Class 9 RD Sharma Solutions Chapter 6 Factorisation of Polynomials Exercise 6.4 | Set 1 (original) (raw)

Last Updated : 23 Jul, 2025

In Class 9 mathematics, the chapter on Factorisation of Polynomials is crucial for understanding algebraic expressions and their manipulation. Exercise 6.4 focuses on practical problems related to factorization enhancing students' skills in breaking down polynomials into simpler components. This exercise helps students grasp the concept of factorization and apply it to solve polynomial equations efficiently.

Factorization of Polynomials

The Factorisation of polynomials involves expressing a polynomial as a product of its factors. This process is essential for solving polynomial equations and simplifying the algebraic expressions. By breaking down a polynomial into its constituent factors one can solve equations find roots and simplify expressions more easily. Common techniques include factoring by the grouping using the special identities and applying the quadratic formula.

**In each of the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x), or not(Question 1-7):

Question 1. f(x) = x3 - 6x2 + 11x - 6, g(x) = x - 3

**Solution:

**Given: f(x) = x3 - 6x2 + 11x - 6, g(x) = x - 3

Here,

x - 3 = 0

x = 3

**To prove: g(x) is a factor of f(x), that is f(3) = 0

On substituting the value of x in f(x), we get

f(3) = 33 - 6 × 32 + 11 × 3 - 6

= 27 - (6 × 9) + 33 - 6

= 27 - 54 + 33 - 6

= 60 - 60

= 0

Since, the result is 0,

Hence, proved that g(x) is a factor of f(x).

Question 2. f(x) = 3x4 + 17x3 + 9x2 - 7x - 10, g(x) = x + 5

**Solution:

**Given: f(x) = 3x4 + 17x3 + 9x2 - 7x - 10, g(x) = x + 5

Here,

x + 5 = 0

x = -5

**To prove: g(x) is a factor of f(x), that is f(-5) = 0

On substituting the value of x in f(x), we get

f(-5) = 3(-5)4+ 17(-5)3+ 9(-5)2- 7(-5) - 10

= 3 × 625 + 17 × (-125) + 9 × 25 + 35 - 10

= 1875 - 2125 + 225 + 35 - 10

= 2135 - 2135

= 0

Since, the result is 0,

Hence, proved that g(x) is a factor of f(x).

Question 3. f(x) = x5 + 3x4 - x3 - 3x2 + 5x + 15, g(x) = x + 3

**Solution:

**Given: f(x) = x5 + 3x4 - x3 - 3x2 + 5x + 15, g(x) = x + 3

Here,

x + 3 = 0

x = -3

**To prove: g(x) is a factor of f(x), that is f(-3) = 0

On substituting the value of x in f(x), we get

f(-3) = (-3)5 + 3(-3)4 - (-3)3 - 3(-3)2 + 5(-3) + 15

= -243 + 243 + 27 - 27 - 15 + 15

= 0

Since, the result is 0,

Hence, proved that g(x) is a factor of f(x).

Question 4. f(x) = x3 - 6x2 - 19x + 84, g(x) = x - 7

**Solution:

**Given: f(x) = x3 - 6x2 - 19x + 84, g(x) = x - 7

Here,

x - 7 = 0

x = 7

**To prove: g(x) is a factor of f(x), that is f(7) = 0

On substituting the value of x in f(x), we get

f(7) = 73 - 6 × 72 - 19 × 7 + 84

= 343 - (6 × 49) - (19 × 7) + 84

= 343 - 294 - 133 + 84

= 427 - 427

= 0

Since, the result is 0,

Hence, proved that g(x) is a factor of f(x).

