# 6.5 Divide monomials  (Page 2/4)

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Simplify: $\frac{{x}^{18}}{{x}^{22}}$ $\frac{{12}^{15}}{{12}^{30}}.$

$\frac{1}{{x}^{4}}$ $\frac{1}{{12}^{15}}$

Simplify: $\frac{{m}^{7}}{{m}^{15}}$ $\frac{{9}^{8}}{{9}^{19}}.$

$\frac{1}{{m}^{8}}$ $\frac{1}{{9}^{11}}$

Notice the difference in the two previous examples:

• If we start with more factors in the numerator, we will end up with factors in the numerator.
• If we start with more factors in the denominator, we will end up with factors in the denominator.

The first step in simplifying an expression using the Quotient Property for Exponents is to determine whether the exponent is larger in the numerator or the denominator.

Simplify: $\frac{{a}^{5}}{{a}^{9}}$ $\frac{{x}^{11}}{{x}^{7}}.$

## Solution

1. Is the exponent of $a$ larger in the numerator or denominator? Since 9>5, there are more $a\text{'}\text{s}$ in the denominator and so we will end up with factors in the denominator.
 Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$ . Simplify.
2. Notice there are more factors of $x$ in the numerator, since 11>7. So we will end up with factors in the numerator.
 Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$ . Simplify.

Simplify: $\frac{{b}^{19}}{{b}^{11}}$ $\frac{{z}^{5}}{{z}^{11}}.$

${b}^{8}$ $\frac{1}{{z}^{6}}$

Simplify: $\frac{{p}^{9}}{{p}^{17}}$ $\frac{{w}^{13}}{{w}^{9}}.$

$\frac{1}{{p}^{8}}$ ${w}^{4}$

## Simplify expressions with an exponent of zero

A special case of the Quotient Property is when the exponents of the numerator and denominator are equal, such as an expression like $\frac{{a}^{m}}{{a}^{m}}$ . From your earlier work with fractions, you know that:

$\frac{2}{2}=1\phantom{\rule{2em}{0ex}}\frac{17}{17}=1\phantom{\rule{2em}{0ex}}\frac{-43}{-43}=1$

In words, a number divided by itself is 1. So, $\frac{x}{x}=1$ , for any $x\phantom{\rule{0.2em}{0ex}}\left(x\ne 0\right)$ , since any number divided by itself is 1.

The Quotient Property for Exponents shows us how to simplify $\frac{{a}^{m}}{{a}^{n}}$ when $m>n$ and when $n by subtracting exponents. What if $m=n$ ?

Consider $\frac{8}{8}$ , which we know is 1.

$\begin{array}{cccccc}& & & \hfill \phantom{\rule{4em}{0ex}}\frac{8}{8}& =\hfill & 1\hfill \\ \text{Write 8 as}\phantom{\rule{0.2em}{0ex}}{2}^{3}.\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\frac{{2}^{3}}{{2}^{3}}& =\hfill & 1\hfill \\ \text{Subtract exponents.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{2}^{3-3}& =\hfill & 1\hfill \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{2}^{0}& =\hfill & 1\hfill \end{array}$

Now we will simplify $\frac{{a}^{m}}{{a}^{m}}$ in two ways to lead us to the definition of the zero exponent. In general, for $a\ne 0$ :

We see $\frac{{a}^{m}}{{a}^{m}}$ simplifies to ${a}^{0}$ and to 1. So ${a}^{0}=1$ .

## Zero exponent

If $a$ is a non-zero number, then ${a}^{0}=1$ .

Any nonzero number raised to the zero power is 1.

In this text, we assume any variable that we raise to the zero power is not zero.

Simplify: ${9}^{0}$ ${n}^{0}.$

## Solution

The definition says any non-zero number raised to the zero power is 1.

1. $\begin{array}{cccc}& & & \hfill \phantom{\rule{4em}{0ex}}{9}^{0}\hfill \\ \text{Use the definition of the zero exponent.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}1\hfill \end{array}$

2. $\begin{array}{cccc}& & & \hfill \phantom{\rule{4em}{0ex}}{n}^{0}\hfill \\ \text{Use the definition of the zero exponent.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}1\hfill \end{array}$

Simplify: ${15}^{0}$ ${m}^{0}.$

1 1

Simplify: ${k}^{0}$ ${29}^{0}.$

1 1

Now that we have defined the zero exponent, we can expand all the Properties of Exponents to include whole number exponents.

What about raising an expression to the zero power? Let’s look at ${\left(2x\right)}^{0}$ . We can use the product to a power rule to rewrite this expression.

$\begin{array}{cccc}& & & \hfill \phantom{\rule{4em}{0ex}}{\left(2x\right)}^{0}\hfill \\ \text{Use the product to a power rule.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{2}^{0}{x}^{0}\hfill \\ \text{Use the zero exponent property.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}1·1\hfill \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}1\hfill \end{array}$

This tells us that any nonzero expression raised to the zero power is one.

