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:

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

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

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

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

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

How do you factor 60x^2 - 124xy + 63y^260x^2 - 124xy + 63y^2?