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# RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials

Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 6 - Factorisation of Polynomials.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

Exercise/Page

## RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 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 .   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  .

## RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 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

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

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 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

## RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 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)

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)

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(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)

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

## RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 6.34

Solution 1

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 2

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 3

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 4

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 5

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 6

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 7

Solution 8

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 9

Solution 10

Solution 11

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).

## RD Sharma Solution for Class 9 Mathematics Chapter 6 - Factorisation of Polynomials Page/Excercise 6.35

Solution 12

Solution 13

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 14

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 15

Solution 16

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 17

Solution 18

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 = 2= 128

Hence, correct option is (c).

Solution 19

Solution 20

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TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 6 - Factorisation of Polynomials for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 6 - Factorisation of Polynomials.

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