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

### In each of the following using the remainder theorem, find the reminder 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

^{3}+4x

^{2}-3x+10, g(x) = x+4

**Solution:**

Given:f(x)=x

^{3}+4x^{2}-3x+10, g(x)=x+4from, 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

^{4}-3x

^{3}-2x

^{2}+x-7, g(x) =x-1

**Solution:**

Given:f(x)= 4x

^{4}-3x^{3}-2x^{2}+x-7, g(x)=x-1from, 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

^{4}-6x

^{3}+2x

^{2}-x+2, g(x)=x+2

**Solution:**

Given: f(x)=2x

^{4}-6x^{3}+2x^{2}-x+2, g(x)=x+2from, 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

^{3}-12x

^{2}+14x-3, g(x)=2x-1

**Solution:**

Given:f(x)=4x

^{3}-12x^{2}+14x-3, g(x)=2x-1from, 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)=x**^{3}-6x^{2}+2x-4, g(x)=1-2x

^{3}-6x

^{2}+2x-4, g(x)=1-2x

**Solution:**

Given:f(x)=x

^{3}-6x^{2}+2x-4, g(x)=1-2xfrom, 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)=x**^{4}-3x^{2}+4, g(x)=x-2

^{4}-3x

^{2}+4, g(x)=x-2

**Solution:**

Given:f(x)=x

^{4}-3x^{2}+4, g(x)=x-2from, 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

^{4}-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)=

^{3}-3x

^{2}+x-5, g(x)=

**Solution:**

Given:f(x)=9x

^{3}-3x^{2}+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 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 .

^{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)=2x

^{3}+ax^{2}+3x-5,p(x)=x^{3}+x^{2}-4x+aThe 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)=2

^{3}+2^{2}-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

^{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)=ax

^{3}+3x^{2}-3,p(x)=2x^{3}-5x+alet,

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

^{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)=ax

^{3}+3x^{2}-13,p(x)=2x^{3}-5x+aEquate 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,

^{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)=x

^{3}+3x^{2}+3x+1by reminder theorem

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

Substitute the value of x in f(x)

=

=

=

3. x = 0Substitute 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 =0x = -π

Substitute the value of x in f(x)

f(-π)=(-π)

^{3}+3(-π)^{2}+3(-π)+1=-π

^{3}+3π^{2}-3π +1

5. 5+2x =0x = -5/2

Substitute the value of x in f(x)

=

Taking L.C.M

=

=