Question 5. f(x) = 3x3 + x2 - 20x + 12, g(x) = 3x - 2

**Solution:

**Given: f(x) = 3x3 + x2 - 20x + 12, g(x) = 3x - 2

Here,

3x - 2 = 0

x = 2/3

**To prove: g(x) is a factor of f(x), that is f(2/3) = 0

On substituting the value of x in f(x), we get

f(\frac23)=3(\frac23)^3+(\frac23)^2-20(\frac23)+12

= 3(\frac8{27})+\frac49-\frac{40}3+12

=\frac89+\frac49-\frac{40}3+12

= \frac{12}9-\frac{40}3+12

Taking L.C.M

= \frac{12-120+108}{9}

=\frac{120-120}{9}

= 0

Since, the result is 0,

Hence, proved that g(x) is a factor of f(x).

Question 6. f(x) = 2x3 - 9x2 + x + 13, g(x) = 3 - 2x

**Solution:

**Given: f(x) = 2x3 - 9x2 + x + 13, g(x) = 3 - 2x

Here,

3 - 2x = 0

x = 3/2

**To prove: g(x) is a factor of f(x), that is f(3/2) = 0

On substituting the value of x in f(x), we get

f(\frac32)=2(\frac32)^3-9(\frac32)^2+(\frac32)+13

=2(\frac{27}8)-9(\frac94)+\frac34+13

=\frac{27}4-\frac{81}4 +\frac{3}{2}+12

=\frac{21-81+6+48}{4}

= \frac{81-81}4

= 0

Since, the result is 0,

Hence, proved that g(x) is a factor of f(x).

Question 7. f(x) = x3 - 6x2 + 11x - 6, g(x) = x2 - 3x + 2

**Solution:

**Given: f(x) = x3 - 6x2 + 11x - 6, g(x) = x2 - 3x + 2

Here,

x2 - 3x + 2 = 0

On factorizing the above, we get

⇒ x2 - 3x + 2 = 0

⇒ x2 - 2x - x + 2 = 0

⇒ x(x - 2) - 1(x - 2)

⇒ (x - 1)(x - 2) are the factors

**To prove: g(x) is a factor of f(x), that is f(1) and f(2) should be 0

Let x = 1

On substituting the value of x in f(x), we get

f(1) = 13 - 6 × 12 + 11 × 1 - 6

= 1 - 6 + 11 - 6

= 12 - 12

= 0

Let x = 2

On substituting the value of x in f(x), we get

f(x) = 23 - 6 × 22 + 11 × 2 - 6

= 8 - (6 × 4) + 22 - 6

= 8 - 24 +22 - 6

= 30 - 30

= 0

Since, the results are 0 g(x) is the factor of f(x)

Question 8. Show that (x - 2), (x + 3) and (x - 4) are the factors of x3 - 3x2 - 10x + 24

**Solution:

**Given:

f(x) = x3 - 3x2 - 10x + 24

Factors given are (x - 2), (x + 3) and (x - 4)

**To prove: g(x) is a factor of f(x), that is f(2), f(-3), f(4) should be 0

Here, x - 2 = 0

Let, x = 2

On substituting the value of x in f(x), we get

f(2) = 23 - 3 × 22 - 10 × 2 + 24

= 8 - (3 × 4) - 20 + 24

= 8 - 12 - 20 + 24

= 32 - 32

= 0

Here, x + 3 = 0

Let, x = -3

On substituting the value of x in f(x), we get

f(-3) = (-3)3 - 3 × (-3)2 - 10 × (-3) + 24

= -27 -3(9) + 30 + 24

= -27- 27 + 30 + 24

= 54 - 54

= 0

Here, x - 4 = 0

Let, x = 4

On substituting the value of x in f(x), we get

f(4) = (4)3 - 3 × (4)2 - 10 × (4) + 24

= 64 - 3(16) - 40 + 24

= 64 - 48 - 40 + 24

= 84 - 84

= 0

Since, the results are 0, g(x) is the factor of f(x)

Question 9. Show that (x + 4), (x - 3) and (x - 7) are the factors of x3 - 6x2 - 19x + 84

**Solution:

**Given:

f(x) = x3 - 6x2- 19x + 84

Factors given are (x + 4), (x - 3) and (x - 7)

