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Find the remainder using remainder theorem find the remainder when $x^3-6x^2+2x-4$ is divided by $1-\frac{3x}{2}$.
Given: $x^3-6x^2+2x-4$ is divided by $1-\frac{3x}{2}$.
To do: To find the remainder using remainder theorem find the remainder when $x^3-6x^2+2x-4$ is divided by $1-\frac{3x}{2}$.
Solution:
Let $f( x)=x^3-6x^2+2x-4$ and $g( x)=1-\frac{3x}{2}$
Let $g( x)=1-\frac{3x}{2}=0$
$\Rightarrow \frac{3x}{2}=1$
$\Rightarrow 3x=2$
$\Rightarrow x=\frac{2}{3}$, on substituting this value in $f( x)$.
$f( \frac{2}{3})=( \frac{2}{3})^3-6( \frac{2}{3})^2+2( \frac{2}{3})-4$
$\Rightarrow f( \frac{2}{3})=( \frac{2\times2\times2}{3\times3\times3})-6( \frac{2\times2}{3\times3})+2( \frac{2}{3})-4$
$f( \frac{2}{3})=( \frac{8}{27})-6( \frac{4}{9})+2( \frac{2}{3})-4$
$f( \frac{2}{3})=( \frac{8}{27})-( \frac{24}{9})+( \frac{4}{3})-4$
$f( \frac{2}{3})=( \frac{8}{27})-( \frac{24\times3}{9\times3})+( \frac{4\times9}{3\times9})-4\times\frac{27}{27}$
$f( \frac{2}{3})=( \frac{8}{27})-( \frac{72}{27})+( \frac{36}{27})-\frac{108}{27}$
$f( \frac{2}{3})=( \frac{8-72+36-108}{27})$
$f( \frac{2}{3})=( \frac{44-180}{27})$
$f( \frac{2}{3})=( \frac{-136}{27})$
Thus, the remainder is $( \frac{-136}{27})$.
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