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Mathematics 23 Online
OpenStudy (anonymous):

Factor 5a3 + 15a2 – 6a – 18

OpenStudy (zehanz):

I used to have problems with this kind of things, until it became clear that they are specially made to be factored ;) Look at the first two terms: 5a²(a+3) Second two terms: -6(a+3) So: 5a²(a+3)-6(a+3) Now you can do still one more step...

OpenStudy (zehanz):

I mean: It is not just a random polynomial, but it is chosen in such a way that factoring is easily possible.

OpenStudy (anonymous):

@zehanz (5a2 – 3)(a + 6) is that right?

OpenStudy (anonymous):

@ZeHanz wait no, its (5a2 – 6)(a + 3) right?

OpenStudy (zehanz):

OK, now you are right!!

OpenStudy (anonymous):

hahaha!!(: thank youu. so it is (5a2 – 6)(a + 3)?

OpenStudy (zehanz):

Yup!

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