Class 9 RD Sharma Solutions - Chapter 6 Factorisation of Polynomials- Exercise 6.5 | Set 1

In Class 9 mathematics, understanding the factorization of polynomials is crucial as it forms the foundation for the more advanced algebraic concepts. Chapter 6 of RD Sharma's textbook focuses on the various methods of factorizing polynomials helping the students simplify complex expressions and solve equations effectively. Exercise 6.5 in this chapter deals with the specific problems designed to reinforce the student's grasp of the factorization techniques.

Factorization of Polynomials

The Factorisation of the polynomials involves expressing a polynomial as a product of its factors. This process is essential in solving polynomial equations and simplifying expressions. There are several methods of factorization including:

• Common Factor Method: The Factorising by taking out the greatest common divisor from the terms.
• Grouping Method: Grouping terms in pairs and factoring out common factors.
• Factorising Quadratic Polynomials: Using techniques like splitting the middle term or completing the square.

Question 1. Using factor theorem, factorize of the polynomials: x³ + 6x² + 11x + 6

Solution:

Given that, polynomial eqn., f(x) = x³ + 6x² + 11x + 6

The constant term in f(x) is 6,

The factors of 6 are ± 1, ± 2, ± 3, ± 6

Let, x + 1 = 0

x = -1

Substitute the value of x in f(x) and we get,

f(-1) = (−1)³ + 6(−1)² + 11(−1) + 6

= - 1 + 6 - 11 + 6 = 12 – 12 = 0

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

Similarly, (x + 2) and (x + 3) are also the factors of f(x)

Since, f(x) is a polynomial having a degree 3, therefore it cannot have more than three linear factors.

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

x³ + 6x² + 11x + 6 = k(x + 1)(x + 2)(x + 3)

Substitute x = 0 on both the sides

0 + 0 + 0 + 6 = k(0 +1)(0 + 2)(0 + 3)

6 = k(1*2*3)

6 = 6k

k = 1

Substitute k value in f(x) = k(x + 1)(x + 2)(x + 3)

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

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

Hence, x³ + 6x² + 11x + 6 = (x + 1)(x + 2)(x + 3)

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

Solution:

Given that, f(x) = x³+ 2x² – x – 2

The constant term in f(x) is -2,

The factors of (-2) are ±1, ± 2,

Let, x – 1 = 0

x = 1

Substitute the value of x in f(x)

f(1) = (1)³ + 2(1)² – 1 – 2

1 + 2 – 1 – 2 = 0

Similarly, the other factors (x + 1) and (x + 2) of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

therefore, f(x) = k(x – 1)(x + 2)(x + 1 )

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

Substitute x = 0 on both the sides

0 + 0 – 0 – 2 = k(-1)(1)(2)

– 2 = - 2k

k = 1

Substitute k value in f(x) = k(x – 1)(x + 2)(x + 1)

f(x) = (1)(x – 1)(x + 2)(x + 1)

f(x) =  (x – 1)(x + 2)(x + 1)

therefore, x³ + 2x² – x – 2 = (x – 1)(x + 2)(x + 1)

Question 3. Using factor theorem, factorize of the polynomials : x³ – 6x² + 3x + 10

Solution:

Given that, f(x) = x³ – 6x² + 3x + 10

The constant term in f(x) is 10,

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

Let, x + 1 = 0

x = -1

Substitute the value of x in f(x)

f(-1) = (−1)³- 6(−1)² + 3(−1) + 10

-1 – 6 – 3 + 10 = 0

Similarly, the other factors (x – 2) and (x – 5) of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

therefore, f(x) = k(x + 1)(x – 2)(x – 5)

Substitute x = 0 on both sides

x³– 6x² + 3x + 10 = k(x + 1)(x – 2)(x – 5)

0 – 0 + 0 + 10 = k(1)(-2)(-5)

10 = k(10)

k = 1

Substitute k = 1 in f(x) = k(x + 1)(x – 2)(x – 5)

f(x) = (1)(x + 1)(x – 2)(x – 5 )

therefore, x³ – 6x² + 3x + 10 = (x + 1)(x – 2)(x – 5)

Question 4. Using factor theorem, factorize of the polynomials : x⁴ –7x³ + 9x² + 7x –10

