How do you factor $2 x^{3} + 3 x^{2} - 16 x - 24$ $2x^3+3x^2-16x-24$ ?

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How do you factor $2 x^{3} + 3 x^{2} - 16 x - 24$ $2x^3+3x^2-16x-24$ ?

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Answer 1 · 2015-05-25T13:05:59

Given $2 x^{3} + 3 x^{2} - 16 x - 24$ $2x^3+3x^2-16x-24$

Regroup as
$2 x^{3} - 16 x + 3 x^{2} - 24$ $color(red)(2x^3-16x) + color(blue)(3x^2-24)$

Extract factor from each pair
$= (2 x) (x^{2} - 8) + (3) (x^{2} - 8)$ $= color(red)((2x)(x^2-8)) + color(blue)((3)(x^2-8)$

Extract the $(x^{2} - 8)$ $(x^2-8)$ common factor
$= (2 x + 3) (x^{2} - 8)$ $=(2x+3)(x^2-8)$

If you are willing to employ irrational constants, this can be further factored (using the difference of squares) as
$= (2 x - 3) (x + 2 \sqrt{2}) (x - 2 \sqrt{2})$ $=(2x-3)(x+2sqrt(2))(x-2sqrt(2))$

Answer 2 · 2015-05-25T13:29:56

Notice that the ratio between the 1st and 2nd term is the same as that between the 3rd and 4th term. So grouping will work...

$2 x^{3} + 3 x^{2} - 16 x - 24 = (2 x^{3} + 3 x^{2}) - (16 x + 24)$ $2x^3+3x^2-16x-24 = (2x^3+3x^2)-(16x+24)$

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

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

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

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

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How do you factor $2 x^{3} + 3 x^{2} - 16 x - 24$ $2x^3+3x^2-16x-24$ ?

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