# How do you factor completely 4r^2 -21+5?

Jun 20, 2016

$\left(2 r + 4\right) \left(2 r - 4\right)$

#### Explanation:

To factor $4 {r}^{2} - 21 + 5$ completely we would begin by combining like terms.

$4 {r}^{2} - 16$

We next would recognize that the binomial is a difference (subtraction) of two squares.

$4 {r}^{2} = \left(2 r\right) \left(2 r\right)$
and
$16 = \left(4\right) \left(4\right)$

The difference of two squares always factors in this pattern

${a}^{2} - {b}^{2}$

(a+b)((a-b)

So our binomial doctors as

$4 {r}^{2} - 16$

$\left(2 r + 4\right) \left(2 r - 4\right)$