How do you factor #60x^2 - 124xy + 63y^2#?

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Answer 1 · 2015-05-28T21:33:15

Let's try a version of the AC Method:

#A=60#, #B=124#, #C=63#

We are looking for a pair of factors of #AC=60xx63# which add up to #B=124#.

#AC=60*63=2^2*3^3*5*7# has quite a few possible pairs of factors.

We can narrow down the search a bit: Since the sum is even, the factors of #2# must be split between them.

So look for a pair of factors of #3^3*5*7# that add to #62#.
Well #3^3=27# and #5*7=35# work.

So the original pair we were looking for is #2*27 = 54# and #2*35=70#

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

#60x^2-124xy+63y^2#

#=60x^2-54xy-70xy+63y^2#

#=(60x^2-54xy)-(70xy-63y^2)#

#=6x(10x-9y)-7y(10x-9y)#

#=(6x-7y)(10x-9y)#