Question
Using factor theorem, show that $g(x)$ is a factor of $p(x)$, when
$p(x)=2 x^3+7 x^2-24 x-45, g(x)=x-3$
$p(x)=2 x^3+7 x^2-24 x-45, g(x)=x-3$
✓
Answer
$f(x)=\left(2 x^3+7 x^2-24 x-45\right)$
By the Factor Theorem, $(x-3)$ will be a factor of $f(x)$ if $f(3)=0$.
Here, $f(3)=2 \times 3^3+7 \times 3^2-24 \times 3-45$
$=54+63-72-45$
$=117-117=0$
$\therefore(x-3)$ is a factor of $\left(2 x^3+7 x^2-24 x-45\right)$.
By the Factor Theorem, $(x-3)$ will be a factor of $f(x)$ if $f(3)=0$.
Here, $f(3)=2 \times 3^3+7 \times 3^2-24 \times 3-45$
$=54+63-72-45$
$=117-117=0$
$\therefore(x-3)$ is a factor of $\left(2 x^3+7 x^2-24 x-45\right)$.
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