# How do you multiply (3y^2+18y+15)/(6y+6)*(y-5)/(y^2-25)?

Aug 18, 2016

$\frac{3 {y}^{2} + 18 y + 15}{6 y + 6} \times \frac{y - 5}{{y}^{2} - 25} = \textcolor{g r e e n}{\frac{1}{2} , y \ne - 5 , - 1 , 5}$

#### Explanation:

For problems like these, you must start by factoring everything.

$3 {y}^{2} + 18 y + 5$ can be factored as:

$3 \left({y}^{2} + 6 y + 5\right) = 3 \left(y + 5\right) \left(y + 1\right)$

$6 y + 6$ can be factored as:

$6 \left(y + 1\right)$

$y - 5$ is as simplified as it can be.

${y}^{2} - 25$ can be factored as $\left(y + 5\right) \left(y - 5\right)$

Putting all of this back together:

$\frac{3 \left(y + 5\right) \left(y + 1\right)}{6 \left(y + 1\right)} \times \frac{y - 5}{\left(y + 5\right) \left(y - 5\right)}$

See what you can cancel out, using the property $\frac{a}{a} = 1$:

$\frac{\cancel{\left(3\right)} \cancel{\left(y + 5\right)} \cancel{\left(y + 1\right)}}{{\cancel{\left(6\right)}}^{2} \cancel{\left(y + 1\right)}} \times \frac{\cancel{y - 5}}{\cancel{\left(y + 5\right)} \cancel{\left(y - 5\right)}}$

We are left with:

$\frac{1}{2}$

Now, before stating the final answer, we need to determine any restrictions on the variable. These will occur when the denominator equals $0$. Hence, they can be found by setting the denominator to $0$ and solving for $x$.

$6 y + 6 = 0 \text{ and } {y}^{2} - 25 = 0$

$y = - 1 \text{ and } y = \pm 5$

Thus, $y \ne - 1 , 5 , - 5$

Hopefully this helps!