How do you factor completely #3a^2+9a-12#?

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Answer 1 · 2016-04-11T12:29:16

#3(a-1)(a+4)#

3 is the factor common to all the terms.
Let's take it out

#3(a^2+3a-4)#

Now, we find the factors of #(a^2+3a-4)#

Use splitting of the middle term to find the answer.

Find a pair of factors of #1xx-4=-4# which sum up to #3#. The pair #-1# and #4# works.

Write the expression, #(a^2+3a-4)# as:

#a^2-a+4a-4#

Take out the common terms

#a(a-1)+4(a-1)#

Separate the like terms and you get

#(a-1)(a+4)#

The factors of #3a^2+9a-12# = #3(a-1)(a+4)#