How do you factor #x^ 2+ 4xy - 12y ^2#?

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Answer 1 · 2017-06-24T00:41:34

#(x+6y)(x-2y)#

I usually start by opening a pair of parentheses:

#()()#

In this case, I know that the 1st terms of each pair of parentheses is #x# since the required product is #x^2#.

#(x)(x)#

In this case, I also know that the last terms of each pair of parentheses contains a #y# since the required product is #y^2#.

#(xy)(xy)#

No we know the middle term has to be #4xy# and the last term needs a #-12#. So we'll use a #6 and a #2# and the signs need to be different:

#(x+6y)(x-2y)#