Question
For what value of m is $x^3-2 m x^2+16$ divisible by $x + 2$ ?

Answer

Let $p(x)=x^3-2 m x^2+16$
Since, $p(x)$ is divisible by $(x+2)$, then remainder $=0$
$P(-2)=0$
$\Rightarrow(-2)^3-2 \mathrm{~m}(-2)^2+16=0$
$\Rightarrow-8-8 \mathrm{~m}+16=0$
$\Rightarrow 8=8 \mathrm{~m}$
$\mathrm{~m}=1$
Hence, the value of $m$ is $1 .$

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