How do you factor completely #6a^2-15a-8a+20#?

1 Answer
Nov 12, 2015

Answer:

#(3a - 4)(2a - 5) = 0#.
#a = 4/3#.
#a = 5/2#.

Explanation:

We start with #6a^2 - 15a - 8a +20#.
First of all, we see we can do #15 - 8a# easily, so
#-15 - 8a# turns into #-23a#. So,
we have #6a^2 - 23a + 20#.

Now, we multiply the factor of #6a^2#, (#6#) by the last number (#20#). We get #120#. We now look for factors that add up to #23#.
For example, #-8# and #-15# add up to #23#, and multiplied, give #120#.

We now put out the equation as:
#((6a - 8)(6a - 15))/6#. We now divide the right bracket and the denominator by 3, to simplify it. We get:
#((6a - 8)(2a - 5))/2#. We now divide the left bracket and the denominator by 2, leaving the denominator as 1. We get:
#(3a - 4)(2a - 5)#.

Hope it Helps! :D .