# How do you solve by factoring and using the principle of zero products 36r^2=16?

##### 1 Answer
Jan 3, 2017

$r = \frac{2}{3}$
$r = - \frac{2}{3}$

#### Explanation:

$36 {r}^{2} = 16$

Make one side equal to zero by subtracting 16 from each side:
$36 {r}^{2} - 16 = 0$

First, I would factor out a $4$ so that I'm working with smaller numbers:
$4 \left(9 {r}^{2} - 4\right) = 0$

Use the property that: ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

$4 \left[{\left(3 r\right)}^{2} - {\left(2\right)}^{2}\right] = 0$

$4 \left(3 r - 2\right) \left(3 r + 2\right) = 0$

Set each factor (apart from constants) equal to zero individually.
$3 r - 2 = 0$
$3 r = 2$
$r = \frac{2}{3}$

$3 r + 2 = 0$
$3 r = - 2$
$r = - \frac{2}{3}$