How do you factor #15q^2-7q -4#?

1 Answer
George C.
Feb 3, 2016

#15q^2-7q-4 = (3q+1)(5q-4)#

Explanation:

Look for a pair of factors of #AC = 15*4 = 60# with difference #B=7#.

The pair #12, 5# works, since #12xx5 = 60# and #12-5 = 7#.

Use that to split the middle term and factor by grouping:

#15q^2-7q-4#

#= 15q^2-12q+5q-4#

#= (15q^2-12q)+(5q-4)#

#= 3q(5q-4)+1(5q-4)#

#= (3q+1)(5q-4)#

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