Factor completely. 6a^2 + 19as + 15s^2

Question

Factor completely.  6a^2 + 19as + 15s^2

chaise · Answer

(2a+3s)(3a+5s)
Do you need help with how this answer was obtained?

anonymous · Answer

Yes

chaise · Answer

It's a little bit of trial and error, and very hard to explain. After a while you notice relationships between numbers.

anonymous · Answer

Yes, that was discussed in class.  Isn't this called the FOIL technique or something like that?

anonymous · Answer

If you multiply the lead coefficient, 6, time the constant, 15, you get 6*15=90.  The poly will be factorable if there are factors of 90 that add up to 19, the coefficient of the middle term.  In this case:
9*10=90
9+10=19
So you can use the 9 and the 10 to rewrite the 19as to be 9as+10as.$$6a^2+9as+10as+15s^2$$This can be factored by"grouping method" in which you factor out the GCF of the first two terms and the GCF of the second two terms:$$3a(2a+3s)+5s(2a+3s)$$Now there are two terms in which the binomial 2a+3s is a GCF, so factor it out to get$$(2a+3s)(3a+5s)$$And you are done.