# How do you multiply (2r+9s)^2?

Aug 29, 2016

#### Answer:

$4 {r}^{2} + 36 r s + 81 {s}^{2}$

#### Explanation:

Consider ${\left(2 r + 9 s\right)}^{2} = \left(2 r + 9 s\right) \left(2 r + 9 s\right)$

We must ensure that each term in the 2nd bracket is multiplied by each term in the first bracket.This can be achieved as follows.

$\left(\textcolor{red}{2 r + 9 s}\right) \left(2 r + 9 s\right)$

$= \textcolor{red}{2 r} \left(2 r + 9 s\right) + \textcolor{red}{9 s} \left(2 r + 9 s\right)$

distribute the brackets : $4 {r}^{2} + 18 r s + 18 r s + 81 {s}^{2}$

and simplifying to obtain : $4 {r}^{2} + 36 r s + 81 {s}^{2}$

$\Rightarrow {\left(2 r + 9 s\right)}^{2} = 4 {r}^{2} + 36 r s + 81 {s}^{2}$
$\text{----------------------------------------------------------}$
Alternatively there is the FOIL method.

$F \to \text{ First terms}$
$O \to \text{ Outer terms}$
$I \to \text{ Inner terms}$
$L \to \text{ Last terms}$

F-multiply the first terms in each bracket together.
O-multiply the outer terms together.
I-multiply the inner terms together.
L-multiply the last terms together.

$\Rightarrow \left(2 r + 9 s\right) \left(2 r + 9 s\right)$

$= \left(2 r \times 2 r\right) + \left(2 r \times 9 s\right) + \left(2 r \times 9 s\right) + \left(9 s \times 9 s\right)$

$= 4 {r}^{2} + 18 r s + 18 r s + 81 {s}^{2} = 4 {r}^{2} + 36 r s + 81 {s}^{2}$