SOLUTION: Factor completely. b^2 � ab � 6a^2 I got (b � 3a)(b + 2a) Did I just maybe get it right?

Question 84487: Factor completely. b^2 � ab � 6a^2
I got (b � 3a)(b + 2a)
Did I just maybe get it right?
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Yes, you got it correct.
.
You can always check your answer by multiplying the two factors you got using the FOIL method.
.
FOIL stands for Firsts ... Outsides ... Insides ... Lasts and it applies to the arrangement
of the terms in your answer. Your answer was:
.
(b � 3a)(b + 2a)
.
The two "Firsts" in each set of parentheses are both b. Multiplying them together gives you
b*b which is b^2.
.
The two "Outsides" in this string are ... b from the first factor and +2a from the second
factor. Multiplying b times +2a results in +2ab.
.
The two "Insides" are -3a from the first factor and b from the second factor. Multiplying
these two gives you -3ab.
.
And finally, the two "Lasts" are -3a from the first set of parentheses and +2a from the
second set. Multiplying -3a times +2a results in -6a^2.
.
Now add your four answers from the FOIL process:
.
b^2 + 2ab - 3ab - 6a^2
.
The two middle terms (both containing ab) combine to give you -ab. Substitute this result
in place of +2ab - 3ab and your product becomes:
.
b^2 - ab - 6a^2
.
This is the same as the trinomial you were asked to factor. Therefore, you have factored
it correctly. Had your FOIL multiplication given you a different answer you might have
expected that your factors were wrong (or you made a mistake in the FOIL multiplication).
.
Hope this helps you to build confidence in what you are doing. Keep up the good work!

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