor g(x), when the polynomial p(x) = 4x3 + 2x2 – 10x + 2 is divided by g(x) and the quotient and remainder obtained are (2x2 + 4x + 1) and 5 respectively.

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-2s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-2s

Question 3:

On dividing the polynomial p(x) = x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were
(x – 2) and (-2x + 4), respectively. Find g(x).

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-3s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-3s

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 x2 + 2x + 1x2 – x + 1 2x2 + x + 5 x3 + 3x2 -x + 1 x3 – x2 – x – 1 x4 – 2x2 + 5x -74 3x + 1 0 2x – 4 4x + 12

Solution :

  1. i. p(x) = [g(x) × q(x)] + r(x)
    = [(x – 2)(x2 – x + 1 )] + 4
    = x3 – x2 + x – 2x2 + 2x – 2 + 4
    = x3 – 3x2 + 3x + 2
  2. p(x) = [g(x) × q(x)] + r(x)
    = [(x + 3)(2x2 + x + 5)] + 3x + 1
    = 2x3 + x2 + 5x + 6x2 + 3x + 15 + 3x + 1
    = 2x3 + 7x2 + 11x + 16
  3. p(x) = [g(x) × q(x)] + r(x)
    = [(2x + 1)(x3 + 3x2 – x + 1)] + 0
    = 2x4 + 6x3 – 2x2 + 2x + x3 + 3x2 – x + 1 +0
    = 2x4 + 7x3 + x2 + x + 1
  4. p(x) = [g(x) × q(x)] + r(x)
    = [(x – 1)(x3 – x2 – x – 1)] + 2x – 4
    = x4 – x3 – x2 – x – x3 + x2 + x + 1 + 2x – 4
    = x4 – 2x3 + 2x – 3
  5. p(x) = [g(x) × q(x)] + r(x)
    = [(x2 + 2x + 1)(x4 – 2x2 + 5x – 7)] + 4x +12
    = x6 – 2x4 + 5x3 – 7x2 + 2x5 – 4x3 + 10x2 – 14x + x4 – 2x2 + 5x – 7 + 4x + 12
    = x6 + 2x5 – x4 + x3 + x2 + 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) = x2 + 7x + 10; g(x) = x – 2

Solution :

p(x) = x2 + 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)
∴ x2 + 7x + 10 = (x – 2)(ax + c) + k
∴ x2 + 7x + 10 = ax2 + cx – 2ax – 2c + k
∴ x2 + 7x + 10 = ax2 + (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) = 2x2 + 14x – 5; g(x) = x + 1

Solution :

p(x) = 2x2 + 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)
∴ 2x2 + 14x – 5 = (x + 1)(ax + c) + k
∴ 2x2 + 14x – 5 = ax2 + cx + ax + c + k
∴ 2x2 + 14x – 5 = ax2 + (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) = x3 + 4x2 – 6x + 2; g(x) = x – 3

Solution :

p(x) = x3 + 4x2 – 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) = ax2 + bx + c and r(x) = k
By using division algorithm, we have
p(x) = [g(x) × q(x)] + r(x)
∴ x3 + 4x2 – 6x + 2 = (x – 3)(ax2 + bx + c) + k
∴ x3 + 4x2 – 6x + 2 = ax3 + bx2 + cx – 3ax2 – 3bx – 3c  + k
∴ x3 + 4x2 – 6x + 2 = ax3 + (b – 3a)x2 + (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) = ax2 + bx + c = 1x2 + 7x + 15 = x2 + 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) = 4x3 + 11x2 – 11x + 8; g(x) = x2 – x + 1

Solution :

p(x) = 4x3 + 11x2 – 11x + 8 ∴ degree of p(x) is 3.
g(x) = x2 – 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)
∴ 4x3 + 11x2 – 11x + 8 = (x2 – x + 1)(ax + c) + k
∴ 4x3 + 11x2 – 11x + 8 = ax3 + cx2 – ax2 – cx + ax + c + k
∴ 4x3 + 11x2 – 11x + 8 = ax3 + (c – a)x2 + (-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 (x3 + 5x2 + 5x + 8) so that the resulting polynomial is exactly divisible by (x2 + 3x – 2)?

