RD SHARMA Solutions for Class 9 Maths Chapter 6 - Factorisation of Polynomials
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.1
Binomial has two terms in it. So binomial of degree 35 can be written as x35 + 7 .
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.2
f(x) = 2x3 - 13x2 + 17x + 12
f(2) = 2(2)3 - 13(2)2 + 17(2) + 12
= 16 - 52 + 34 + 12
= 10
(ii)
f(-3) = 2(-3)3 - 13(-3)2 + 17(-3) + 12
= -54 - 117 - 51 + 12
= - 210
(iii)
f(0) = 2(0)3 - 13(0)2 + 17(0) + 12
= 0 - 0 + 0 + 12
=12
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.3
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.4
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.5
Let p(x) = x3 + 13x2 + 32x + 20
The factors of 20 are 1,
2,
4,
5 ... ...
By hit and trial method
p(- 1) = (- 1)3 + 13(- 1)2 + 32(- 1) + 20
= - 1 + 13 - 32 + 20
= 33 - 33 = 0
As p(-1) is zero, so x + 1 is a factor of this polynomial p(x).
Let us find the quotient while dividing x3 + 13x2 + 32x + 20 by (x + 1)
By long division
Dividend = Divisor

x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0
= (x + 1) (x2 + 10x + 2x + 20)
= (x + 1) [x (x + 10) + 2 (x + 10)]
= (x + 1) (x + 10) (x + 2)
= (x + 1) (x + 2) (x + 10)
Let p(x) = x3 - 3x2 - 9x - 5
Factors of 5 are 1,
5.
By hit and trial method
p(- 1) = (- 1)3 - 3(- 1)2 - 9(- 1) - 5
= - 1 - 3 + 9 - 5 = 0
So x + 1 is a factor of this polynomial
Let us find the quotient while dividing x3 + 3x2 - 9x - 5 by x + 1
By long division

= (x + 1) (x2 - 5 x + x - 5)
= (x + 1) [(x (x - 5) +1 (x - 5)]
= (x + 1) (x - 5) (x + 1)
= (x - 5) (x + 1) (x + 1)
Let p(y) = 2y3 + y2 - 2y - 1
p(1) = 2 ( 1)3 + (1)2 - 2( 1) - 1
= 2 + 1 - 2 - 1= 0
So, y - 1 is a factor of this polynomial
By long division method,
= (y - 1) (2y2 +3y + 1)
= (y - 1) (2y2 +2y + y +1)
= (y - 1) [2y (y + 1) + 1 (y + 1)]
= (y - 1) (y + 1) (2y + 1)
Chapter 6 - Factorisation of Polynomials Exercise 6.34
Let p(x) = x2 + 3ax - 2a be the given polynomial.
x - 2 is a factor of p(x).
Thus, p(2) = 0
(2)2 + 3a × 2 - 2a = 0
4 + 4a = 0
a = -1
Hence, correct option is (d).
Since, p(x) = x3 + 6x2 + 4x + k is exactly divisible by x + 2,
(x + 2) is a factor of p(x).
So, p(-2) = 0
i.e (-2)3 + 6(-2)2 + 4(-2) + k = 0
-8 + 24 - 8 + k = 0
24 - 16 + k = 0
8 + k = 0
k = -8
Hence, correct option is (c).
Let p(x) = x3 - 3x2a + 2a2x + b
(x - a) is a factor of p(x).
So, p(a) = 0
a3 - 3a2.a + 2a2.a + b = 0
a3 - 3a3 + 2a3 + b = 0
3a3 - 3a3 + b = 0
b = 0
Hence, correct option is (a).
Let p(x) = x140 + 2x151 + k
Since p(x) is divisible by (x + 1),
(x + 1) is a factor of p(x).
So, p(-1) = 0
(-1)140 + 2(-1)151 + k = 0
1 + 2(-1) + k = 0
1 - 2 + k = 0
k - 1 = 0
k = 1
Hence, correct option is (a).
If x + 2 is a factor of x2 + mx + 14,
then at x = -2,
x2 + mx + 14 = 0
i.e. (-2)2 + m(-2) + 14 = 0
4 - 2m + 14 = 0
2m = 18
m = 9
Hence, correct option is (c).
x - 3 is a factor of x2 - ax - 15,
then at x = 3,
x2 - ax - 15 = 0
i.e. (3)2 - a(3) - 15 = 0
9 - 3a - 15 = 0
a = -2
Hence, correct option is (a).
x + 1 is a factor of p(x) = 2x2 + kx
Then, p(-1) = 0
i.e. 2(-1)2 + k(-1) = 0
2 - k = 0
k = 2
Hence, correct option is (d).
If (x + 2) and (x - 1) are factors of polynomial x3 + 10x2 + mx + n,
then x = -2, x = +1 will satisfy the polynomial.
Let p(x) = x3 + 10x2 + mx + n
Then, p(-2) = 0
(-2)3 + 10(-2)2 + m(-2) + n = 0
-8 + 40 - 2m + n = 0
32 - 2m + n = 0 ....(1)
And, p(1) = 0
(1)3 + 10(1)2 + m(1) + n = 0
1 + 10 + m + n = 0
11 + m + n = 0 ....(2)
Substracting equation (1) from equation (2), we get
-21 + 3m = 0
3m = 21
m = 7
Substituting m = 7 in equation (2),
11 + 7 + n = 0
18 + n = 0
n = -18
Hence, correct option is (c).
Chapter 6 - Factorisation of Polynomials Exercise 6.35
If x + 1 is a factor of xn + 1,
then, at x = -1, xn + 1 = 0
(-1)n + 1 = 0
(-1)n = -1
(-1)n will be equal to -1 if and only if n is an odd integer.
If n is even, then (-1)n = 1
So, n should be an odd integer.
Hence, correct option is (a).
Correct option (c)
(3x - 1)7 = a7x7 + a6x6 + ......... + a1x + a0 ....(1)
Putting x = 1 in equation (1), we have
[3(1) - 1]7 = a7 + a6 + ..... + a1 + a0
So, a7 + a6 + a5 + ..... + a1 + a0 = 27 = 128
Hence, correct option is (c).
Let p(x) = x3 - 2x2 + ax - b, r(x) = x - 6 and q(x) = x2 - 2x - 3
Then q(x) is a factor of [p(x) - r(x)]
{because if p(x) is divided by q(x), remainder is r(x). So, [p(x) - r(x)] will be exactly divided by q(x)}
Now, q(x) = x2 - 2x - 3 = (x - 3)(x + 1)
If q(x) is a factor of [p(x) - r(x)] then (x - 3) and (x + 1) are also factors of [p(x) - r(x)]
So, at x = 3 and x = -1, p(x) - r(x) will be zero.
Now p(3) - r(3) = 0
i.e. (3)3 - 2(3)2 + a(3) - b - (3 - 6) = 0
i.e. 27 - 18 + 3a - b + 3 = 0
i.e. 3a - b + 12 = 0 ....(1)
And, p(-1) - r(-1) = 0
i.e. (-1)3 - 2(-1)2 + a(-1) - b - (-1 - 6) = 0
i.e. -1 - 2 - a - b + 7 = 0
i.e -a - b + 4 = 0 ....(2)
Subtracting equation (2) from equation (1), we get
4a + 8 = 0
a = -2
From (2), -(-2) - b + 4 = 0
b = 6
Hence, correct option is (c).
x4 + x2 - 20
= x4 + 5x2 - 4x2 - 20
=x2(x2 + 5) - 4(x2 + 5)
= (x2 + 5)(x2 - 4)
So, other factor is x2 - 4.
Hence, correct option is (a).
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