RD Sharma Class 9 Mathematics Chapter 6 Exercise 6.5 Factorization of Polynomials is provided here. In this exercise, students will learn how to perform factorisation of a polynomial with integral coefficient using factor theorem. The solutions covered in RD Sharma Class 9 are detailed with easy explanations so that the students can easily comprehend and understand the concepts in detail.

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#### Exercise 6.5 Page No: 6.32

**Using factor theorem, factorize each of the following polynomials:**

**Question 1: x ^{3} + 6x^{2} + 11x + 6**

**Solution**:

Let f(x) = x^{3} + 6x^{2} + 11x + 6

Step 1: Find the factors of constant term

Here constant term = 6

Factors of 6 are Â±1, Â±2, Â±3, Â±6

Step 2: Find the factors of f(x)

Let x + 1 = 0

â‡’ x = -1

Put the value of x in f(x)

f(-1) = (âˆ’1)^{3} + 6(âˆ’1)^{2} + 11(âˆ’1) + 6

= -1 + 6 -11 + 6

= 12 â€“ 12

= 0

So, (x + 1) is the factor of f(x)

Let x + 2 = 0

â‡’ x = -2

Put the value of x in f(x)

f(-2) = (âˆ’2)^{3} + 6(âˆ’2)^{2} + 11(âˆ’2) + 6 = -8 + 24 â€“ 22 + 6 = 0

So, (x + 2) is the factor of f(x)

Let x + 3 = 0

â‡’ x = -3

Put the value of x in f(x)

f(-3) = (âˆ’3)^{3} + 6(âˆ’3)^{2} + 11(âˆ’3) + 6 = -27 + 54 â€“ 33 + 6 = 0

So, (x + 3) is the factor of f(x)

Hence, f(x) = (x + 1)(x + 2)(x + 3)

**Question 2: x ^{3} + 2x^{2} â€“ x â€“ 2**

**Solution**:

Let f(x) = x^{3} + 2x^{2} â€“ x â€“ 2

Constant term = -2

Factors of -2 are Â±1, Â±2

Let x â€“ 1 = 0

â‡’ x = 1

Put the value of x in f(x)

f(1) = (1)^{3} + 2(1)^{2} â€“ 1 â€“ 2 = 1 + 2 â€“ 1 â€“ 2 = 0

So, (x â€“ 1) is factor of f(x)

Let x + 1 = 0

â‡’ x = -1

Put the value of x in f(x)

f(-1) = (-1)^{3} + 2(-1)^{2} â€“ 1 â€“ 2 = -1 + 2 + 1 â€“ 2 = 0

(x + 1) is a factor of f(x)

Let x + 2 = 0

â‡’ x = -2

Put the value of x in f(x)

f(-2) = (-2)^{3} + 2(-2)^{2} â€“ (-2) â€“ 2 = -8 + 8 + 2 â€“ 2 = 0

(x + 2) is a factor of f(x)

Let x – 2 = 0

â‡’ x = 2

Put the value of x in f(x)

f(2) = (2)^{3} + 2(2)^{2} â€“ 2 â€“ 2 = 8 + 8 â€“ 2 â€“ 2 = 12 â‰ 0

(x – 2) is not a factor of f(x)

Hence f(x) = (x + 1)(x- 1)(x+2)

**Question 3: x ^{3} â€“ 6x^{2} + 3x + 10**

**Solution**:

Let f(x) = x^{3} â€“ 6x^{2} + 3x + 10

Constant term = 10

Factors of 10 are Â±1, Â±2, Â±5, Â±10

Let x + 1 = 0 or x = -1

f(-1) = (-1)^{3} â€“ 6(-1)^{2} + 3(-1) + 10 = 10 â€“ 10 = 0

f(-1) = 0

Let x + 2 = 0 or x = -2

f(-2) = (-2)^{3} â€“ 6(-2)^{2} + 3(-2) + 10 = -8 â€“ 24 â€“ 6 + 10 = -28

f(-2) â‰ 0

Let x – 2 = 0 or x = 2

f(2) = (2)^{3} â€“ 6(2)^{2} + 3(2) + 10 = 8 â€“ 24 + 6 + 10 = 0

f(2) = 0

Let x – 5 = 0 or x = 5

f(5) = (5)^{3} â€“ 6(5)^{2} + 3(5) + 10 = 125 â€“ 150 + 15 + 10 = 0

f(5) = 0

Therefore, (x + 1), (x â€“ 2) and (x-5) are factors of f(x)

Hence f(x) = (x + 1) (x â€“ 2) (x-5)

**Question 4: x ^{4} â€“ 7x^{3} + 9x^{2} + 7x- 10**

**Solution:**

Let f(x) = x^{4} â€“ 7x^{3} + 9x^{2} + 7x- 10

Constant term = -10

Factors of -10 are Â±1, Â±2, Â±5, Â±10

Let x – 1 = 0 or x = 1

f(1) = (1)^{4} â€“ 7(1)^{3} + 9(1)^{2} + 7(1) â€“ 10 = 1 â€“ 7 + 9 + 7 -10 = 0

f(1) = 0

Let x + 1 = 0 or x = -1

f(-1) = (-1)^{4} â€“ 7(-1)^{3} + 9(-1)^{2} + 7(-1) â€“ 10 = 1 + 7 + 9 – 7 -10 = 0

f(-1) = 0

Let x – 2 = 0 or x = 2

f(2) = (2)^{4} â€“ 7(2)^{3} + 9(2)^{2} + 7(2) â€“ 10 = 16 â€“ 56 + 36 + 14 â€“ 10 = 0

f(2) = 0

Let x – 5 = 0 or x = 5

f(5) = (5)^{4} â€“ 7(5)^{3} + 9(5)^{2} + 7(5) â€“ 10 = 625 â€“ 875 + 225 + 35 â€“ 10 = 0

f(5) = 0

Therefore, (x – 1), (x + 1), (x â€“ 2) and (x-5) are factors of f(x)