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-6s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-6s

Question 7:

What should be added to the polynomial (7x3 + 4x2 – x – 10) So that the resulting polynomial is exactly divisible by (2x2 + 3x – 2)?

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-7s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-7s

Question 8:

What should be added to (x4 – 1) so that it is exactly divisible by (x2 + 2x +1)?

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-8s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-8s

Question 9:

What should be subtracted from (x6 – a6), so that it is exactly divisible by (x – a)?

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-9s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-9s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-10(i)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-10(i)s

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 :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-10(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-10(ii)s

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 – Polynomials-Ex-8.2-10(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.2-10(iii)s

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) = x3 + 3x2 – 5x + 8, g(x) = x – 3

Solution :

By remainder theorem, the required remainder is equal to p(3).
p(x) = x3 + 3x2 – 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) = x3 + 3x2 – 5x + 8, g(x) = x – 3
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(i)s

Question 1(ii):

p(x) = 4x3 – 10x2 + 12x – 3, g(x) = x + 1

Solution :

By remainder theorem, the required remainder is equal to p(-1).
p(x) = 4x3 – 10x2 + 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) = 4x3 – 10x2 + 12x – 3, g(x) = x + 1
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(ii)s

Question 1(iii):

p(x) = 2x4 – 5x2 + 15x – 6, g(x) = x – 2

Solution :

By remainder theorem, the required remainder is equal to p(2).
p(x) = 2x4 – 5x2 + 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) = 2x4 – 5x2 + 15x – 6, g(x) = x – 2
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(iii)s

Question 1(iv):

p(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(iv)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(iv)s

Question 1(v):

p(x) = 7x3 – x2 + 2x – 1, g(x) = 1 – 2x

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(v)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(v)s

Question 1(vi):

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(vi)
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(vi)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-1(vi)s

Question 2(i):

Find the remainder using reminder theorem when
(2x2 + 3x2 + x + 1) is divided by
x – 1

Solution :

Let p(x) = 2x3 + 3x2 + 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
(2x2 + 3x2 + x + 1) is divided by
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-2(ii)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-2(ii)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-2(ii)s

Question 2(iii):

Find the remainder using reminder theorem when
(2x2 + 3x2 + x + 1) is divided by
3 + 2x

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-2(iii)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-2(iii)s

Question 2(iv):

Find the remainder using reminder theorem when
(2x2 + 3x2 + x + 1) is divided by x

Solution :

Let p(x) = 2x3 + 3x2 + 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
(2x2 + 3x2 + x + 1) is divided by
x+1/3

Solution :

Let p(x) = 2x3 + 3x2 + x + 1
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-2(v)s

Question 3:

If the polynomials (2x3 + ax2 + 3x – 5) and (x3 + x2 – 4x – a) leave the same remainder when divided by
(x – 1), find the value of a.

Solution :

Let p(x) = 2x3 + ax2 + 3x – 5 and g(x) = x3 + x2 – 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
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-3s

Question 4:

If the polynomials (ax3 + 4x2 – 3x + 4) and (x3 + x2 – 4x + a) leave the same remainder when divided by (x – 3),
find the value of a.

Solution :

Let p(x) = ax3 + 4x2 – 3x + 4 and g(x) = x3 + x2 – 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
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-4s

Question 5:

If p(x) = x4 – 2x3 + 3x2 – 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) = x4 – 2x3 + 3x2 – 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) = x4 – 2x3 + 3x2 – 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 (2x3 – 5x2 + x + a) and (ax3 + 2x2 – 3) when divided by (x-2) leave the remainder R1 and R2 respectively. Find the value of ‘a’ in each of the following cases if
i. R1 = R2
ii. 2R1 + R2 = 0
iii R1+ R2 = 0
iv R1 – 2R2 = 0

