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 $(-35) / 8$