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

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

can you try to factor it from there?

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

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

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

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

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

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

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

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

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

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?

yeah i get what you're saying, thanks

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