How do you factor $x^{3} + 2 x^{2} - 8 x - 16$ $x^3+2x^2-8x-16$ ?

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How do you factor $x^{3} + 2 x^{2} - 8 x - 16$ $x^3+2x^2-8x-16$ ?

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Answer 1 · 2016-08-28T15:35:04

$x^{3} + 2 x^{2} - 8 x - 16 = (x - 2 \sqrt{2}) (x + 2 \sqrt{2}) (x + 2)$ $x^3+2x^2-8x-16=(x-2sqrt(2))(x+2sqrt(2))(x+2)$

Explanation:

Given:

$x^{3} + 2 x^{2} - 8 x - 16$ $x^3+2x^2-8x-16$

Notice that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this cubic will factor by grouping:

$x^{3} + 2 x^{2} - 8 x - 16$ $x^3+2x^2-8x-16$

$= (x^{3} + 2 x^{2}) - (8 x + 16)$ $=(x^3+2x^2)-(8x+16)$

$= x^{2} (x + 2) - 8 (x + 2)$ $=x^2(x+2)-8(x+2)$

$= (x^{2} - 8) (x + 2)$ $=(x^2-8)(x+2)$

$= (x^{2} - {(2 \sqrt{2})}^{2}) (x + 2)$ $=(x^2-(2sqrt(2))^2)(x+2)$

$= (x - 2 \sqrt{2}) (x + 2 \sqrt{2}) (x + 2)$ $=(x-2sqrt(2))(x+2sqrt(2))(x+2)$

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How do you factor $x^{3} + 2 x^{2} - 8 x - 16$ $x^3+2x^2-8x-16$ ?

1 Answer

Explanation:

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How do you factor $x^{3} + 2 x^{2} - 8 x - 16$ $x^3+2x^2-8x-16$ ?