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

**Question 11. Using factor theorem, factorize of the polynomials : x**^{3} – 10x^{2} – 53x – 42

^{3}– 10x

^{2}– 53x – 42

**Solution:**

Given that,

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free classeswhich will definitely help them in making a wise career choice in the future.f(x) = x

^{3}–10x^{2}– 53x – 42The 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,

x

^{3}– 10x^{2}– 53x – 42 = (x + 1) (x2 – 11x – 42)Now,

x

^{2}– 11x – 42 = x^{2}– 14x + 3x – 42x(x – 14) + 3(x – 14)

(x + 3)(x – 14)

Hence, x^{3}– 10x^{2}– 53x – 42 = (x + 1) (x + 3)(x – 14)

**Question 12. Using factor theorem, factorize of the polynomials : y**^{3} – 2y^{2} – 29y – 42

^{3}– 2y

^{2}– 29y – 42

**Solution:**

Given that, f(x) = y

^{3}– 2y^{2}– 29y – 42The 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,

y

^{3}– 2y^{2}– 29y – 42 = (y + 2) (y2 – 4y – 21)Now,

y

^{2}– 4y – 21 = y^{2}– 7y + 3y – 21y(y – 7) +3(y – 7)

(y – 7)(y + 3)

Hence, y^{3}– 2y^{2}– 29y – 42 = (y + 2) (y – 7)(y + 3)

**Question 13. Using factor theorem, factorize of the polynomials : 2y**^{3} – 5y^{2} – 19y + 42

^{3}– 5y

^{2}– 19y + 42

**Solution:**

Given that, f(x) = 2y

^{3}– 5y^{2}– 19y + 42The 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) + 4216 – 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,

2y

^{3}– 5y^{2}– 19y + 42 = (y – 2) (2y^{2}– y – 21)Now,

2y

^{2}– y – 21The factors are (y + 3) (2y – 7)

Hence, 2y^{3}– 5y^{2}-19y + 42 = (y – 2) (y + 3) (2y – 7)

**Question 14. Using factor theorem, factorize of the polynomials : x**^{3} + 13x^{2} + 32x + 20

^{3}+ 13x

^{2}+ 32x + 20

**Solution:**

Given that, f(x) = x

^{3}+ 13x^{2}+ 32x + 20The 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,

x

^{3}+ 13x^{2}+32x + 20 = (x + 1)( x2 + 12x + 20)Now,

x

^{2}+ 12x + 20 = x^{2}+ 10x + 2x + 20x(x + 10) + 2(x + 10)

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

Hence, x^{3}+ 13x^{2}+ 32x + 20 = (x + 1)(x + 10)(x + 2)

**Question 15. Using factor theorem, factorize of the polynomials : x**^{3} – 3x^{2} – 9x – 5

^{3}– 3x

^{2}– 9x – 5

**Solution:**

Given that, f(x) = x

^{3}– 3x^{2}– 9x – 5The 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,

x

^{3}– 3x^{2}– 9x – 5 = (x + 1)( x^{2}– 4x – 5)Now,

x

^{2}– 4x – 5 = x^{2}– 5x + x – 5x(x – 5) + 1(x – 5)

The factors are (x – 5) and (x + 1)

Hence, x^{3}– 3x^{2}– 9x – 5 = (x + 1)(x – 5)(x + 1)

**Question 16. Using factor theorem, factorize of the polynomials : 2y**^{3} + y^{2} – 2y – 1

^{3}+ y

^{2}– 2y – 1

**Solution:**

Given that, f(y) = 2y

^{3}+ y^{2}– 2y – 1The 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) – 12 + 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,

2y

^{3}+ y^{2}– 2y – 1 = (y – 1) (2y^{2}+ 3y + 1)Now,

2y

^{2}+ 3y + 1 = 2y^{2}+ 2y + y + 12y(y + 1) + 1(y + 1)

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

Hence, 2y^{3}+ y^{2}– 2y – 1 = (y – 1) (2y + 1) (y + 1)

**Question 17. Using factor theorem, factorize of the polynomials : x**^{3} – 2x^{2} – x + 2

^{3}– 2x

^{2}– x + 2

**Solution:**

Given that, f(x) = x

^{3}– 2x^{2}– x + 2The 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) + 21 – 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,

x

^{3}– 2x^{2}– y + 2 = (x – 1) (x^{2}– x – 2)Now,

x

^{2}– x – 2 = x^{2}– 2x + x – 2x(x – 2) + 1(x – 2)

(x – 2)(x + 1) are the factors

Hence, x^{3}– 2x^{2}– y + 2 = (x – 1)(x + 1)(x – 2)

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

**1. x ^{3} + 13x^{2} + 31x – 45 given that x + 9 is a factor**

**2. 4x ^{3} + 20x^{2} + 33x + 18 given that 2x + 3 is a factor**

**Solution:**

1. x^{3}+ 13x^{2}+ 31x – 45Given that, x + 9 is a factor

Let’s assume, f(x) = x

^{3}+ 13x^{2}+ 31x – 45divide f(x) with (x + 9) to get other factors

By using long division method we get,

x

^{3}+ 13x^{2}+ 31x – 45 = (x + 9)( x2 + 4x – 5)Now,

x

^{2}+ 4x – 5 = x^{2}+ 5x – x – 5x(x + 5) -1(x + 5)

(x + 5) (x – 1) are the factors

Hence, x^{3}+ 13x^{2}+ 31x – 45 = (x + 9)(x + 5)(x – 1)

2. 4x^{3}+ 20x^{2}+ 33x + 18Given that, 2x + 3 is a factor

let’s assume, f(x) = 4x

^{3}+ 20x^{2}+ 33x + 18divide f(x) with (2x + 3) to get other factors

By using long division method we get,

4x

^{3}+ 20x^{2}+ 33x + 18 = (2x + 3) (2×2 + 7x + 6)Now,

2x

^{2}+ 7x + 6 = 2x^{2}+ 4x + 3x + 62x(x + 2) + 3(x + 2)

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

Hence, 4x^{3}+ 20x^{2}+ 33x + 18 = (2x + 3)(2x + 3)(x + 2)