# (n+5) (n+4) = ?? someone help thank yoh

Mar 3, 2018

The result is ${n}^{2} + 9 n + 20$.

#### Explanation:

You can use the distributive property twice. First, distribute $\left(n + 5\right)$ onto $n$, and then onto $4$, like this:

$\textcolor{w h i t e}{=} \textcolor{b l u e}{\left(n + 5\right)} \textcolor{red}{\left(n + 4\right)}$

$= \textcolor{b l u e}{\left(n + 5\right)} \textcolor{red}{n} + \textcolor{b l u e}{\left(n + 5\right)} \textcolor{red}{4}$

$= \textcolor{red}{n} \textcolor{b l u e}{\left(n + 5\right)} + \textcolor{red}{4} \textcolor{b l u e}{\left(n + 5\right)}$

Now, use the distributive in each of these smaller parts:

$\textcolor{w h i t e}{=} \textcolor{red}{n} \textcolor{b l u e}{\left(n + 5\right)} + \textcolor{red}{4} \textcolor{b l u e}{\left(n + 5\right)}$

$= \textcolor{red}{n} \textcolor{b l u e}{n} + \textcolor{red}{n} \textcolor{b l u e}{5} + \textcolor{red}{4} \textcolor{b l u e}{\left(n + 5\right)}$

$= \textcolor{p u r p \le}{{n}^{2}} + \textcolor{b l u e}{5} \textcolor{red}{n} + \textcolor{red}{4} \textcolor{b l u e}{\left(n + 5\right)}$

$= \textcolor{p u r p \le}{{n}^{2}} + \textcolor{b l u e}{5} \textcolor{red}{n} + \textcolor{red}{4} \textcolor{b l u e}{n} + \textcolor{red}{4} \cdot \textcolor{b l u e}{5}$

$= \textcolor{p u r p \le}{{n}^{2}} + \textcolor{b l u e}{5} \textcolor{red}{n} + \textcolor{red}{4} \textcolor{b l u e}{n} + \textcolor{p u r p \le}{20}$

Lastly, combine the like terms:

$\textcolor{w h i t e}{=} \textcolor{p u r p \le}{{n}^{2}} + \textcolor{b l u e}{5} \textcolor{red}{n} + \textcolor{red}{4} \textcolor{b l u e}{n} + \textcolor{p u r p \le}{20}$

$= \textcolor{p u r p \le}{{n}^{2}} + \textcolor{p u r p \le}{9 n} + \textcolor{p u r p \le}{20}$

This is the result. (It is called a quadratic.)

Mar 3, 2018

$\textcolor{red}{{n}^{2}} + \textcolor{b l u e}{9} \textcolor{red}{n} + \textcolor{b l u e}{20}$

#### Explanation:

To solve this, we must multiply each variable in one bracket by each variable in the other brackets.

This is called distributing:

$\left(\textcolor{red}{n} + \textcolor{b l u e}{5}\right) \left(\textcolor{red}{n} + \textcolor{b l u e}{4}\right)$

becomes:

$\left(\textcolor{red}{n} \cdot \textcolor{red}{n}\right) + \left(\textcolor{red}{n} \cdot \textcolor{b l u e}{4}\right) + \left(\textcolor{red}{n} \cdot \textcolor{b l u e}{5}\right) + \left(\textcolor{b l u e}{5} \cdot \textcolor{b l u e}{4}\right)$

$= \textcolor{red}{{n}^{2}} + \textcolor{b l u e}{4} \textcolor{red}{n} + \textcolor{b l u e}{5} \textcolor{red}{n} + \textcolor{b l u e}{20}$

Simplifying:

$\to \textcolor{red}{{n}^{2}} + \textcolor{b l u e}{9} \textcolor{red}{n} + \textcolor{b l u e}{20}$

Thus, solved.