# How do you factor 16 = 38r - 12r^2?

Aug 7, 2017

$2 \left(2 r - 1\right) \left(3 r - 8\right) = 0$

#### Explanation:

Set it equal to $0$
Take out the common factor of $2$ first.

$12 {r}^{2} - 38 r + 16 = 0$

$2 \left(6 {r}^{2} - 19 r + 8\right) = 0$

Find factors of $6 \mathmr{and} 8$ whose products ADD to $- 19$

As a hint, we see that $19$ is odd. Therefore the factors of $8$ cannot be $2 \mathmr{and} 4$, because their products will be even, so use $1 \mathmr{and} 8$

$\text{ } 6 \mathmr{and} 8$
$\text{ "darr" } \downarrow$
$\text{ "2" "1" } \rightarrow 3 \times 1 = 3$
$\text{ "3" "8" } \rightarrow 2 \times 8 = \underline{16}$
$\textcolor{w h i t e}{w w w w w w w w w w w w w w w w} 19$

The signs in the brackets will both be negative.

$2 \left(2 r - 1\right) \left(3 r - 8\right) = 0$

These are the factors which will lead to solutions of:

$r = \frac{1}{2} \mathmr{and} r = \frac{8}{3}$