# Math Help - Help with simplification

1. ## Help with simplification

I have the following problem that needs to be simplified. I already have the simplification results, but I don't understand how they got it. For example, how does there and up being 2 h's, when there is only a single h to begin with. Any help on how to simplify this problem one step at a time would be appreciated.

Step 1: $((3-4(-2+h)^2)-(3-4(-2)^2) / h)$

Step 2: $((16h-4h^2) / h)$

Step 3: $(16-4h)$

2. You have

$\frac{16h-4h^2}{h}$

now factoring out $h$ in the numerator

$\frac{h(16-4h)}{h}$

Which cancels with the h in the denominator

$16-4h$

3. But how did they get from Step 1 to Step 2? I have edited my original post to include the step numbers. I understand the last part which you explained, but I don't understand the simplification process which resulted in Step 2.

4. Hello, softwareguy!

How did they get from Step 1 to Step 2?

Step 1: $\frac{[3-4(-2+h)^2]-[3-4(-2)^2]}{h}$

Step 2: $\frac{16h-4h^2}{h}$

Step 3: $16-4h$

We have: . $\frac{[3-4(4-4h + h^2)] - [3-4(4)]}{h}$

. . . . . . $=\;\frac{(3-16 + 16h - 4h^2) - (-13)}{h}$

. . . . . . $=\;\frac{-13 + 16h - 4h^2 + 13}{h}$

. . . . . . $=\; \frac{16h-4h^2}{h}$

. . . . . . $=\;\frac{{\color{red}\rlap{/}}h(16-4h)}{{\color{red}\rlap{/}}h}$

. . . . . . $=\;16-4h$

5. Thanks! I totally forgot about having to FOIL out the $(-2+h)^2$ equation. That was blocking my ability to proceed.