How do you factor the trinomial #16a^2-22ab-3b^2#?

1 Answer
George C.
Dec 13, 2015

Use an AC Method: Look for a pair of factors of #16xx3 = 48# with difference #22#, hence find:

#16a^2-22ab-3b^2 = (8a+b)(2a-3b)#

Explanation:

Multiply the first coefficient #A=16# by the third coefficient (ignoring sign) #C=3# to get #48#.

Then since the sign of the third coefficient is negative, look for a pair of factors of #AC=48# whose difference is #B=22#.

The pair #(B1, B2) = (24, 2)# works, so use that to split the middle term:

#16a^2-24ab+2ab-3b^2#

We can then factor by grouping to find:

#16a^2-24ab+2ab-3b^2#

#=(16a^2-24ab)+(2ab-3b^2)#

#=8a(2a-3b)+b(2a-3b)#

#=(8a+b)(2a-3b)#

Alternatively, just write down the pairs #(A, B1)# and #(A, B2)# with appropriate signs and divide by highest common factors to find pairs of coefficients as follows:

#(A, B1) -> (16, -24) -> (2, -3) -> (2a-3b)# (dividing through by #8#)

#(A, B2) -> (16, 2) -> (8, 1) -> (8a+b)# (dividing through by #2#)

The slight complication is the choosing of signs for the second term so that the resulting #+-B1 +- B2 = -22#

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