# How do you multiply (2y + 11) ( 2y - 11)?

May 28, 2017

$4 {y}^{2} - 121$

#### Explanation:

If you didn't already know, we can use the $\text{F.O.I.L}$ method.
$\text{F.O.I.L}$ stands for:

$\textcolor{red}{\text{F}}$: First, as in you multiply the first two terms in each $\left(\right)$
$\textcolor{b l u e}{\text{O}}$: Outer (multiply the outer terms)
$\textcolor{g r e e n}{\text{I}}$: Inner (multiply the inner terms)
$\textcolor{\mathmr{and} a n \ge}{\text{L}}$: Last (multiply the last terms

$\left(\textcolor{red}{2 y} + 11\right) \left(\textcolor{red}{2 y} - 11\right) : \textcolor{red}{4 {y}^{2}}$

$\left(\textcolor{b l u e}{2 y} + 11\right) \left(2 y \textcolor{b l u e}{- 11}\right) : \textcolor{b l u e}{- 22 y}$

$\left(2 y + \textcolor{g r e e n}{11}\right) \left(\textcolor{g r e e n}{2 y} - 11\right) : \textcolor{g r e e n}{22 y}$

$\left(2 y + \textcolor{\mathmr{and} a n \ge}{11}\right) \left(2 y \textcolor{\mathmr{and} a n \ge}{- 11}\right) : \textcolor{\mathmr{and} a n \ge}{- 121}$

Gathering this information, we can rewrite the expression and combine like terms:

$4 {y}^{2} \textcolor{red}{- 22 y + 22 y} - 121$

$4 {y}^{2} \cancel{\textcolor{red}{- 22 y + 22 y}} - 121$

The middle terms cancel and so we are left with this as our final answer:

$4 {y}^{2} - 121$