# Using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division

Question:

Using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division

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

Solution:

Here, $f(x)=x^{3}-6 x^{2}+2 x-4$

g(x) = 1 - 2x

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

Let, g(x) = 0

⟹ 1 - 2x = 0

⟹ -2x = -1

⟹ 2x = 1

⟹ x = 1/2

Substitute the value of x in f(x)

$f(1 / 2)=(1 / 2)^{3}-6(1 / 2)^{2}+2(1 / 2)-4$

= 1/8 - 8(1/4) + 2(1/2) - 4

= 1/8 - (1/2) + 1 - 4

= 1/8 - (1/2) - 3

Taking L.C.M

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

$=\frac{1-36}{8}$

$=\frac{-35}{8}$

Therefore, the remainder is $(-35) / 8$