Solution :

Let p(x) = 2x3 – 5x2 + x + a and g(x) = ax3 + 2x2 – 3
By the remainder theorem, the two remainders are p(2) and g(2).
Now, p(2) = R1 ….(given)
∴ 2(2)3 – 5(2)2 + 2 + a = R1
∴ 16 – 20 + 2 + a = R1
∴ -2 + a = R1
Also, g(2) = R2 ….(given)
∴ a(2)3 + 2(2)2 – 3 = R2
∴ 8a + 8 – 3 = R2
∴ 8a + 5 = R2
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.3-6s

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) = x3 – 3x2 + 6x – 20, g(x) = x – 2

Solution :

p(x) = x3 – 3x2 + 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) = 2x4 + x3 + 4x2 – x – 7, g(x) = x + 2

Solution :

p(x) = 2x4 + x3 + 4x2 – 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) = 3x4 + 17x3 + 9x2 – 7x – 10, g(x) = x + 5

Solution :

p(x) = 3x4 + 17x3 + 9x2 – 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) = x3 – 6x2 – 19x + 84, g(x) = x – 7

Solution :

p(x) = x3 – 6x2 – 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):

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-1(v)
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-1(v)

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-1(v)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-1(v)s

Question 1(vi):

p(x) = 3x3 + x2 – 20x + 12, g(x) = 3x – 2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-1(vi)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-1(vi)s

Question 1(vii):

p(x) = x3 – 6x2 + 11x – 6, g(x) = x2 – 3x + 2

Solution :

p(x) = x3 – 6x2 + 11x – 6,
g(x) = x2 – 3x + 2
= x2 – 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) = 2x4 + 3x3 – 2x2 – 9x – 12, g(x) = x2 – 3

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-1(viii)s
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-1(viii)s

Question 2:

Find the value of a if (x – 5) is a factor of (x3 – 3x2 + ax – 10).

Solution :

Let p(x) = x3 – 3x2 + 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
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-2s

Question 3:

For what value of a, (x + 2) is a factor of (4x4 + 2x3 – 3x2 + 8x + 5a).

Solution :

Let p(x) = 4x4 + 2x3 – 3x2 + 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
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-3s

Question 4:

if (x3 + ax2 – bx + 10) is divisible by x2 – 3x + 2, find the values of a and b.

Solution :

Let p(x) = x3 + ax2 – bx + 10 and
g(x) = x2 – 3x + 2
= x2 – 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 (x4 + ax3 + 2x2 – 3x + b) is divisible by (x2 – 1).

Solution :

Let p(x) = x4 + ax3 + 2x2 – 3x + b and
g(x) = x2 – 1 = x2 – (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:

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-6
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-6

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-6s
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 x3 – 2x2 – 12x – 9 so that it is exactly divisible by x2 + x – 6?

Solution :

Let p(x) = x3 – 2x2 – 12x – 9 and
g(x) = x2 + 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).
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-7s
∴ 3x + 27 must be added to p(x) so that it is exactly divisible by g(x).

Question 8:

What must be added to 2x3 + 3x2 – 22x + 12 so that the result is exactly divisible by (2x2 + 5x – 14)?

Solution :

Let p(x) = 2x3 + 3x2 – 22x + 12 and
g(x) = 2x2 + 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).
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-8s
∴ 3x + 2 must be added to p(x) so that it is exactly divisible by g(x).

Question 9:

What must be subtracted from x3 – 6x2 – 5x + 80 so that the result is exactly divisible by x2 + x – 12?

Solution :

Let p(x) = x3 – 6x2 – 5x + 8 and g(x) = x2 + 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).
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.4-9s

∴ 14x – 4 must be subtracted from p(x) so that it is exactly divisible by x2 + x – 12.

