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

1 Answer

Sep 27, 2016

$(8 x^{3} - 12 x + 2 x - 3 = (2 x - 3) (4 x^{2} + 1)$ $(8x^3-12x+2x-3=color(green)((2x-3)(4x^2+1))$

Explanation:

Notice that the ratio of the coefficients of the terms $8 x^{3}$ $8x^3$ and $2 x$ $2x$ : $8 : 2 = 4 : 1$ $8:2=4:1$
is the same as the ratio of the coefficients of the terms $- 12 x^{2}$ $-12x^2$ and $- 3$ $-3$ : $- 12 : - 3 = 4 : 1$ $-12:-3 =4:1$

This hints hat we should group the original expression as
$XXX (8 x^{3} + 2 x) - (12 x^{2} + 3)$ $color(white)("XXX")(color(red)(8x^3+2x))-(color(blue)(12x^2+3))$

$XXX = 2 x (4 x^{2} + 1)) - 3 (4 x^{2} + 1)$ $color(white)("XXX")=color(red)(2x(4x^2+1)))-color(blue)(3(4x^2+1))$

then extracting the common factor $(4 x^{2} + 1)$ $(4x^2+1)$
$XXX = (2 x - 3) (4 x^{2} + 1)$ $color(white)("XXX")=(2x-3)(4x^2+1)$

(Note that since $4 x^{2} \geq 0$ $4x^2 >= 0$ for $AAx in RR$ ,
$color(white)("XXX")4x^2+1$ can not be equal to $0$
$color(white)("XXX")$ and $(4x^2+1)$ has no Real factors.

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