How do you factor completely #15a^2 + 55a - 20#?

1 Answer
Oct 26, 2016

Answer:

#15a^2+55a-20 = 5(3a-1)(a+4)#

Explanation:

First separate out the common scalar factor #5# to find:

#15a^2+55a-20 = 5(3a^2+11a-4)#

Next use an AC Method to factor #3a^2+11a-4#:

Find a pair of factors of #AC = 3*4 = 12# which differ by #B=11#.

The pair #12, 1# works, in that #12*1 = 12# and #12-1 = 11#.

Use this pair to split the middle term and factor by grouping:

#3a^2+11a-4 = 3a^2+12a-a-4#

#color(white)(3a^2+11a-4) = (3a^2+12a)-(a+4)#

#color(white)(3a^2+11a-4) = 3a(a+4)-1(a+4)#

#color(white)(3a^2+11a-4) = (3a-1)(a+4)#

So:

#15a^2+55a-20 = 5(3a-1)(a+4)#