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

Question

4764 views around the world

You can reuse this answer
Creative Commons License

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)#