Show that $(x + 2)$ is a factor of the polynomial $(x^{3} - 4 x^{2} - 2 x + 20)$

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Solution

Given Polynomial:
$x^{3} - 4 x^{2} - 2 x + 20$
Lets split the middle terms as,
$- 4 x^{2} = 2 x^{2} - 6 x^{2}, - 2 x = - 12 x + 10 x$
$\Rightarrow x^{3} - 4 x^{2} - 2 x + 20$
$= x^{3} + 2 x^{2} - 6 x^{2} - 12 x + 10 x + 20$
$= x^{2} (x + 2) - 6 x (x + 2) + 10 (x + 2)$
$= (x + 2) (x^{2} - 6 x + 10)$
$∴ x^{3} - 4 x^{2} - 2 x + 20 = (x + 2) (x^{2} - 6 x + 10)$
Hence, $(x + 2)$ is a factor of the polynomial $x^{3} - 4 x^{2} - 2 x + 20$
(OR)
Let $p (x) = x^{3} - 4 x^{2} - 2 x + 20$
By factor theorem, $(x + 2)$ is a factor of $p (x)$ if $p (- 2) = 0$ .
$∴$ It is sufficient to show that $(x + 2)$ is a factor of $p (x)$
Now, $p (x) = x^{3} - 4 x^{2} - 2 x + 20$
$∴ p (- 2) = (- 2)^{3} - 4 (- 2)^{2} - 2 (- 2) + 20 = - 8 - 16 + 4 + 20 = 0$
$∴ (x + 2)$ is a factor of $p (x) = x^{3} - 4 x^{2} - 2 x + 20$

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