Simplify: ${\left(5b\right)}^{0}$ ${\left(-4{a}^{2}b\right)}^{0}.$

## Solution

1. $\begin{array}{cccc}& & & \hfill \phantom{\rule{4em}{0ex}}{\left(5b\right)}^{0}\hfill \\ \text{Use the definition of the zero exponent.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}1\hfill \end{array}$

2. $\begin{array}{cccc}& & & \hfill \phantom{\rule{4em}{0ex}}{\left(-4{a}^{2}b\right)}^{0}\hfill \\ \text{Use the definition of the zero exponent.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}1\hfill \end{array}$

Simplify: ${\left(11z\right)}^{0}$ ${\left(-11p{q}^{3}\right)}^{0}.$

$1$ $1$

Simplify: ${\left(-6d\right)}^{0}$ ${\left(-8{m}^{2}{n}^{3}\right)}^{0}.$

$1$ $1$

## Simplify expressions using the quotient to a power property

Now we will look at an example that will lead us to the Quotient to a Power Property.

$\begin{array}{cccc}& & & \hfill \phantom{\rule{4em}{0ex}}{\left(\frac{x}{y}\right)}^{3}\hfill \\ \text{This means:}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\frac{x}{y}·\frac{x}{y}·\frac{x}{y}\hfill \\ \text{Multiply the fractions.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\frac{x·x·x}{y·y·y}\hfill \\ \text{Write with exponents.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\frac{{x}^{3}}{{y}^{3}}\hfill \end{array}$

Notice that the exponent applies to both the numerator and the denominator.

We see that ${\left(\frac{x}{y}\right)}^{3}$ is $\frac{{x}^{3}}{{y}^{3}}$ .

$\begin{array}{cccc}\text{We write:}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{\left(\frac{x}{y}\right)}^{3}\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}\frac{{x}^{3}}{{y}^{3}}\hfill \end{array}$

Priam has pennies and dimes in a cup holder in his car. The total value of the coins is $4.21 . The number of dimes is three less than four times the number of pennies. How many pennies and how many dimes are in the cup? Cecilia Reply Arnold invested$64,000 some at 5.5% interest and the rest at 9% interest how much did he invest at each rate if he received $4500 in interest in one year Heidi Reply List five positive thoughts you can say to yourself that will help youapproachwordproblemswith a positive attitude. You may want to copy them on a sheet of paper and put it in the front of your notebook, where you can read them often. Elbert Reply Avery and Caden have saved$27,000 towards a down payment on a house. They want to keep some of the money in a bank account that pays 2.4% annual interest and the rest in a stock fund that pays 7.2% annual interest. How much should they put into each account so that they earn 6% interest per year?
324.00
Irene
1.2% of 27.000
Irene
i did 2.4%-7.2% i got 1.2%
Irene
I have 6% of 27000 = 1620 so we need to solve 2.4x +7.2y =1620
Catherine
I think Catherine is on the right track. Solve for x and y.
Scott
next bit : x=(1620-7.2y)/2.4 y=(1620-2.4x)/7.2 I think we can then put the expression on the right hand side of the "x=" into the second equation. 2.4x in the second equation can be rewritten as 2.4(rhs of first equation) I write this out tidy and get back to you...
Catherine
Darrin is hanging 200 feet of Christmas garland on the three sides of fencing that enclose his rectangular front yard. The length is five feet less than five times the width. Find the length and width of the fencing.
Mario invested $475 in$45 and $25 stock shares. The number of$25 shares was five less than three times the number of $45 shares. How many of each type of share did he buy? Jawad Reply let # of$25 shares be (x) and # of $45 shares be (y) we start with$25x + $45y=475, right? we are told the number of$25 shares is 3y-5) so plug in this for x. $25(3y-5)+$45y=$475 75y-125+45y=475 75y+45y=600 120y=600 y=5 so if #$25 shares is (3y-5) plug in y.
Joshua
will every polynomial have finite number of multiples?
a=# of 10's. b=# of 20's; a+b=54; 10a + 20b=$910; a=54 -b; 10(54-b) + 20b=$910; 540-10b+20b=$910; 540+10b=$910; 10b=910-540; 10b=370; b=37; so there are 37 20's and since a+b=54, a+37=54; a=54-37=17; a=17, so 17 10's. So lets check. $740+$170=$910. David Reply . A cashier has 54 bills, all of which are$10 or $20 bills. The total value of the money is$910. How many of each type of bill does the cashier have?
whats the coefficient of 17x
the solution says it 14 but how i thought it would be 17 im i right or wrong is the exercise wrong
Dwayne
17
Melissa
wow the exercise told me 17x solution is 14x lmao
Dwayne
thank you
Dwayne
A private jet can fly 1,210 miles against a 25 mph headwind in the same amount of time it can fly 1,694 miles with a 25 mph tailwind. Find the speed of the jet
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Mckenzie
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Mr Hernaez runs his car at a regular speed of 50 kph and Mr Ranola at 36 kph. They started at the same place at 5:30 am and took opposite directions. At what time were they 129 km apart?
90 minutes