Class 9 RD Sharma Solutions Chapter 6 Factorisation of Polynomials Exercise 6.3 (original) (raw)

Last Updated : 23 Jul, 2025

The Factorisation of polynomials is a key algebraic skill taught in Class 9. This chapter focuses on breaking down polynomials into simpler components that when multiplied together give the original polynomial. Understanding factorization is crucial for solving algebraic equations and simplifying expressions.

Factorisation of Polynomials

Factorisation involves expressing a polynomial as a product of its factors. This process simplifies polynomials and helps in solving algebraic problems more efficiently. In Exercise 6.3 of RD Sharma's Class 9 book students practice various techniques for factorizing polynomials including factoring by grouping using the identities and solving quadratic equations.

In each of the following using the remainder theorem, find the remainder when f(x) is divided by g(x)and verify by actual division:(1-8)

**Question 1. f(x) = x 3 +4x 2 -3x+10, g(x) = x+4

**Solution:

Given:f(x)=x3+4x2-3x+10, g(x)=x+4

from, the remainder theorem when f(x) is divided by g(x) =x-(-4) the remainder will be equal to f(-4).

Let, g(x)=0

⇒ x+4=0

⇒ x = -4

Substitute the value of x in f(x)

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

= -64+(4*16)+12+10

= -64 +64 +12+10

= 22

Therefore, the remainder is 22.

**Question 2. f(x)=4x 4 -3x 3 -2x 2 +x-7, g(x) =x-1

**Solution:

Given:f(x)= 4x4-3x3-2x2+x-7, g(x)=x-1

from, the remainder theorem when f(x) is divided by g(x) = x-(1) the remainder will be equal to f(1)

Let, g(x)=0

⇒ x-1=0

⇒ x=1

Substitute the value of x in f(x)

f(1)= 4(1)4-3(1)3-2(1)2+1-7

= 4-3-2+1-7

= 5-12

= -7

Therefore, the reminder is 7.

**Question 3. f(x)=2x 4 -6x 3 +2x 2 -x+2, g(x)=x+2

**Solution:

Given: f(x)=2x4-6x3+2x2-x+2, g(x)=x+2

from, the remainder theorem when f(x) is divided by g(x) = x-(-2) the remainder will be equal to f(-2)

Let, g(x)=0

⇒ x+2=0

⇒ x=-2

Substitute the value of x in f(x)

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

= (2*16)-(6*(-8))+(2*4)+2+2

= 32+48+8+2+2

= 92

Therefore, the reminder is 92.

**Question 4. f(x)=4x 3 -12x 2 +14x-3, g(x)=2x-1

**Solution:

Given:f(x)=4x3-12x2+14x-3, g(x)=2x-1

from, the remainder theorem when f(x) is divided by g(x) = 2(x-1/2) the remainder will be equal to f(1\2)

Let, g(x)=0

⇒ 2x-1=0

⇒ x=-1/2

Substitute the value of x in f(x)

f(\frac12)=4(\frac12)^3-12(\frac12)^2+14(\frac12)-3

=4(\frac18)-12(\frac14)+4(\frac12)-3

=\frac12 -3 +7-3

=\frac12+1

=\frac32

Therefore, the reminder is\frac32

**Question 5. f(x)=x 3 -6x 2 +2x-4, g(x)=1-2x

**Solution:

Given:f(x)=x3-6x2+2x-4, g(x)=1-2x

from, the remainder theorem when f(x) is divided by g(x) = -2(x-1/2) the remainder will be equal to f(1\2)

Let, g(x)=0

⇒ 1-2x=0

⇒ x=1/2

substitute the value of x in f(x)

f(\frac12)=(\frac12)^3-6(\frac12)^2+2(\frac12)-4

=\frac18-8(\frac14)+2(\frac12)-4

=\frac18-(\frac12)+1-4

=\frac18-\frac12-3

Taking L.C.M

=\frac{1-4+8-32}{8}

=\frac{1-36}8

=\frac{-35}{8}

Therefore, the remainder is\frac{-35}{8}

**Question 6. f(x)=x 4 -3x 2 +4, g(x)=x-2

**Solution:

Given:f(x)=x4-3x2+4, g(x)=x-2

from, the remainder theorem when f(x) is divided by g(x) = x-(2) the remainder will be equal to f(2)

Let, g(x)=0

⇒ x-2=0

⇒ x=2

Substitute the value of x in f(x)

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

= 16-3(4) + 4

= 16 - 12 + 4

= 20 - 12

= 8

Therefore, the remainder is 8

**Question 7. f(x)=9x 3 -3x 2 **+x-5, g(x)=**x-\frac23

**Solution:

Given:f(x)=9x3-3x2+x-5, g(x)=x-\frac23

from, the remainder theorem when f(x) is divided by g(x) = x-(\frac23 ) the remainder will be equal to f(\frac23 )

Let, g(x)=0

⇒ x-2/3=0

⇒ x=2/3

substitute the value of x in f(x)

f(\frac23)=9(\frac23)^3-3(\frac23)^2+(\frac23)-5

=9(\frac{8}{27})-3(\frac49)+\frac23-5

=\frac83-\frac43+\frac23-5

=\frac{10-19}{3}

= -3

Therefore, the remainder is -3

****Question 8. f(x) =**3x^4+2x^3-\frac{x^3}{3}-\frac{x}{9}+\frac{2}{27} ****, g(x) =**x+\frac23

**Solution:

Given:f(x)=3x^4+2x^3-\frac{x^3}{3}-\frac{x}{9}+\frac{2}{27} ,g(x)=x+\frac23

from, the remainder theorem when f(x) is divided by g(x) = x-(-\frac23) the remainder will be equal to f(-\frac23 )

substitute the value of x in f(x)

f(-\frac23)=3(-\frac23)^4+2(-\frac23)^3- \frac{(-\frac23)^3}{3}-\frac{-\frac{2}{3}}{9}+ \frac{2}{27}

