Problem with simplifying $\frac{(3+h)^2-9}{(3+h)-3}$

Question

I need help simplifying $$ {(3+h)^2-9\over (3+h)-3}. $$

The answer is $6+h$.

I keep getting $h$.

Answer 1

$$ \frac{(3+h)^2-9}{(3+h)-3}=\frac{(3+h)(3+h)-9}{h}=\frac{(h^2+6h+9)-9}{h}=\frac{h^2+6h}{h}=\;? $$

Answer 2

Hint $$ a^2 - b^2 = (a-b)(a+b)$$

Answer 3

You got h, because you assumed that $(3+h) ^2 - 9 = ((3+h) - 3)((3+h) - 3)$

But it's $((3+h) - 3)((3+h) + 3)$.

After the simplification you'll get $h+6$.

Important: You have to point that $h$ is not equal to $0$.