How do you factor #8x²+20x-48#?

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Answer 1 · 2016-04-26T21:25:29

#8x^2+20x-48 = 4(2x-3)(x+4)#

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

#8x^2+20x-48 = 4(2x^2+5x-12)#

Next use an AC method to factor #2x^2+5x-12#:

Look for a pair of factors of #AC=2*12=24# with difference #B=5#.

The pair #8, 3# works: #8xx3=24# and #8-3=5#.

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

#2x^2+5x-12#

#=2x^2+8x-3x-12#

#=(2x^2+8x)-(3x+12)#

#=2x(x+4)-3(x+4)#

#=(2x-3)(x+4)#

Hence:

#8x^2+20x-48 = 4(2x-3)(x+4)#