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:

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

How do you factor 8x²+20x-488x²+20x-48?