# How do you find the product of (y^2-1)/(y^2-49)*(y-7)/(y+1)?

Feb 9, 2017

$\frac{y - 1}{y + 7}$

#### Explanation:

$\frac{{y}^{2} - 1}{{y}^{2} - 49} \cdot \frac{y - 7}{y + 1}$

difference of two squares rule:

$\left(a + b\right) \cdot \left(a - b\right) = {a}^{2} - {b}^{2}$

in the same way:

$\left(y + 1\right) \left(y - 1\right) = {y}^{2} - {1}^{2} = {y}^{2} - 1$

$\left(y + 7\right) \left(y - 7\right) = {y}^{2} - {7}^{2} = {y}^{2} - 49$

$\therefore \frac{{y}^{2} - 1}{{y}^{2} - 49} \cdot \frac{y - 7}{y + 1}$

$= \frac{\left(y + 1\right) \left(y - 1\right)}{\left(y + 7\right) \left(y - 7\right)} \cdot \frac{y - 7}{y + 1}$

cancel out fractions:

$\frac{\left(y + 1\right) \left(y - 1\right)}{y + 1} = \left(y - 1\right)$
$\frac{\left(y + 7\right) \left(y - 7\right)}{y - 7} = \left(y + 7\right)$

$\therefore \frac{{y}^{2} - 1}{{y}^{2} - 49} \cdot \frac{y - 7}{y + 1}$

$= \frac{y - 1}{y + 7} \cdot \frac{1}{1}$

$= \frac{y - 1}{y + 7}$

Feb 9, 2017

$= \frac{y - 1}{y + 7}$

#### Explanation:

We need to remember two things here. First, for $a , b , c , d \in \mathbb{R}$
and $b , d \ne 0$, we have:

$\frac{a}{b} \cdot \frac{c}{d} = \frac{a c}{b d}$

Then, we must know how to spot and factor a difference of squares:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$ for any $a , b \in \mathbb{R}$.

Using this relationship, the first fraction is converted to:

$\frac{{y}^{2} - 1}{{y}^{2} - 49} = \frac{\left(y - 1\right) \left(y + 1\right)}{\left(y - 7\right) \left(y + 7\right)}$

Then, multiply that with the other fraction:

$\frac{\left(y - 1\right) \left(y + 1\right)}{\left(y - 7\right) \left(y + 7\right)} \cdot \frac{y - 7}{y + 1} = \frac{\left(y - 1\right) \cancel{\left(y + 1\right)} \cancel{\left(y - 7\right)}}{\cancel{\left(y - 7\right)} \left(y + 7\right) \cancel{\left(y + 1\right)}} = \frac{y - 1}{y + 7}$

Of course, we have the restrictions $y \ne \pm 7$ and $y \ne - 1$, otherwise the denominator would have been $0$.

Remember these two things to help you spot and factor a difference of squares:

1) See if you can transform the terms into squares, for example:

${x}^{2} = {\left(x\right)}^{2}$ (obviously)
$81 {x}^{2} = {\left(9 x\right)}^{2}$
$256 = {\left(16\right)}^{2}$
etc.

2) Remember that $1 = {1}^{2}$, so if one of the terms is a $1$, you can still factor. For example:

${x}^{2} - 1 = {x}^{2} - {1}^{2} = \left(x - 1\right) \left(x + 1\right)$
$9 {y}^{4} - 1 = {\left(3 {y}^{2}\right)}^{2} - {1}^{2} = \left(3 {y}^{2} - 1\right) \left(3 {y}^{2} + 1\right)$
etc.