# How do you find the product of (9x + 4)^2?

Apr 12, 2017

$81 {x}^{2} + 72 x + 16$

#### Explanation:

To simplify this expression, or any expression, a good start would be writing it out completely and getting rid of the exponent. This will make it a lot easier to multiply.

${\left(9 x + 4\right)}^{2}$ will become $\left(9 x + 4\right) \left(9 x + 4\right)$

Now we can begin to multiply by using the FOIL method.

Take the $\textcolor{b l u e}{\text{first term in the first binomial}}$ and multiply it with $\textcolor{g r e e n}{\text{every term in the second binomial}}$.

Then take the $\textcolor{red}{\text{second term in the binomial}}$ and multiply it with $\textcolor{g r e e n}{\text{every term in the second binomial}}$.

$\left(9 x + 4\right) \left(9 x + 4\right)$

$\left(\textcolor{b l u e}{9 x} + 4\right) \left(\textcolor{g r e e n}{9 x} + 4\right)$ $\textcolor{\mathmr{and} a n \ge}{\to} 9 x \cdot 9 x \textcolor{\mathmr{and} a n \ge}{\to} \textcolor{red}{81 {x}^{2}}$

 (color(blue)(9x) + 4)(9x   color(green)( + 4))  $\textcolor{\mathmr{and} a n \ge}{\to} 9 x \cdot 4 \textcolor{\mathmr{and} a n \ge}{\to} \textcolor{red}{36 x}$

 (9x  color(red)( + 4))(color(green)(9x) + 4)  $\textcolor{\mathmr{and} a n \ge}{\to} 4 \cdot 9 x \textcolor{\mathmr{and} a n \ge}{\to} \textcolor{red}{36 x}$

 (9x  color(red)( + 4))(9x   color(green)( + 4))  $\textcolor{\mathmr{and} a n \ge}{\to} 4 \cdot 4 \textcolor{\mathmr{and} a n \ge}{\to} \textcolor{red}{16}$

Now all we have to do is add the terms that we got and simplify.

$81 {x}^{2} + 36 x + 36 x + 16$
$81 {x}^{2} + 72 x + 16$

As you can see, when we simplify our initial expression, we get our answer which is $81 {x}^{2} + 72 x + 16$.