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

1 Answer
George C.
May 3, 2016

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

Explanation:

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)#

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