How do you divide $\frac{x^{4} - 16}{x + 2}$ $(x^4 - 16) /( x + 2)$ ?

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How do you divide $\frac{x^{4} - 16}{x + 2}$ $(x^4 - 16) /( x + 2)$ ?

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Answer 1 · 2015-10-26T16:38:39

The answer is $(x + 2) (x^{2} + 4)$ $(x + 2)(x^2 + 4)$ = $x^{3} - 2 x^{2} + 4 x - 8$ $x^3 - 2x^2 +4x -8$

Explanation:

Factorise $(x^{4} - 16)$ $(x^4 - 16)$ as it is the difference of 2 squares
You then get
$\frac{(x^{2} - 4) (x^{2} + 4)}{x + 2}$ $((x^2 - 4)(x^2 + 4))/(x + 2)$

$x^{2} - 4$ $x^2 - 4$ is also the difference of 2 squares and will factorise to $(x - 2) (x + 2)$ $(x - 2)(x + 2)$
so this gives you $\frac{(x - 2) (x + 2) (x^{2} + 4)}{x + 2}$ $((x - 2)(x + 2)(x^2 +4))/(x + 2)$
The $(x + 2)$ $(x + 2)$ terms cancel out to give you
$(x - 2) (x^{2} + 4)$ $(x - 2)(x^2 + 4)$

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How do you divide $\frac{x^{4} - 16}{x + 2}$ $(x^4 - 16) /( x + 2)$ ?

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Explanation:

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How do you divide x4−16x+2(x^4 - 16) /( x + 2)?

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How do you divide $\frac{x^{4} - 16}{x + 2}$ $(x^4 - 16) /( x + 2)$ ?