Class 9 RD Sharma Solutions Chapter 6 Factorisation of Polynomials Exercise 6.5 | Set 2 (original) (raw)

Last Updated : 23 Jul, 2025

Factorisation is a key concept in algebra especially important for simplifying polynomial expressions. It involves breaking down a polynomial into a product of the simpler polynomials or factors. This process is crucial for solving polynomial equations and understanding polynomial functions in various mathematical contexts. Chapter 6 of RD Sharma’s Class 9 textbook delves into the factorisation of polynomials offering the exercises to practice these fundamental techniques. In Exercise 6.5 | Set 2 students are tasked with factorizing various polynomial expressions to reinforce their understanding and application of these concepts.

Factorization of Polynomials

The Factorisation of polynomials involves expressing a polynomial as a product of its factors. These factors can be constants, variables, or more complex polynomials. Key techniques include:

• Common Factor Method: Extracting the greatest common divisor (GCD) from the all terms of the polynomial.
• Grouping Method: Grouping terms with the common factors and then factoring out the common binomials.
• Special Products: Recognizing and applying formulas for the special polynomial products such as the difference of the squares perfect square trinomials and sum or difference of cubes.

**Question 11. Using factor theorem, factorize of the polynomials: x 3 – 10x 2 – 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, 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

**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, 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

**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, 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

**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, 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

**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, 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

**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, 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

**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, 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 – 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, x 3 + 13x 2 + 31x – 45 = (x + 9)(x + 5)(x – 1)

**2. 4x 3 + 20x 2 + 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) (2x2 + 7x + 6)

Now,

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

2x(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)

**Read More:

• Factorization of Polynomial
• What are Polynomials?

Conclusion

The Factorisation of polynomials is an essential skill in algebra that simplifies complex expressions and aids in solving the polynomial equations. By mastering techniques such as the common factor method, grouping and recognizing special products students can tackle the various polynomial problems with ease. The Practice exercises like those in RD Sharma's Chapter 6 Exercise 6.5 | Set 2 are crucial for the reinforcing these techniques and ensuring a solid understanding of the polynomial factorisation.