# How do you find the product (2q+5r)(2q-5r)?

Mar 9, 2018

$\left(2 q + 5 r\right) \left(2 q - 5 r\right) = \textcolor{red}{4 {q}^{2} - 25 {r}^{2}}$

#### Explanation:

We know (hopefully you remember ...I know I do) that
$\textcolor{w h i t e}{\text{XXX}} \left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ ...this is called "the difference of squares"

In this case using
$\textcolor{w h i t e}{\text{XXX}} 2 q$ in place of $a$, and
$\textcolor{w h i t e}{\text{XXX}} 5 r$ in place of $b$
we have
$\left(2 q + 5 r\right) \left(2 q - 5 r\right) = {\left(2 q\right)}^{2} - {\left(5 r\right)}^{2}$

$\textcolor{w h i t e}{\text{XXXXXXXXXXX}} = 4 {q}^{2} - 25 {r}^{2}$

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Of course, if you don't remember "the difference of squares", you could apply the distributive property:
$\left(2 q + 5 r\right) \left(2 q - 5 r\right)$
$\textcolor{w h i t e}{\text{XXX}} = 2 q \left(2 q - 5 r\right) + 5 r \left(2 q - 5 r\right)$
$\textcolor{w h i t e}{\text{XXX}} = {\left(2 q\right)}^{2} - \left(2 q\right) \left(5 r\right) + \left(5 r\right) \left(2 q\right) - {\left(5 r\right)}^{2}$
$\textcolor{w h i t e}{\text{XXX}} = {\left(2 q\right)}^{2} - {\left(5 r\right)}^{2}$
$\textcolor{w h i t e}{\text{XXX}} = 4 {q}^{2} - 25 {r}^{2}$