Solution:

Given that, f(x) = x⁴–7x³ + 9x² + 7x – 10

The constant term in f(x) is 10,

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

Let, x – 1 = 0

x = 1

Substitute the value of x in f(x)

f(x) = 14 – 7(1)³ + 9(1)² + 7(1) – 10

1 – 7 + 9 + 7 – 10

10 – 10 = 0

(x – 1) is the factor of f(x)

Similarly, the other factors are (x + 1), (x – 2), (x – 5)

Since, f(x) is a polynomial of degree 4. So, it cannot have more than four linear factor.

therefore, f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)

x⁴ –7x³ + 9x² + 7x – 10 = k(x – 1)(x + 1)(x – 2)(x – 5)

Put x = 0 on both sides

0 – 0 + 0 – 10 = k(-1)(1)(-2)(-5)

– 10 = k(-10)

k = 1

Substitute k = 1 in f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)

f(x) = (1)(x – 1)(x + 1)(x – 2)(x – 5)

(x – 1)(x + 1)(x – 2)(x – 5)

therefore, x⁴ – 7x³ + 9x² + 7x – 10 = (x – 1)(x + 1)(x – 2)(x – 5)

Question 5. Using factor theorem, factorize of the polynomials : x⁴ – 2x³ – 7x² + 8x + 12

Solution:

Given that,

f(x) = x⁴ – 2x³ –7x² + 8x + 12

The constant term f(x) is equal is 12,

The factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12,

Let, x + 1 = 0

x = -1

Substitute the value of x in f(x)

f(-1) = (−1)4 – 2(−1)3–7(−1)2 + 8(−1)+12

1 + 2 – 7 – 8 + 12 = 0

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

Similarly, (x + 2), (x – 2), (x – 3) are also the factors of f(x)

Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors.

f(x) = k(x + 1)(x + 2)(x – 3)(x – 2)

x⁴ – 2x³ –7x² + 8x + 12 = k(x + 1)(x + 2)(x – 3)(x – 2)

Substitute x = 0 on both sides,

0 – 0 – 0 + 12 = k(1)(2)(- 2)(- 3)

12 = 12K

k = 1

Substitute k = 1 in f(x) = k(x – 2)(x + 1)(x + 2)(x – 3)

f(x) = (x – 2)(x + 1)(x + 2)(x – 3)

Hence, x⁴ – 2x³ – 7x² + 8x + 12 = (x – 2)(x + 1)(x + 2)(x – 3)

Question 6. Using factor theorem, factorize of the polynomials : x⁴ + 10x³ + 35x² + 50x + 24

Solution:

Given that, f(x) = x⁴ + 10x³ + 35x² + 50x + 24

The constant term in f(x) is equal to 24,

The factors of 24 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24,

Let, x + 1 = 0

x = -1

Substitute the value of x in f(x)

f(-1) = (-1)⁴ + 10(-1)³ + 35(-1)² + 50(-1) + 24

1-10 + 35 - 50 + 24 = 0

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

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

Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors.

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

x⁴ + 10x³ + 35x² + 50x + 24 = k(x + 1)(x + 2)(x + 3)(x + 4)

Substitute x = 0 on both sides

0 + 0 + 0 + 0 + 24 = k(1)(2)(3)(4)

24 = k(24)

k = 1

Substitute k = 1 in f(x) = k(x + 1)(x + 2)(x + 3)(x + 4)

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

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

Hence, x⁴ + 10x³ + 35x² + 50x + 24 = (x + 1)(x + 2)(x + 3)(x + 4)

Question 7. Using factor theorem, factorize of the polynomials : 2x⁴–7x³–13x² + 63x – 45

Solution:

Given that, f(x) = 2x⁴–7x³–13x² + 63x – 45

The factors of constant term - 45 are ± 1, ± 3, ± 5, ± 9, ± 15, ± 45,

The factors of the coefficient of x4 is 2. 

