How do you factor completely $40x^2 – 2xy – 24y^2$ ?

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Answer 1 · 2016-05-03T22:47:23

$40x^2-2xy-24y^2=2(4x+3y)(5x-4y)$

First separate out the common scalar factor $2$ :

$40x^2-2xy-24y^2$

$=2(20x^2-xy-12y^2)$

Since the remaining factor is homogeneous of degree $2$ we can attempt to use an AC method to factor it:

Look for a pair of factors of $AC = 20*12 = 240$ with difference $B=1$

The pair $16, 15$ works, so we can use that to split the middle term and factor by grouping:

$20x^2-xy-12y^2$

$=(20x^2-16xy)+(15xy-12y^2)$

$=4x(5x-4y)+3y(5x-4y)$

$=(4x+3y)(5x-4y)$

How do you factor completely 40x^2 – 2xy – 24y^240x^2 – 2xy – 24y^2?