How do you factor $2 x^{3} + 3 x^{2} - 8 x - 12$ $2x^3 + 3x^2 - 8x-12$ ?

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How do you factor $2 x^{3} + 3 x^{2} - 8 x - 12$ $2x^3 + 3x^2 - 8x-12$ ?

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Answer 1 · 2016-06-29T02:55:48

$(x - 2) (2 x^{2} + 7 x + 6)$ $(x - 2)(2x^2 + 7x + 6)$

Explanation:

$y = 2 x^{3} + 3 x^{2} - 8 x - 12$ $y = 2x^3 + 3x^2 - 8x - 12$ .
Note that f(2) = 0, then, one factor is (x - 2).
After division, we get:
$y = (x - 2) (2 x^{3} + 7 x + 6)$ $y = (x - 2)(2x^3 + 7x + 6)$
The trinomial in parentheses can't be factored because its D < 0.
Therefor,
$y = (x - 2) (2 x^{3} + 7 x + 6)$ $y = (x - 2)(2x^3 + 7x + 6)$

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How do you factor $2 x^{3} + 3 x^{2} - 8 x - 12$ $2x^3 + 3x^2 - 8x-12$ ?

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How do you factor $2 x^{3} + 3 x^{2} - 8 x - 12$ $2x^3 + 3x^2 - 8x-12$ ?