How do you factor completely $8w^3 - 4w^2 - 60w$ ?

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Answer 1 · 2016-07-31T08:28:01

$4w(2w+5)(w-3)$

Always look for a common factor first. In this case it is $4w$

$8w^3 - 4w^2 - 60w = 4w(2w^2 -w -15)$

Now, find factors of 2 and 15 which subtract to give 1.
There must be more negatives.

find the cross products and subtract.
$" 2 5 "rArr 1xx5 = 5$
$" 1 3 "rArr2 xx3 = 6" 6-5 = 1"$

But we need $-1," 6 must be negative and 5 must be positive"$

= $4w(2w+5)(w-3)$

How do you factor completely 8w^3 - 4w^2 - 60w8w^3 - 4w^2 - 60w?