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OpenStudy (anonymous):
Start with greatest common factor GCF
OpenStudy (anonymous):
i did that 3a
OpenStudy (anonymous):
3a(3a^2+5a-12a)
OpenStudy (anonymous):
nice... now I would use ac method
OpenStudy (anonymous):
ac would be (3)(-12) to get the number that you must find factors for that might add up to the middle term.
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OpenStudy (anonymous):
yea i did -9 and 4
OpenStudy (anonymous):
-36
ignoring signs for a moment....
1 and 36... the sum or difference doesn't get 5
2 and 18... the sum or difference doesn't get 5
3 and 12... the sum or difference doesn't get 5
OpenStudy (anonymous):
yeah
OpenStudy (anonymous):
now split the middle term into -4a + 9a to make the 5a
OpenStudy (anonymous):
thats what i did. i got stuck at the last part.
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OpenStudy (anonymous):
then it is a grouping problem
OpenStudy (anonymous):
\[3a(3a ^{2}-4a+9a-12)\]
OpenStudy (anonymous):
\[3a(a(3a-4)+3(3a-4))\]
OpenStudy (anonymous):
3a-4 is common to both of the inside terms, pull it out in front of the others
OpenStudy (anonymous):
which leaves behind a+3
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OpenStudy (anonymous):
i got that but i dnt understand how to put it as the answer
OpenStudy (anonymous):
3a(3a-4)(a+3) should be the fully factored form
OpenStudy (anonymous):
If it is set equal to zero... you can solve
OpenStudy (anonymous):
ok yes that looks more like it. I was having a hard time grouping.
OpenStudy (anonymous):
thank you
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