How do you factor 16p² -4pq - 30q²?

1 Answer
George C.
Jun 3, 2015

#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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