By remainder Theorem find the remainder, when | vedclass

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Question

We have $g(x)=2 x-1$

$\Rightarrow \quad 2 x-1=0$

$\Rightarrow \quad 2 x-1 \Rightarrow x=\frac{1}{2}$

Remainder $=p\left(\frac{1}{2}\right)$

Vedclass · Accepted Answer

$\frac{-3}{2}$

Vedclass · Answer

We have $g(x)=2 x-1$

$\Rightarrow \quad 2 x-1=0$

$\Rightarrow \quad 2 x-1 \Rightarrow x=\frac{1}{2}$

Remainder $=p\left(\frac{1}{2}\right)$

$=4\left(\frac{1}{2}\right)^{3}-12\left(\frac{1}{2}\right)^{2}+14\left(\frac{1}{2}\right)-3$

$=4\left(\frac{1}{8}\right)-12\left(\frac{1}{4}\right)+7-3$

$=\frac{1}{2}-3+7-6=\frac{1}{2}-2=\frac{-3}{2}$

By remainder Theorem find the remainder, when $p(x)$ is divided by $g(x),$ where

$p(x)=4 x^{3}-12 x^{2}+14 x-3, g(x)=2 x-1$

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