Hence possible rational roots of f(x) are ± 1, ± 3, ± 5, ± 9, ± 15, ± 45, ± 1/2, ± 3/2, ± 5/2, ± 9/2, ± 15/2, ± 45/2

Let, x – 1 = 0

x = 1

f(1) = 2(1)⁴ – 7(1)³ – 13(1)² + 63(1) – 45

2 – 7 – 13 + 63 – 45 = 0

Let, x – 3 = 0

x = 3

f(3) = 2(3)⁴ – 7(3)³ – 13(3)² + 63(3)  - 45

162 – 189 – 117 + 189 – 45 = 0

therefore, (x – 1) and (x – 3) are the roots of f(x)

x2 – 4x + 3 is the factor of f(x)

Divide f(x) with x2 – 4x + 3 to get other three factors,

By using long division we get,

2x⁴ – 7x³ – 13x² + 63x – 45 = (x²– 4x + 3) (2x²+ x – 15)

2x⁴ - 7x³– 13x² + 63x – 45 = (x – 1) (x – 3) (2x²+ x – 15)

Now,

2x² + x – 15 = 2x² + 6x – 5x –15

2x(x + 3) – 5 (x + 3)

(2x – 5) (x + 3)

Hence, 2x⁴ – 7x³ – 13x² + 63x – 45 = (x – 1)(x – 3)(x + 3)(2x – 5)

Question 8. Using factor theorem, factorize of the polynomials : 3x³ - x² – 3x + 1

Solution:

Given that, f(x) = 3x³ - x² – 3x + 1

The factors of constant term 1 is ± 1,

The factors of the coefficient of x2 = 3,

The possible rational roots are ±1, 1/3,

Let, x – 1 = 0

x = 1

f(1) = 3(1)³ - (1)² - 3(1) + 1

3 – 1 – 3 + 1 = 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,

3x³- x² – 3x + 1 = (x – 1)( 3x² + 2x – 1)

Now,

3x² + 2x -1 = 3x² + 3x – x - 1

3x(x + 1) -1(x + 1)

(3x – 1)(x + 1)

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

Question 9. Using factor theorem, factorize of the polynomials : x³- 23x² + 142x - 120

Solution:

Given that, f(x) = x³- 23x² + 142x - 120

The constant term in f(x) is -120,

The factors of -120 are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 8, ± 10, ± 12, ± 15, ± 20, ± 24, ± 30, ± 40, ± 60, ± 120,

Let, x - 1 = 0

x = 1

f(1) = (1)³- 23(1)² + 142(1) - 120

1 - 23 + 142 - 120 = 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 we get,

x³ – 23x² + 142x – 120 = (x  – 1) (x² – 22x + 120)

Now,

x² – 22x + 120 = x² – 10x – 12x + 120

x(x – 10) – 12(x – 10)

(x – 10) (x – 12)

Hence, x³ – 23x² + 142x – 120 = (x – 1) (x – 10) (x – 12)

Question 10. Using factor theorem, factorize of the polynomials : y³ – 7y + 6

Solution:

Given that, f(y) = y³ – 7y + 6

The constant term in f(y) is 6,

The factors are ± 1, ± 2, ± 3, ± 6,

Let, y – 1 = 0

y = 1

f(1) = (1)3 – 7(1) + 6

1 – 7 + 6 = 0

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

Similarly, (y – 2) and (y + 3) are also the factors

Since, f(y) is a polynomial which has degree 3, it cannot have more than 3 linear factors

f(y) = k(y – 1)(y – 2)(y + 3)

y³ – 7y + 6 = k(y – 1)( y – 2)(y + 3)      -----------------(i)

Substitute k = 0 in eqn. 1

0 – 0 + 6 = k(-1)(-2)(3)

6 = 6k

k = 1

y³ – 7y + 6 = (1)(y – 1)( y – 2)(y + 3)

y³ – 7y + 6 = (y – 1)( y – 2)(y + 3)

Hence, y³–7y + 6 = (y – 1)( y – 2)(y + 3)

Read More:

• Factorization of Polynomial
• What are Polynomials?

Conclusion

The Factorisation of polynomials is an essential algebraic skill that simplifies complex expressions and solves polynomial equations. Exercise 6.5 | Set 1 from RD Sharma's Class 9 textbook offers practice in applying the various factorisation techniques reinforcing the student's understanding and problem-solving abilities. Mastery of these techniques not only aids in the academic success but also builds a strong foundation for the future mathematical concepts.