How do you factor $6a^2-50ab+16b^2$ ?

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Answer 1 · 2015-06-13T13:33:10

$6a^2-50ab+16b^2 = 2(3a-b)(a-8b)$

$6a^2-50ab+16b^2$

$=2(3a^2-25ab+8b^2)$

Use a version of AC Method:

Let $A=3$ , $B=25$ , $C=8$

Look for a factorization of $AC=3*8=24$ into two factors whose sum is $B=25$ . The pair $B1=1$ , $B2=24$ works.

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

$(A, B1) = (3, 1) -> (3, 1) -> (3a-b)$
$(A, B2) = (3, 24) -> (1, 8) -> (a-8b)$

Hence:

$3a^2-25ab+8b^2 = (3a-b)(a-8b)$

and

$6a^2-50ab+16b^2 = 2(3a-b)(a-8b)$

How do you factor 6a^2-50ab+16b^26a^2-50ab+16b^2?