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

• Last Updated : 30 Apr, 2021

Question 11. Using factor theorem, factorize of the polynomials : x3 â€“ 10x2 â€“ 53x â€“ 42

Solution:

Given that,

f(x) = x3â€“10x2 â€“ 53x â€“ 42

The constant in f(x) is – 42,

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

Let’s assume, x + 1 = 0

x = – 1

f(-1) = (âˆ’1)3 â€“10(âˆ’1)2 â€“ 53(âˆ’1) â€“ 42

-1 â€“ 10 + 53 â€“ 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 â€“ 10x2 â€“ 53x â€“ 42 = (x + 1) (x2 â€“ 11x â€“ 42)

Now,

x2 â€“ 11x â€“ 42 = x2 â€“ 14x + 3x â€“ 42

x(x â€“ 14) + 3(x â€“ 14)

(x + 3)(x â€“ 14)

Hence, x3 â€“ 10x2 â€“ 53x â€“ 42 = (x + 1) (x + 3)(x â€“ 14)

Question 12. Using factor theorem, factorize of the polynomials : y3 â€“ 2y2 â€“ 29y â€“ 42

Solution:

Given that, f(x) = y3 â€“ 2y2 â€“ 29y â€“ 42

The constant in f(x) is – 42,

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

Let’s assume, y + 2 = 0

y = â€“ 2

f(-2) =  (âˆ’2)3 â€“ 2(âˆ’2)2â€“29(âˆ’2) â€“ 42

-8 -8 + 58 â€“ 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 â€“ 2y2 â€“ 29y â€“ 42 = (y + 2) (y2 â€“ 4y â€“ 21)

Now,

y2 â€“ 4y â€“ 21 = y2 â€“ 7y + 3y â€“ 21

y(y â€“ 7) +3(y â€“ 7)

(y â€“ 7)(y + 3)

Hence, y3 â€“ 2y2 â€“ 29y â€“ 42 = (y + 2) (y â€“ 7)(y + 3)

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

Solution:

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

The constant in f(x) is + 42,

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

Let’s assume, y â€“ 2 = 0

y = 2

f(2) = 2(2)3 â€“ 5(2)2 â€“ 19(2) + 42

16 â€“ 20 â€“ 38 + 42 = 0

therefore, (y â€“ 2) is the factor of f(y)

Now, divide f(y) with (y â€“ 2) to get other factors

By using long division method we get,

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

Now,

2y2 â€“ y â€“ 21

The factors are (y + 3) (2y â€“ 7)

Hence, 2y3 â€“ 5y2 -19y + 42 = (y â€“ 2) (y + 3) (2y â€“ 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 Â± 1, Â± 2, Â± 4, Â± 5, Â± 10, Â± 20,

Let’s assume, x + 1 = 0

x = -1

f(-1) =  (âˆ’1)3+13(âˆ’1)2 + 32(âˆ’1) + 20

-1 + 13 â€“ 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 â€“ 3x2 â€“ 9x â€“ 5

Solution:

Given that, f(x) = x3 â€“ 3x2 â€“ 9x â€“ 5

The constant in f(x) is -5,

The factors of -5 are Â±1, Â±5,

Let’s assume, x + 1 = 0

x = -1

f(-1) = (âˆ’1)3 – 3(âˆ’1)2 – 9(-1) – 5

-1 â€“ 3 + 9 â€“ 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 â€“ 3x2 â€“ 9x â€“ 5 = (x + 1)( x2 â€“ 4x â€“ 5)

Now,

x2 â€“ 4x â€“ 5 = x2 â€“ 5x + x â€“ 5

x(x â€“ 5) + 1(x â€“ 5)

The factors are (x â€“ 5) and (x + 1)

Hence, x3 â€“ 3x2 â€“ 9x â€“ 5 = (x + 1)(x â€“ 5)(x + 1)

Question 16. Using factor theorem, factorize of the polynomials : 2y3 + y2 â€“ 2y â€“ 1

Solution:

Given that, f(y) = 2y3 + y2 â€“ 2y â€“ 1

The constant term is 2,

The factors of 2 are Â± 1, Â± 1/2,

Let’s assume, y â€“ 1= 0

y = 1

f(1) = 2(1)3 +(1)2 â€“ 2(1) â€“ 1

2 + 1 â€“ 2 â€“ 1 = 0

therefore, (y â€“ 1) is the factor of f(y)

Divide f(y) with (y â€“ 1) to get other factors

By using long division method we get,

2y3 + y2 â€“ 2y â€“ 1 = (y â€“ 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 â€“ 2y â€“ 1 = (y â€“ 1) (2y + 1) (y + 1)

Question 17. Using factor theorem, factorize of the polynomials : x3 â€“ 2x2 â€“ x + 2

Solution:

Given that, f(x) = x3 â€“ 2x2 â€“ x + 2

The constant term is 2,

The factors of 2 are Â±1, Â± 1/2,

Let’s assume, x â€“ 1= 0

x = 1

f(1) = (1)3 â€“ 2(1)2 â€“ (1) + 2

1 â€“ 2 â€“ 1 + 2 = 0

therefore, (x â€“ 1) is the factor of f(x)

Divide f(x) with (x â€“ 1) to get other factors

By using long division method we get,

x3 â€“ 2x2 â€“ y + 2 = (x â€“ 1) (x2 â€“ x â€“ 2)

Now,

x2 â€“ x â€“ 2 = x2 â€“ 2x + x  â€“ 2

x(x â€“ 2) + 1(x â€“ 2)

(x â€“ 2)(x + 1)  are the factors

Hence, x3 â€“ 2x2 â€“ y + 2 = (x â€“ 1)(x + 1)(x â€“ 2)

Question 18. Factorize each of the following polynomials :

1. x3 + 13x2 + 31x â€“ 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 â€“ 45

Given that, x + 9 is a factor

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

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

By using long division method we get,

x3 + 13x2 + 31x â€“ 45 = (x + 9)( x2 + 4x â€“ 5)

Now,

x2 + 4x â€“ 5 = x2 + 5x â€“ x  â€“ 5

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

(x + 5) (x â€“ 1) are the factors

Hence, x3 + 13x2 + 31x â€“ 45 = (x + 9)(x + 5)(x â€“ 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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