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• RD Sharma Class 9 Solutions

# Class 9 RD Sharma Solutions – Chapter 6 Factorisation of Polynomials- Exercise 6.5 | Set 2

### Question 11. Using factor theorem, factorize of the polynomials : x3 – 10x2 – 53x – 42

Solution:

Given that,

f(x) = x3–10x2 – 53x – 42

The constant in f(x) is – 42,

The factors of – 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42,

Let’s assume, x + 1 = 0

x = – 1

f(-1) = (−1)3 –10(−1)2 – 53(−1) – 42

-1 – 10 + 53 – 42 = 0

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

Now, divide f(x) with (x + 1) to get other factors

By using long division method we get,

x3 – 10x2 – 53x – 42 = (x + 1) (x2 – 11x – 42)

Now,

x2 – 11x – 42 = x2 – 14x + 3x – 42

x(x – 14) + 3(x – 14)

(x + 3)(x – 14)

Hence, x3 – 10x2 – 53x – 42 = (x + 1) (x + 3)(x – 14)

### Question 12. Using factor theorem, factorize of the polynomials : y3 – 2y2 – 29y – 42

Solution:

Given that, f(x) = y3 – 2y2 – 29y – 42

The constant in f(x) is – 42,

The factors of -42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42,

Let’s assume, y + 2 = 0

y = – 2

f(-2) =  (−2)3 – 2(−2)2–29(−2) – 42

-8 -8 + 58 – 42 = 0

therefore, (y + 2) is the factor of f(y)

Now, divide f(y) with (y + 2) to get other factors

By using long division method we get,

y3 – 2y2 – 29y – 42 = (y + 2) (y2 – 4y – 21)

Now,

y2 – 4y – 21 = y2 – 7y + 3y – 21

y(y – 7) +3(y – 7)

(y – 7)(y + 3)

Hence, y3 – 2y2 – 29y – 42 = (y + 2) (y – 7)(y + 3)

### Question 13. Using factor theorem, factorize of the polynomials : 2y3 – 5y2 – 19y + 42

Solution:

Given that, f(x) = 2y3 – 5y2 – 19y + 42

The constant in f(x) is + 42,

The factors of 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42,

Let’s assume, y – 2 = 0

y = 2

f(2) = 2(2)3 – 5(2)2 – 19(2) + 42

16 – 20 – 38 + 42 = 0

therefore, (y – 2) is the factor of f(y)

Now, divide f(y) with (y – 2) to get other factors

By using long division method we get,

2y3 – 5y2 – 19y + 42 = (y – 2) (2y2 – y – 21)

Now,

2y2 – y – 21

The factors are (y + 3) (2y – 7)

Hence, 2y3 – 5y2 -19y + 42 = (y – 2) (y + 3) (2y – 7)

### Question 14. Using factor theorem, factorize of the polynomials : x3 + 13x2 + 32x + 20

Solution:

Given that, f(x) = x3 + 13x2 + 32x + 20

The constant in f(x) is 20,

The factors of 20 are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20,

Let’s assume, x + 1 = 0

x = -1

f(-1) =  (−1)3+13(−1)2 + 32(−1) + 20

-1 + 13 – 32 + 20 = 0

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

Divide f(x) with (x + 1) to get other factors

By using long division method we get,

x3 + 13x2 +32x + 20 = (x + 1)( x2 + 12x + 20)

Now,

x2 + 12x + 20 = x2 + 10x + 2x + 20

x(x + 10) + 2(x + 10)

The factors are (x + 10) and (x + 2)

Hence, x3 + 13x2 + 32x + 20 = (x + 1)(x + 10)(x + 2)

### Question 15. Using factor theorem, factorize of the polynomials : x3 – 3x2 – 9x – 5

Solution:

Given that, f(x) = x3 – 3x2 – 9x – 5

The constant in f(x) is -5,

The factors of -5 are ±1, ±5,

Let’s assume, x + 1 = 0

x = -1

f(-1) = (−1)3 – 3(−1)2 – 9(-1) – 5

-1 – 3 + 9 – 5 = 0

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

Divide f(x) with (x + 1) to get other factors

By using long division method we get,

x3 – 3x2 – 9x – 5 = (x + 1)( x2 – 4x – 5)

Now,

x2 – 4x – 5 = x2 – 5x + x – 5

x(x – 5) + 1(x – 5)

The factors are (x – 5) and (x + 1)

Hence, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x + 1)

### Question 16. Using factor theorem, factorize of the polynomials : 2y3 + y2 – 2y – 1

Solution:

Given that, f(y) = 2y3 + y2 – 2y – 1

The constant term is 2,

The factors of 2 are ± 1, ± 1/2,

Let’s assume, y – 1= 0

y = 1

f(1) = 2(1)3 +(1)2 – 2(1) – 1

2 + 1 – 2 – 1 = 0

therefore, (y – 1) is the factor of f(y)

Divide f(y) with (y – 1) to get other factors

By using long division method we get,

2y3 + y2 – 2y – 1 = (y – 1) (2y2 + 3y + 1)

Now,

2y2 + 3y + 1 = 2y2 + 2y + y + 1

2y(y + 1) + 1(y + 1)

(2y + 1) (y + 1) are the factors

Hence, 2y3 + y2 – 2y – 1 = (y – 1) (2y + 1) (y + 1)

### Question 17. Using factor theorem, factorize of the polynomials : x3 – 2x2 – x + 2

Solution:

Given that, f(x) = x3 – 2x2 – x + 2

The constant term is 2,

The factors of 2 are ±1, ± 1/2,

Let’s assume, x – 1= 0

x = 1

f(1) = (1)3 – 2(1)2 – (1) + 2

1 – 2 – 1 + 2 = 0

therefore, (x – 1) is the factor of f(x)

Divide f(x) with (x – 1) to get other factors

By using long division method we get,

x3 – 2x2 – y + 2 = (x – 1) (x2 – x – 2)

Now,

x2 – x – 2 = x2 – 2x + x  – 2

x(x – 2) + 1(x – 2)

(x – 2)(x + 1)  are the factors

Hence, x3 – 2x2 – y + 2 = (x – 1)(x + 1)(x – 2)

### Question 18. Factorize each of the following polynomials :

1. x3 + 13x2 + 31x – 45 given that x + 9 is a factor

2. 4x3 + 20x2 + 33x + 18 given that 2x + 3 is a factor

Solution:

1. x3 + 13x2 + 31x – 45

Given that, x + 9 is a factor

Let’s assume, f(x) = x3 + 13x2 + 31x – 45

divide f(x) with (x + 9) to get other factors

By using long division method we get,

x3 + 13x2 + 31x – 45 = (x + 9)( x2 + 4x – 5)

Now,

x2 + 4x – 5 = x2 + 5x – x  – 5

x(x + 5) -1(x + 5)

(x + 5) (x – 1) are the factors

Hence, x3 + 13x2 + 31x – 45 = (x + 9)(x + 5)(x – 1)

2. 4x3 + 20x2 + 33x + 18

Given that, 2x + 3 is a factor

let’s assume, f(x) =  4x3 + 20x2 + 33x + 18

divide f(x) with (2x + 3) to get other factors

By using long division method we get,

4x3 + 20x2 + 33x + 18 = (2x + 3) (2×2 + 7x + 6)

Now,

2x2 + 7x + 6 = 2x2 + 4x + 3x + 6

2x(x + 2) + 3(x + 2)

(2x + 3)(x + 2) are the factors

Hence, 4x3 + 20x2 + 33x + 18 = (2x + 3)(2x + 3)(x + 2)

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