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
$f(x) = (x^3 - 8)$
By the Factor Theorem, $(x - 2)$ will be a factor of $f(x)$ if $f(2) = 0.$
Here, $f(2) = (2)^3 - 8$
$= 8 - 8 = 0$
$\therefore (x - 2)$ is a factor of $(x^3 - 8).$
By the Factor Theorem, $(x - 2)$ will be a factor of $f(x)$ if $f(2) = 0.$
Here, $f(2) = (2)^3 - 8$
$= 8 - 8 = 0$
$\therefore (x - 2)$ is a factor of $(x^3 - 8).$
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