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
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.
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
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
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)
Solution :
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
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(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.
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):
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.
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
Question 5(vii):
Solution :
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)
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 :
Question 1(ii):
p(x) = 2x2 – 9x + 9; g(x) = x – 3
Solution :
Question 1(iii):
p(x) = x3 + 4x2 – 5x + 6; g(x) = x + 1
Solution :
Question 1(iv):
p(x) = 4x3 – 6x2 + x + 14; g(x) = 2x – 3
Solution :
Question 1(v):
p(x) = x4 – 3x2 – 4; g(x) = x + 2
Solution :
Question 1(vi):
p(x) = x3 – 1; g(x) = x – 1
Solution :
Question 1(vii):
p(x) = x6 – 64; g(x) = x – 2
Solution :
Question 1(viii):
p(x) = x7 – a7; g(x) = x – a
Solution :
Question 1(ix):
p(x) =x4 – 4x2 + 12x + 9; g(x) = x2 + 2x – 3
Solution :
Question 1(x):
p(x) = a4 + 4b4; g(x) = a2 + 2ab + b2
Solution :
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 :
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 :
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 + 1 | x2 – x + 1 2x2 + x + 5 x3 + 3x2 -x + 1 x3 – x2 – x – 1 x4 – 2x2 + 5x -7 | 4 3x + 1 0 2x – 4 4x + 12 |
Solution :
- 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 - 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 - 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 - 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 - 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 :
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 :
Question 8:
What should be added to (x4 – 1) so that it is exactly divisible by (x2 + 2x +1)?
Solution :
Question 9:
What should be subtracted from (x6 – a6), 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) = 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
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
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
Question 1(iv):
p(x) = 4x3 – 12x2 + 14x – 3, g(x) = 2x – 1
Solution :
Question 1(v):
p(x) = 7x3 – x2 + 2x – 1, g(x) = 1 – 2x
Solution :
Question 1(vi):
Solution :
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
Solution :
Question 2(iii):
Find the remainder using reminder theorem when
(2x2 + 3x2 + x + 1) is divided by
3 + 2x
Solution :
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
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
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
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 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):
Solution :
Question 1(vi):
p(x) = 3x3 + x2 – 20x + 12, g(x) = 3x – 2
Solution :
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 :
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
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
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:
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 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).
∴ 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).
∴ 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).
∴ 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:
∴ 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
∴ 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:
∴ q(x) = 4x2 – 24x + 39 and r(x) = -114
Question 1(iv):
(2x3 – x2 + 22x – 24) ÷ (x + 2)
Solution :
Dividend: 2x3 – x2 + 22x – 24
Divisor: x + 2
Synthetic division:
∴ 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:
∴ 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:
∴ 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
Question 1(viii):
(8x4 – 2x2 + 6x – 5) ÷ (4x + 1)
Solution :
Dividend: 8x4 – 2x2 + 6x – 5
Question 1(ix):
(2x4 – 7x3 – 13x2 + 63x – 48) ÷ (2x – 1)
Solution :
Dividend: 2x4 – 7x3 – 13x2 + 63x – 48
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
∴ 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
All Chapter KSEEB Solutions For Class 10 Maths
—————————————————————————–
All Subject KSEEB Solutions For Class 10
*************************************************
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.
If these solutions have helped you, you can also share kseebsolutionsfor.com to your friends.
Best of Luck!!