Basic algebra problem I am in a class higher than this, so I am embarrassed to be having a problem with this:
The problem:
Multiply: $(z-7)(z+3)(z-4)$
I am going to label each group to make my explanations more clear.

.................A..........B...........C

Multiply: $(z-7)(z+3)(z-4)$
My work:
(A)(B) = $z^2-4z-21$
(B)(C) =$ z^2-z-12$
(A)(C) = $z^2-11z+28$
Thus: $z^2 -4z-21+z^2-z-12+z^2-11z+28 $
So I get: $z^6-16z+5$
 A: Just multiply the first two, take the simplified result, and multiply that by the third. Just as you would with $2\times 3\times 4$.
The answer to your problem is a third degree polynomial, not a sixth degree polynomial.
A: You want $ABC$. Instead you've calculated $AB+BC+AC$ which most certainly is not $ABC$. 
Instead take $AB$ and multiply by $C$:
$$(AB)C = (z^2-4z-21)(z-4) = (z^2-4z-21)z+(z^2-4z-21)(-4)$$ $$= z^3-4z^2-21z-4z^2+16z+84
=z^3-8z^2-5z+84$$
Of course, $(BC)A$ or $(AC)B$ will yield the same answer since polynomial multiplication is associative and commutative. :)
A: Here's a simple way of doing it:
$1. \,\,(z−7)(z+3)(z−4)$
$2. \,\,(z-7)(z^2-1z-12)$ [It was $(z-7)(z^2-4z+3z-12)$, not anymore]
$3. \,\,(z^3-z^2-12z-7z^2+7z+84)$
$4. \,\,(z^3-8z^2-5z+84)$
This should be your answer: $(z^3-8z^2-5z+84)$
English explanation:
1. The first step is to write down your problem.
2. Multiply the last two pairs and ignore the first pair. You should first get $(z-7)(z^2-4z+3z-12)$ by doing FOIL (first, outside, inside, last).
3. Multiply the first pair with the new pair you just created. Simply multiply z with each value in the new pair, and then multiply 7 with each value in the new pair.
4. Simplify.