**To prove: g(x) is a factor of f(x), that is f(4), f(3), f(-7) should be 0

Here, x + 4 = 0

Let, x = -4

On substituting the value of x in f(x), we get

f(-4) = (-4)3 - 6(-4)2 - 19(-4) + 84

= -64 - (6 × 16) - 19 × (-4) + 84

= -64 - 96 + 76 + 84

= 160 - 160

= 0

Here, x - 3 = 0

Let, x = 3

On substituting the value of x in f(x), we get

f(3) = (3)3 - 6(3)2 - 19(3) + 84

= 27 - 6(9) - 19 × 3 + 84

= 27 - 54 - 57 + 84

= 111 - 111

= 0

Here, x - 7 = 0

Let, x = 7

On substituting the value of x in f(x), we get

f(7) = (7)3 - 6(7)2 - 19(7) + 84

= 343 - 6(49) - 19 × 7 + 84

= 343 - 294 - 133 + 84

= 427 - 427

= 0

Since, the results are 0, g(x) is the factor of f(x)

Question 10. For what value of a is (x - 5) a factor of x3 - 3x2 + ax - 10

**Solution:

Here, f(x) = x3 - 3x2 + ax - 10

By factor theorem

If (x - 5) is the factor of f(x) then, f(5) = 0

⇒ x - 5 = 0

⇒ x = 5

On substituting the value of x in f(x), we get

f(5) = 53 - 3 × 52 + a × 5 - 10

= 125 - (3 × 25) + 5a -10

= 125 - 75 + 5a - 10

= 5a + 40

Equate f(5) to zero

f(5) = 0

⇒ 5a + 40 = 0

⇒ 5a = -40

⇒ a = -40/5

⇒ a = -8

When a = -8, then (x - 5) will be factor of f(x)

Question 11. Find the value of such that (x - 4) is a factor of 5x3 - 7x2 - ax - 28

**Solution:

**Given: f(x) = 5x3- 7x2- ax - 28

Using factor theorem

(x - 4) is the factor of f(x), then f(4) = 0

⇒ x - 4 = 0

⇒ x = 4

On substituting the value of x in f(x), we get

f(4) = 5(4)3 - 7(4)2 - a × 4 - 28

= 5(64) - 7(16) - 4a - 28

= 320 - 112 - 4a - 28

= 180 - 4a

Equate f(4) to zero, to find a

f(4) = 0

⇒ 180 - 4a = 0

⇒ a = 180/4

⇒ a = 45

When a = 45, then (x - 4) will be factor of f(x)

Question 12. Find the value of a, if (x + 2) is a factor of 4x4 + 2x3 - 3x2 + 8x + 5a

**Solution:

**Given: f(x) = 4x4+ 2x3- 3x2+ 8x + 5a

Using factor theorem

If (x + 2) is the factor of f(x), then f(-2) should be zero

⇒ x + 2 =0

⇒ x = -2

On substituting the value of x in f(x), we get

f(-2) = 4(-2)4+ 2(-2)3- 3(-2)2+ 8(-2) + 5a

= 4(16) + 2 (-8) - 3(4) - 16 + 5a

= 64 - 16 - 12 -16 + 5a

= 5a + 20

Equate f(-2) to 0

f(-2) = 4(-2)2+ 2(-2)3- 3(-2)2+ 8(-2) + 5a

= 4(16) + 2(-8) - 3(4) - 16 + 5a

= 64 - 16 - 12 - 16 + 5a

= 5a + 20

Equate f(-2) to 0

f(-2) = 0

⇒ 5a + 20 = 0

⇒ 5a = - 20

⇒ a = -4

When a = -4, then (x + 2) is the factor of f(x)

**Read More:

• Factorization of Polynomial
• What are Polynomials?

Conclusion

Exercise 6.4 in RD Sharma’s Class 9 textbook provides the comprehensive practice on factorising polynomials using the various techniques. By solving these problems, students gain a deeper understanding of the polynomial factorisation and improve their algebraic manipulation skills. Mastery of these techniques is essential for the progressing in the higher algebra and mathematics.