# How do you factor 16y^2 - 9?

Mar 3, 2018

See a solution process below:

#### Explanation:

This is a special form of the quadratic:

${\textcolor{red}{a}}^{2} - {\textcolor{b l u e}{b}}^{2} = \left(\textcolor{red}{a} + \textcolor{b l u e}{b}\right) \left(\textcolor{red}{a} - \textcolor{b l u e}{b}\right)$

Let: ${a}^{2} = 16 {y}^{2}$ then:

$\sqrt{{a}^{2}} = \sqrt{16 {y}^{2}}$

$a = 4 y$

Let: ${b}^{2} = 9$ then:

$\sqrt{{b}^{2}} = \sqrt{9}$

$b = 3$

Substituting into the rule gives:

$16 {y}^{2} - 9 \implies {\textcolor{red}{\left(4 y\right)}}^{2} - {\textcolor{b l u e}{3}}^{2} \implies \left(\textcolor{red}{4 y} + \textcolor{b l u e}{3}\right) \left(\textcolor{red}{4 y} - \textcolor{b l u e}{3}\right)$

Mar 3, 2018

$\left(4 y + 3\right) \left(4 y - 3\right)$

#### Explanation:

$\textcolor{red}{{A}^{2} - {B}^{2} = \left(A + B\right) \left(A - B\right)}$
$16 {y}^{2} - 9 = {\left(4 y\right)}^{2} - {\left(3\right)}^{2}$
Applying above formula,
$16 {y}^{2} - 9 = \left(4 y + 3\right) \left(4 y - 3\right)$