## Calculus 8th Edition

$\frac{(x^2 -9)}{(x-3)} = \frac{(x+3)(x-3)}{(x-3)} = \frac{(x+3)}{1} = x+3$ When you factor out the numerator one of the factors cancels out the denominator, making the expression equal to the other factor. Thus, $\frac{(x^2 -9)}{(x-3)}= x+3$ Which makes the answer true.