How do you factor #16c ^ { 3} + 52c ^ { 2} - 48c#?

1 Answer
George C.
Dec 31, 2016

#16c^3+52c^2-48c = 4c(4c-3)(c+4)#

Explanation:

Given:

#16c^3+52c^2-48c#

First note that all of the terms are divisible by #4c#, so separate that out as a factor first:

#16c^3+52c^2-48c = 4c(4c^2+13c-12)#

To factor the remaining quadratic, use an AC method:

Find a pair of factors of #AC=4*12=48# which differ by #B=13#

The pair #16, 3# works.

Use this pair to split the middle term, then factor by grouping:

#4c^2+13c-12 = 4c^2+16c-3c-12#

#color(white)(4c^2+13c-12) = (4c^2+16c)-(3c+12)#

#color(white)(4c^2+13c-12) = 4c(c+4)-3(c+4)#

#color(white)(4c^2+13c-12) = (4c-3)(c+4)#

Putting it all together:

#16c^3+52c^2-48c = 4c(4c-3)(c+4)#

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