How do you factor completely #6a^3+3a^2-18a#?

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Answer 1 · 2018-07-04T21:08:34

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

Given that

#6a^3+3a^2-18a#

#=3a(2a^2+a-6)#

#=3a(2a^2+4a-3a-6)#

#=3a(2a(a+2)-3(a+2))#

#=3a((a+2)(2a-3))#

#=3a(a+2)(2a-3)#

Answer 2 · 2018-07-04T21:10:25

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

To start, we notice that all terms have a #3a# in common, so we can factor that out to get

#3a(2a^2+a-6)#

Next, we can factor by grouping. We can rewrite #a# as #4a-3a#. We now have

#3a(color(blue)(2a^2+4a)+color(purple)(-3a-6))#

We can factor a #2a# out of the blue term and a #-3# out of the purple term. We now have

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

Both terms have an #a+2# in common, so we can factor that out. We will get

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

Hope this helps!