# Plus and Minus of Powers

When multiplying equal variables you get powers: $x\cdot x={x}^{2}$. Do not fall for the temptation of writing $x\cdot x=2x$. This is wrong! Because $x+x=2x$.

The rule below explains how you multiply equal variables.

Rule

Rule

### PowersofDifferentDegreesinTermswithVariables

You’ve just learned that different variables can’t be simplified or combined. Only equal variable combinations like $a$ and $a$ or $abc$ and $abc$ can be simplified. If you have an expression where you have powers with the same variables, but they’re different degrees, they can’t be simplified either. The factors ${a}^{2}$ and ${a}^{3}$ use the same variables, but of different degrees, so they’re not the same variable combination and can’t be combined. On the other hand, ${a}^{2}$ and ${a}^{2}$ have both the same variable and the same degree, and can therefore be simplified.

If there are numbers—called constants—in front of the variables, these should be multiplied as well. It is important to remember that you multiply the numbers together and the variables together, separately. When dividing, you cancel out equal variables just like when canceling numbers in a fraction.

Example 1

Simplify $a\cdot b+a\cdot a$

By using the rule above and rules for calculations with variables, you get

 $a\cdot b+a\cdot a=ab+{a}^{2}$

Example 2

Simplify

$\begin{array}{llll}\hfill & a\cdot a+a\cdot a\cdot a+2\cdot b\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+2a\cdot 3a\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$a\cdot a+a\cdot a\cdot a+2\cdot b+2a\cdot 3a$

By using the rules you’ve learned so far, you get

$\begin{array}{llll}\hfill & a\cdot a+a\cdot a\cdot a+2\cdot b\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+\phantom{\rule{-8.03003pt}{0ex}}\underset{2\cdot 3\cdot a\cdot a=6{a}^{2}}{\underset{⏟}{2a\cdot 3a}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{2}+{a}^{3}+2b+6{a}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{3}+7{a}^{2}+2b\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill a\cdot a+a\cdot a\cdot a+2\cdot b+\underset{2\cdot 3\cdot a\cdot a=6{a}^{2}}{\underset{⏟}{2a\cdot 3a}}& ={a}^{2}+{a}^{3}+2b+6{a}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{3}+7{a}^{2}+2b\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$