Question
Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when
$p(x)=x^3-8, g(x)=x-2$
$p(x)=x^3-8, g(x)=x-2$
✓
Answer
$f(x)=\left(x^3-8\right)$
By the Factor Theorem, $(x-2)$ will be a factor of $f(x)$ if $f(2)=0$.
Here, $f(2)=(2)^3-8$
$=8-8=0$
$\therefore(\mathrm{x}-2)$ is a factor of $\left(\mathrm{x}^3-8\right)$.
By the Factor Theorem, $(x-2)$ will be a factor of $f(x)$ if $f(2)=0$.
Here, $f(2)=(2)^3-8$
$=8-8=0$
$\therefore(\mathrm{x}-2)$ is a factor of $\left(\mathrm{x}^3-8\right)$.
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