How do you factor $32 x^{3} - 4$ $32x^3 - 4$ ?

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Answer 1 · 2016-02-07T13:32:30

$4 (2 x - 1) (4 x^{2} + 2 x + 1)$ $4(2x-1)(4x^2+2x+1)$

the given : $32 x^{3} - 4$ $32x^3-4$

common monomial factor $= 4$ $=4$
factor out $4$ $4$ first

$4 (8 x^{3} - 1)$ $4(8x^3-1)$

use the "difference of two cubes " form

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$

for $8 x^{3} - 1$ $8x^3-1$

so that

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

final answer

$32 x^{3} - 4 = 4 \cdot (2 x - 1) (4 x^{2} + 2 x + 1)$ $32x^3-4=4*(2x-1)(4x^2+2x+1)$

How do you factor 32x^3 - 432x3−432x^3 - 4?