# How do you factor the polynomials 48tu-90t+32u-60?

Apr 28, 2017

$48 t u - 90 t + 32 u - 60 = 2 \left(3 t + 2\right) \left(8 u - 15\right)$

#### Explanation:

Given:

$48 t u - 90 t + 32 u - 60$

Note that the ratio of the first and second terms is the same as the ratio of the third and fourth terms. So this quadrinomial will factor by grouping.

First separate out the common scalar factor $2$, since that is the greatest common factor of the coefficients.

$48 t u - 90 t + 32 u - 60 = 2 \left(24 t u - 45 t + 16 u - 30\right)$

$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(\left(24 t u - 45 t\right) + \left(16 u - 30\right)\right)$

$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(3 t \left(8 u - 15\right) + 2 \left(8 u - 15\right)\right)$

$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(3 t + 2\right) \left(8 u - 15\right)$

As an alternative, we could swap the middle two terms before grouping, which may make the arithmetic seem a little easier:

$48 t u - 90 t + 32 u - 60 = 48 t u + 32 u - 90 t - 60$

$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(24 t u + 16 u - 45 t - 30\right)$

$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(\left(24 t u + 16 u\right) - \left(45 t + 30\right)\right)$

$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(8 u \left(3 t + 2\right) - 15 \left(3 t + 2\right)\right)$

$\textcolor{w h i t e}{48 t u - 90 t + 32 u - 60} = 2 \left(8 u - 15\right) \left(3 t + 2\right)$