# How Do You Multiply Parentheses?

Here you’ll learn how to multiply two parentheses by each other. The most important thing to remember is that each term in each of the parentheses needs to be multiplied by all the terms in the other set of parentheses.

Rule

### MultiplyingTwoParenthesesbyEachOther

$\begin{array}{cc}\left(a+b\right)\left(c+d\right)& \\ =& \\ ac+ad+bc+bd& \end{array}$

(a + b)(c + d) = ac + ad + bc + b d

Example 1

Expand the parentheses $\left(x+2\right)\left(x+3\right)$

$\begin{array}{llll}\hfill \left(x+2\right)\left(x+3\right)& =x\cdot x+x\cdot 3+2\cdot x+2\cdot 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+3x+2x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+5x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \left(x+2\right)\left(x+3\right)& =x\cdot x+x\cdot 3+2\cdot x+2\cdot 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+3x+2x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+5x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The order of multiplication does not matter as long as you perform all the operations. When you write how many you have of each variable, the number goes in front of the variable. When you present your answer, you write the terms of the highest degree first, and then the rest in descending order.

In this case, that means ${x}^{2}$ goes first, as that is the term with the highest order. If you had a term with ${x}^{3}$, that would go in front of ${x}^{2}$.

What do you think will happen if there are negative terms inside the parentheses?

The order of operations is simple. When you multiply two terms by each other, you should always focus on SIGN, NUMBER, and VARIABLE. If you do that for each term, you’ll end up with the correct answer.

Example 2

Expand the parentheses $\phantom{\rule{-0.17em}{0ex}}\left(x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)$

$\begin{array}{llll}\hfill & \phantom{=}\phantom{\rule{-0.17em}{0ex}}\left(x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x\cdot x+x\cdot 3+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+3x-2x-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+x-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(x+3\right)& =x\cdot x+x\cdot 3+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot x+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+3x-2x-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+x-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 3

Expand the parentheses $\left(x-2\right)\left(x-3\right)$

$\begin{array}{llll}\hfill & \phantom{=}\phantom{\rule{-0.17em}{0ex}}\left(x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x\cdot x+x\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot x+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}-3x-2x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}-5x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)& =x\cdot x+x\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot x+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}-3x-2x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}-5x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 4

Expand the parentheses $\phantom{\rule{-0.17em}{0ex}}\left(-x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)$

$\begin{array}{llll}\hfill & \phantom{=}\phantom{\rule{-0.17em}{0ex}}\left(-x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(-x\right)\cdot x+\phantom{\rule{-0.17em}{0ex}}\left(-x\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot x+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-{x}^{2}+3x-2x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-{x}^{2}+x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(-x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(x-3\right)& =\phantom{\rule{-0.17em}{0ex}}\left(-x\right)\cdot x+\phantom{\rule{-0.17em}{0ex}}\left(-x\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot x+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-{x}^{2}+3x-2x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-{x}^{2}+x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 5

Expand the parentheses $\phantom{\rule{-0.17em}{0ex}}\left(-x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(-x-3\right)$

$\begin{array}{llll}\hfill & \phantom{=}\phantom{\rule{-0.17em}{0ex}}\left(-x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(-x-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\phantom{\rule{-0.17em}{0ex}}\left(-x\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+\phantom{\rule{-0.17em}{0ex}}\left(-x\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+3x+2x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+5x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(-x-2\right)\phantom{\rule{-0.17em}{0ex}}\left(-x-3\right)& =\phantom{\rule{-0.17em}{0ex}}\left(-x\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-x\right)+\phantom{\rule{-0.17em}{0ex}}\left(-x\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+\phantom{\rule{-0.17em}{0ex}}\left(-2\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(-3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+3x+2x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={x}^{2}+5x+6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$