Find the remainder when x^{3}+3 x^{2}+3 | vedclass

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Question

$\because$ The zero of $x-\frac{1}{2}$ is $\frac{1}{2}$                    $\left[\because x -\

Vedclass · Accepted Answer

$\frac{27}{8}$

Vedclass · Answer

$\because$ The zero of $x-\frac{1}{2}$ is $\frac{1}{2}$                    $\left[\because x -\frac{1}{2}=0 \Rightarrow x =\frac{1}{2}ight]$

and  $p ( x )= x ^{3}+3 x ^{2}+3 x +1$

$	herefore $  For divisor $= x -\frac{1}{2},$ remainder is given as

$p \left(\frac{1}{2}ight)=\left(\frac{1}{2}ight)^{3}+3\left(\frac{1}{2}ight)^{2}+3\left(\frac{1}{2}ight)+1=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$

$=\frac{1+6+12+8}{8}=\frac{27}{8}$

Thus, the required remainder $=\frac{27}{8}$.

Find the remainder when $x^{3}+3 x^{2}+3 x+1$ is divided by $x-\frac{1}{2}$

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