How do you factor completely: #18x^2 − 21x − 15#?

1 Answer
George C.
Jul 13, 2015

#18x^2-21x-15 = 3(6x^2-7x-5) = 3(2x+1)(3x-5)#

Explanation:

First notice the common scalar factor #3# and separate that out:

#18x^2-21x-15 = 3(6x^2-7x-5)#

Then factor #6x^2-7x-5# using a variant of the AC Method.

Let #A=6#, #B=7#, #C=5#.

Look for a pair of factors of #AC=30# whose difference is #B=7#.

The pair #B1=10#, #B2=3# works.

Next for each of the pairs #(A, B1)#, #(A, B2)# divide by the HCF (highest common factor) to give a pair of coefficients of a linear factor, choosing suitable signs:

#(A, B1) = (6, 10) -> (3, 5) -> (3x-5)#
#(A, B2) = (6, 3) -> (2, 1) -> (2x+1)#

Hence:

#18x^2-21x-15 = 3(6x^2-7x-5) = 3(2x+1)(3x-5)#

Related questions
See all questions in Factoring Completely
Impact of this question
2446 views around the world
You can reuse this answer
Creative Commons License