=\frac{16}{27} -\frac{16}{27}-\frac{4}{27}+\frac{2}{27}+\frac{2}{27}

=\frac{4}{27}-\frac{4}{27}

= 0

Therefore, the remainder is 0

**Question 9. If the polynomial2x 3 +ax 2 +3x-5 andx 3 +x 2 -4x+a leave the same reminder when divided by x-2, Find the value of a .

**Solution:

Given:f(x)=2x3+ax2+3x-5,p(x)=x3+x2-4x+a

The remainder are f(2) and p(2) when f(x) and p(x) are divided by x-2

We know that,

f(2) = p(2) (given in problem)

we need to calculate f(2) and p(2)

for, f(2)

substitute (x=2) in f(x)

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

= 16+4a+1

= 4a+17 ---------- 1

for, p(2)

Substitute (x=2) in p(x)

p(2)=23+22-4(2)+a

= 8+4-8+a

= 4+a ----------- 2

Since, f(2) = p(2)

Equate eq1 and eq2

⇒ 4a+17 = 4+a

⇒ 3a = -13

⇒ a = -13/3

The value of a = -13/3

**Question 10. If the polynomialsax 3 +3x 2 -3 and2x 3 -5x+a when divided by (x-4) leave the reminders as R1 and R2 respectively. Find the values of a in each of the following cases, if

**1. R1 = R2

**2. R1+R2=0

**3. 2R1-R2=0

**Solution:

The polynomials are f(x)=ax3+3x2-3,p(x)=2x3-5x+a

let,

R1 is the reminder when f(x) is divided by x-4

⇒ R1=f(4)

⇒ R1=a(4)3 + 3(4)2 -3

= 64a + 48 - 3

= 64a + 45 ----------------- 1

Now, let

R2 is the reminder when p(x) is divided by x-4

⇒ R2=p(4)

⇒ R2=2(4)3-5(4)+a

= 128-20+a

= 108 +a --------------------- 2

1. Given, R1 = R2

⇒ 64a + 45 = 108 +a

⇒ 63a=63

⇒ a =1

2. Given, R1+R2 =0

⇒ 64a + 45 + 108 +a = 0

⇒ 65a + 153 = 0

⇒ a = -153/65

3. Given, 2R1-R2 =0

⇒2( 64a + 45)- (108 +a) =0

⇒ 128a + 90 - 108 -a =0

⇒ 127a - 18 =0

⇒ a =\frac{18}{127}

**Question 11. If the polynomialsax 3 +3x 2 -13 and2x 3 -5x+a when divided by (x-2) leave the same reminder, find the value of a.

**Solution:

Given:f(x)=ax3+3x2-13,p(x)=2x3-5x+a

Equate x-2 to zero

⇒ x=2

Substitute the value of x in f(x) and p(x)

f(2)=a(2)3+3(2)2-13

= 8a+12-13

= 8a-1 -------------- 1

p(2)=2(2)3-5(2)+a

= 16-10+a

= 6 + a ------------- 2

f(2) = p(2)

⇒ 8a-1 = 6+a

⇒ 7a = 7

⇒ a =1

The value of a is 1

**Question 12. Find the reminder whenf(x)=(x) 3 +3(x) 2 +3(x)+1 is divided by,

**1. x+1

**2. x - 1/2

**3. x

**4. x+π

**5. 5+2x

**Solution:

Given:f(x)=x3+3x2+3x+1

by reminder theorem

**1. **x+1 = 0

x=-1

Substitute the value of x in f(x)

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

= -1+3-3+1

=0

**2. **x-1/2 =0

x = 1/2

Substitute the value of x in f(x)

f(\frac12)=(\frac12)^3+3(\frac12)^2+3(\frac12)+1

=\frac18+3(\frac12)^2+3(\frac12)+1

=\frac{1+6+12+8}{8}

=\frac{27}8

**3. x = 0

Substitute the value of x in f(x)

f(0)=(0)3+3(0)2+3(0)+1

= 0 + 0+0+1

= 1

**4. x+π =0

x = -π

Substitute the value of x in f(x)

f(-π)=(-π)3+3(-π)2+3(-π)+1

=-π3+3π2-3π +1

**5. 5+2x =0

x = -5/2

Substitute the value of x in f(x)

f(-\frac52 )=(-\frac52 )^3+3(-\frac52 )^2+3(-\frac52 )+1

=\frac{-125}{8}+3\frac{25}{4}+3\frac{-5}{2}+1

Taking L.C.M

=\frac{-125+150-50+8}{8}

=\frac{-27}{8}

**Read More:

• Factorization of Polynomial
• What are Polynomials?

Conclusion

The Factorisation of polynomials is a fundamental algebraic technique that simplifies expressions and aids in solving the equations. By mastering the techniques in Exercise 6.3 of RD Sharma's Class 9 book students develop a strong foundation in the algebra that will support their studies in the higher mathematics.