RD SHARMA Solutions for Class 9 Maths Chapter 6 - Factorisation of Polynomials
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.1
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Degree of a polynomial is the highest power of variable in the polynomial.
Binomial has two terms in it. So binomial of degree 35 can be written as x35 + 7 .
Binomial has two terms in it. So binomial of degree 35 can be written as x35 + 7 .
Monomial has only one term in it. So monomial of degree 100 can be written as 7x100.
Concept Insight: Mono, bi and tri means one, two and three respectively. So, monomial is a polynomial having one term similarly for binomials and trinomials. Degree is the highest exponent of variable. The answer is not unique in such problems . Remember that the terms are always separated by +ve or -ve sign and not with .
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.2
Solution 1
(i)
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
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
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.3
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.4
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Chapter 6 - Factorisation of Polynomials Exercise Ex. 6.5
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
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
We know that
Dividend = Divisor
Quotient + Remainder
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)
Dividend = Divisor
Quotient + Remainderx3 + 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)
Solution 12
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
Now, Dividend = Divisor 
Quotient + Remainder
x3 - 3x2 - 9x - 5 = (x + 1) (x2 - 4 x - 5) + 0
= (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)
Quotient + Remainder x3 - 3x2 - 9x - 5 = (x + 1) (x2 - 4 x - 5) + 0= (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)
Solution 13
Let p(y) = 2y3 + y2 - 2y - 1
By hit and trial method
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,
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,
p(y) = 2y3 + y2 - 2y - 1
= (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)
= (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)
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Chapter 6 - Factorisation of Polynomials Exercise 6.34
Solution 12
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).
Solution 13
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).
Solution 14
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).
Solution 15
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).
Solution 16
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).
Solution 17
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).
Solution 18
Solution 19
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).
Solution 20
Solution 21
Solution 22
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
Solution 27
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).
Solution 28
Solution 29
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).
Solution 30
Solution 31
Solution 23
Solution 24
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).
Solution 25
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).
Solution 26
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