# Class 9 RD Sharma Solutions – Chapter 6 Factorisation of Polynomials- Exercise 6.3

• Last Updated : 11 Feb, 2021

### Question 1. f(x) = x3+4x2-3x+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)=4x4-3x3-2x2+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)=2x4-6x3+2x2-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)=4x3-12x2+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)

=

=

=

=

Therefore, the reminder is

### Question 5. f(x)=x3-6x2+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)

=

=

=

Taking L.C.M

=

=

=

Therefore, the remainder is

### Question 6. f(x)=x4-3x2+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)=9x3-3x2+x-5, g(x)=

Solution:

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

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

Let, g(x)=0

⇒ x-2/3=0

⇒ x=2/3

substitute the value of x in f(x)

=

=

=

= -3

Therefore, the remainder is -3

### Question 8. f(x) =, g(x) =

Solution:

Given:,

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

substitute the value of x in f(x)

=

=

= 0

Therefore, the remainder is 0

### Question 9. If the polynomial2x3+ax2+3x-5 andx3+x2-4x+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 polynomialsax3+3x2-3 and2x3-5x+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 =

### Question 11. If the polynomialsax3+3x2-13 and2x3-5x+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)

=

=

=

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)

=

Taking L.C.M

=

=

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