# How do you multiply 1/(x+9)*(7x^3+49x^2)/(x+7) and state the excluded values?

Oct 30, 2016

$\frac{7 {x}^{2}}{x + 9}$

#### Explanation:

$\frac{1}{x + 9} \cdot \frac{7 {x}^{3} + 49 {x}^{2}}{x + 7}$

We can state the excluded values immediately because the denominators may not be 0.

$x \ne - 9 \mathmr{and} x \ne - 7$

Factorise first to see there are common factors which can cancel

. $\frac{1}{x + 9} \times \frac{7 {x}^{2} \cancel{\left(x + 7\right)}}{\cancel{\left(x + 7\right)}}$

=$\frac{7 {x}^{2}}{x + 9}$