How do you factor $20m^2+13mn+2n^2$ ?

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Answer 1 · 2015-06-03T06:33:17

$20m^2+13mn+2n^2$ is homogeneous in that all of the terms are of degree $2$ , so it can be factored like $20x^2+13x+2$ ,

Use AC Method to factor.

$A=20$ , $B=13$ , $C=2$

Look for a factorization of $AC=20*2=40$ into a pair of factors whose sum is $B=13$ .

The pair $B1=5$ , $B2=8$ works.

Then for each of the pairs $(A, B1)$ and $(A, B2)$ , divide that pair by the HCF (highest common factor) to get the coefficients of a factor...

$(A, B1) = (20, 5), HCF=5 -> (4, 1) -> (4m+n)$
$(A, B2) = (20, 8), HCF=4 -> (5, 2) -> (5m+2n)$

So $20m^2+13mn+2n^2 = (4m+n)(5m+2)$

How do you factor 20m^2+13mn+2n^220m^2+13mn+2n^2?