How do you factor #11m^2 + 14m -16#?

1 Answer
George C.
Jun 4, 2015

I will use a version of the AC Method.

Given #f(m) = 11m^2+14m-16#

#A=11#, #B=14#, #C=16#

Look for a factorization of #AC=11xx16=176# into a pair of factors whose difference is #B=14#.

The pair #B1=22#, #B2=8# works.

Then for each of the pairs #(A, B1)# and #(A, B2)#, divide by the HCF (highest common factor) to derive the coefficients of a factor of #f(m)#, choosing signs appropriately...

#(A, B1) = (11, 22) -> (1, 2) -> (m+2)#
#(A, B2) = (11, 8) -> (11, 8) -> (11m-8)#

Hence:

#11m^2+14m-16 = (m+2)(11m-8)#

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