How do you factor 16p² -4pq - 30q²?

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How do you factor 16p² -4pq - 30q²?

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Answer 1 · 2015-06-03T19:24:01

$16p^2-4pq-30q^2$ is homogeneous of order $2$ , so is amenable to the AC Method for quadratics.

$16p^2-4pq-30q^2 = 2(8p^2-2pq-15q^2)$

$A=8$ , $B=2$ , $C=15$

Look for a factorization of $AC=8*15=120$ into two factors whose difference is $B=2$ .

The pair $B1=10$ , $B2=12$ works.

Now, for each of the pairs $(A, B1)$ and $(A, B2)$ divide by the HCF (highest common factor) to yield coefficients of factors of our quadratic, choosing signs for the terms appropriately.

$(A, B1) = (8, 10) -> (4, 5) -> (4p+5q)$
$(A, B2) = (8, 12) -> (2, 3) -> (2p-3q)$

Hence

$16p^2-4pq-30q^2 = 2(4p+5q)(2p-3q)$

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How do you factor 16p² -4pq - 30q²?

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