Hence f(x) = (x – 1) (x + 1) (x â€“ 2) (x-5)

**Question 5: x ^{4} â€“ 2x^{3} â€“ 7x^{2} + 8x + 12**

**Solution**:

f(x) = x^{4} â€“ 2x^{3} â€“ 7x^{2} + 8x + 12

Constant term = 12

Factors of 12 are Â±1, Â±2, Â±3, Â±4, Â±6, Â±12

Let x – 1 = 0 or x = 1

f(1) = (1)^{4} â€“ 2(1)^{3} – 7(1)^{2} + 8(1) + 12 = 1 â€“ 2 â€“ 7 + 8 + 12 = 12

f(1) â‰ 0

Let x + 1 = 0 or x = -1

f(-1) = (-1)^{4} â€“ 2(-1)^{3} – 7(-1)^{2} + 8(-1) + 12 = 1 + 2 â€“ 7 – 8 + 12 = 0

f(-1) = 0

Let x +2 = 0 or x = -2

f(-2) = (-2)^{4} â€“ 2(-2)^{3} – 7(-2)^{2} + 8(-2) + 12 = 16 + 16 â€“ 28 – 16 + 12 = 0

f(-2) = 0

Let x – 2 = 0 or x = 2

f(2) = (2)^{4} â€“ 2(2)^{3} – 7(2)^{2} + 8(2) + 12 = 16 – 16 â€“ 28 + 16 + 12 = 0

f(2) = 0

Let x – 3 = 0 or x = 3

f(3) = (3)^{4} â€“ 2(3)^{3} – 7(3)^{2} + 8(3) + 12 = 0

f(3) = 0

Therefore, (x + 1), (x + 2), (x â€“ 2) and (x-3) are factors of f(x)

Hence f(x) = (x + 1)(x + 2) (x â€“ 2) (x-3)

**Question 6: x ^{4} + 10x^{3} + 35x^{2} + 50x + 24**

**Solution**:

Let f(x) = x^{4} + 10x^{3} + 35x^{2} + 50x + 24

Constant term = 24

Factors of 24 are Â±1, Â±2, Â±3, Â±4, Â±6, Â±8, Â±12, Â±24

Let x + 1 = 0 or x = -1

f(-1) = (-1)^{4} + 10(-1)^{3} + 35(-1)^{2} + 50(-1) + 24 = 1 â€“ 10 + 35 â€“ 50 + 24 = 0

f(1) = 0

(x + 1) is a factor of f(x)

Likewise, (x + 2),(x + 3),(x + 4) are also the factors of f(x)

Hence f(x) = (x + 1) (x + 2)(x + 3)(x + 4)

**Question 7: 2x ^{4} â€“ 7x^{3} â€“ 13x^{2} + 63x â€“ 45**

**Solution:**

Let f(x) = 2x^{4} â€“ 7x^{3} â€“ 13x^{2} + 63x â€“ 45

Constant term = -45

Factors of -45 are Â±1, Â±3, Â±5, Â±9, Â±15, Â±45

Here coefficient of x^4 is 2. So possible rational roots of f(x) are

Â±1, Â±3, Â±5, Â±9, Â±15, Â±45, Â±1/2,Â±3/2,Â±5/2,Â±9/2,Â±15/2,Â±45/2

Let x – 1 = 0 or x = 1

f(1) = 2(1)^{4} – 7(1)^{3} – 13(1)^{2} + 63(1) â€“ 45 = 2 â€“ 7 â€“ 13 + 63 â€“ 45 = 0

f(1) = 0

f(x) can be written as,

f(x) = (x-1) (2x^{3} â€“ 5x^{2} -18x +45)

or f(x) =(x-1)g(x) â€¦(1)

Let x – 3 = 0 or x = 3

f(3) = 2(3)^{4} – 7(3)^{3} – 13(3)^{2} + 63(3) â€“ 45 = = 162 â€“ 189 â€“ 117 + 189 â€“ 45= 0

f(3) = 0

Now, we are available with 2 factors of f(x), (x â€“ 1) and (x â€“ 3)

Here g(x) = 2x^{2} (x-3) + x(x-3) -15(x-3)

Taking (x-3) as common

= (x-3)(2x^{2} + x â€“ 15)

=(x-3)(2x^{2}+6x â€“ 5x -15)

= (x-3)(2x-5)(x+3)

= (x-3)(x+3)(2x-5) â€¦.(2)

From (1) and (2)

f(x) =(x-1) (x-3)(x+3)(2x-5)

## RD Sharma Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials Exercise 6.5

RD Sharma SolutionsÂ Class 9 Maths Chapter 6 Factorization of Polynomials Exercise 6.5 is based on factorisation of polynomials by using the factor theorem. Get a clear idea of factorisation of polynomials by looking at the examples which are solved in a step-by-step manner.