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!

How do you factor completely 6a^3+3a^2-18a6a^3+3a^2-18a?