Question 10:

In each of the following two polynomials, find the value of a, if x + a is a factor.
i. x3 + ax2 – 2x + a + 4
ii. x4 – a2x2 + 3x – a

Solution :

i. Let, p(x) = x3 + ax2 – 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
∴ -a3 + a3 + 2a + a + 4 = 0
∴ 3a + 4 = 0
∴ 3a = -4
∴ a=-4/3
ii. Let, p(x) = x4 – a2x2 + 3x – a
Since (x + a) is a factor of p(x), p(-a) = 0
∴ (-a)4 – a2(-a)2 + 3(-a) – a = 0
∴ a4 – a4 – 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)] (x3 + x2 – 3x + 5) ÷ (x – 1)

Solution :

Dividend: x3 + x2 – 3x + 5
Divisor: x – 1
Synthetic division:
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-1(i)
∴ q(x) = x2 + 2x – 1 and r(x) = 4

Question 1(ii):

(3x3 – 2x2 + 7x – 5) ÷ (x + 3)

Solution :

Dividend: 3x3 – 2x2 + 7x – 5
Divisor: x + 3
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-1(ii)s
∴ q(x) = 3x2 – 11x + 40 and r(x) = -125

Question 1(iii):

( 4x3 – 16x2 – 9x – 36 ) ÷ (x + 2)

Solution :

Dividend: 4x3 – 16x2 – 9x – 36
Divisor: x + 2
Synthetic division:
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-1(iii)s
∴ q(x) = 4x2 – 24x + 39 and r(x) = -114

Question 1(iv):

(2x3 – x2 + 22x – 24) ÷ (x + 2)
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-1(v)s

Solution :

Dividend: 2x3 – x2 + 22x – 24
Divisor: x + 2
Synthetic division:
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-1(iv)s
∴ q(x) = 2x2 – 5x + 32 and r(x) = -88

Question 1(v):

(6x4 – 29x3 + 40x2 – 12) ÷ (x – 3)

Solution :

Dividend: 6x4 – 29x3 + 40x2 – 12
Divisor: x – 3
Synthetic division:
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-1(v)s

∴ q(x) = 6x3 – 11x2 + 7x + 21 and r(x) = 51

Question 1(vi):

(8x4 – 27x2 + 6x + 9) ÷ (x + 1)

Solution :

Dividend: 8x4 – 27x2 + 6x + 9
Divisor: x + 1
Synthetic division:
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-1(vi)s
∴ q(x) = 8x3 – 8x2 – 19x + 25 and r(x) = -16

Question 1(vii):

(3x3 – 4x2 – 10x + 6) ÷ (3x – 2)

Solution :

Dividend: 3x3 – 4x2 – 10x + 6
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-1(vii)s

Question 1(viii):

(8x4 – 2x2 + 6x – 5) ÷ (4x + 1)

Solution :

Dividend: 8x4 – 2x2 + 6x – 5
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-1(viii)s

Question 1(ix):

(2x4 – 7x3 – 13x2 + 63x – 48) ÷ (2x – 1)

Solution :

Dividend: 2x4 – 7x3 – 13x2 + 63x – 48
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-1(ix)s

Question 2:

If the quotient obtained on dividing (x4 + 10x3 + 35x2 + 50x + 29) by (x + 4) is (x3 – ax2 + bx + 6) then find a, b and also the remainder.

Solution :

Dividend: p(x) = x4 + 10x3 + 35x2 + 50x + 29
Divisor: g(x) = x + 4
Quotient: q(x) = x3 – ax2 + bx + 6 and
Synthetic division
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-2s
∴ q(x) = x3 + 6x2 + 11x + 6 and r(x) = 5
But q(x) = x3 – ax2 + bx + 6
On comparing,
a = -6 and b = 11
Remainder, r(x) = 5

Question 3:

If the quotient obtained on dividing (8x4 – 2x2 + 6x – 7) by (2x + 1) is (4x3 + px2 – qx + 3) then find p, q and also the remainder.

Solution :

Dividend: p(x) = 8x4 – 2x2 + 6x – 7
KSEEB SSLC Solutions for Class 10 Maths – Polynomials-Ex-8.5-3s

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All Subject KSEEB Solutions For Class 10

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