Q. 14.1( 285 Votes )

# Find the remainder when is divided by

(i)

(ii)

(iii)

(iv)

(v)

Answer :

**Remainder Theorem:**If f(x) is a polynomial and it is divided by another polynomial g(x), the remainder of this division equals to the value f(a), where a is the solution of polynomial g(x) = 0

for example: if x

^{2}+ 2 is divided by x - 1, then to find remainder.

put x - 1 = 0

x = 1 and now putting this value in x

^{2}+ 1, we get 1 + 1 =2 as the remainder.

**(i)**f(x) = x

^{3}+ 3 x

^{2}+ 3 x + 1

Now let g(x) = x + 1

So for finding remainder, put g(x) = 0

x + 1 = 0

x = -1

So f(- 1) will be the remainder when f(x) is divided by g(x)

f (- 1) = (- 1)

^{3}+ 3 (- 1)

^{2}+ 3 (- 1) + 1

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

f (- 1) = 0

Hence, Remainder = 0

**(ii)**f(x) = x

^{3}+ 3 x

^{2}+ 3 x + 1

Now let g(x) = x - 1/2

So for finding remainder, put g(x) = 0

x - 1/2 = 0

x = 1/2

so, f(1/2) will be the remainder when f(x) is divided by g(x)

f(1/2) = (1/2)

^{3}+ 3 (1/2)

^{2}+ 3 (1/2) + 1

Hence, Remainder = 27/8

**(iii)**f(x) = x

^{3}+ 3 x

^{2}+ 3 x + 1

Now let g(x) = x

So for finding remainder, put g(x) = 0

x = 0

So, f(0) will be the remainder when f(x) is divided by g(x)

f(0) = (0)

^{3}+ 3 (0)

^{2}+ 3 (0) + 1

f(0) = 1

Hence, Remainder = 1

**(iv)**f(x) = x

^{3}+ 3 x

^{2}+ 3 x + 1

Now let g(x) = x + Π

So for finding remainder, put g(x) = 0

x + Π = 0

x = - Π

So, f(- Π) will be the remainder when f(x) is divided by g(x)

f(- Π) = (- Π)

^{3}+ 3 (- Π)

^{2}+ 3 (- Π) + 1

f(- Π) = - Π

^{3}+ 3 Π

^{2}- 3Π + 1

- Π

^{3}+ 3 Π

^{2}- 3Π + 1 will be the remainder

**(v)**f(x) = x

^{3}+ 3 x

^{2}+ 3 x + 1

Now let g(x) = 5 + 2x

So for finding the remainder, g(x) = 0

5 + 2x = 0

x = -5/2

So, f(-5/2) will be the remainder when f(x) is divided by g(x)

f(-5/2) = -27/8 will be the remainder .

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is divided by (x-a)

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