Mathematics
OpenStudy (anonymous):

factor by grouping 4z^2+12z-az-3a

OpenStudy (anonymous):

group the two equations into common terms (\[(4z^2+12z) + (Za-3a)\]

OpenStudy (anonymous):

can you try to factor it from there?

OpenStudy (anonymous):

well i got to 4z(z+3)-a(z-3), but I don't know where to go from there

OpenStudy (anonymous):

when you took out -a in the second term, this would leave the equation as (z+3)

hero (hero):

4z^2+12z-az-3a 4z(z+3)-a(z+3) (z+3)(4z-a)

OpenStudy (anonymous):

so what you should have is 4z(z+3) -a (z+3)

OpenStudy (anonymous):

then you can group the other two terms together, (4z-a)(z+3), just like hero said

OpenStudy (anonymous):

oh ok, now I get it, I always forget to change the sign on that

hero (hero):

Change the sign? You shouldn't need to change any signs if you factor properly

OpenStudy (anonymous):

just think of it as you're dividing the entire equation by what you're factoring out, so in this case you're dividing (-za-3a) by -a

OpenStudy (anonymous):

if I'm grouping 4z^2+12z-az-3a, I would think it would be (4z^2+12z)-(az-3a)

OpenStudy (anonymous):

yeah, try to keep the negatives connected to the terms instead of the equations so its the first equation set (4z^2+12z) PLUS (-za-3a), you're adding these two different equations together. what helps me is that i always change all the subtractions to positives, and change the terms to negative so like 4-x is really 4+ (-x) maybe that'll help you?

OpenStudy (anonymous):

yeah i get what you're saying, thanks