How do you factor $8x^3-4x^2-2x+1$ ?

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Answer 1 · 2016-03-26T08:45:18

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

Factor by grouping and using the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

with $a=2x$ and $b=1$ as follows:

$8x^3-4x^2-2x+1$

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

$=4x^2(2x-1)-1(2x-1)$

$=(4x^2-1)(2x-1)$

$=((2x)^2-1^2)(2x-1)$

$=(2x-1)(2x+1)(2x-1)$

How do you factor 8x^3-4x^2-2x+18x^3-4x^2-2x+1?