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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Updated on: 10-Oct-2022

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