How do you factor completely $16x^3-32x^2-25x+50$ ?

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Answer 1 · 2016-04-26T23:30:32

$16x^3-32x^2-25x+50=(4x-5)(4x+5)(x-2)$

Factor by grouping, then using the difference of squares identity:

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

with $a=4x$ and $b=5$ as follows:

$16x^3-32x^2-25x+50$

$=(16x^3-32x^2)-(25x-50)$

$=16x^2(x-2)-25(x-2)$

$=(16x^2-25)(x-2)$

$=((4x)^2-5^2)(x-2)$

$=(4x-5)(4x+5)(x-2)$

How do you factor completely 16x^3-32x^2-25x+5016x^3-32x^2-25x+50?