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# 12.3: Monomials and Powers

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## Introduction

The Laboratory

One of the places that the students were able to visit when they went downtown was a laboratory at the city college. Downtown, the city college had some of its classrooms and one of the classrooms was a laboratory.

“This is a good friend of mine Professor Smith,” Mr. Travis said introducing the students to a woman with blonde hair and a wide smile.

“Welcome,” Professor Smith said. “Are you enjoying your trip downtown?”

Many students responded yes and then were drawn over to one of the laboratory tables where a lot of work was taking place.

“What is happening here?” Sam asked.

“Well, I started with a very small sample of cobalt. I actually had 10 grams of it and I took a third of a third of a third of a third of it,” She explained.

The students began figuring the math out in their heads. Can you figure it out? How many grams did the sample end up being?

What You Will Learn

By the end of this lesson you will be able to complete the following skills.

• Multiply monomials by expanding the expression, regrouping factors and multiply coefficients.
• Recognize and apply the Power of a Product Property to numeric and variable monomial expressions.
• Recognize and apply the Power of a Quotient Property to numeric and variable monomial expressions.
• Recognize and apply the Power of a Power Property to numeric and variable monomial expressions.

Teaching Time

I. Multiply Monomials by Expanding the Expression, Regrouping Factors and Multiply Coefficients

You have already seen exponents, but let’s review the definitions of some words so that we can get started. First, let’s look at some of the parts of a term.

In the monomial above, the 7 is called the coefficient, the $x$ is the variable, and the 3 is the exponent.

We can say that the monomial $7x^3$ has a power of 3 or is to the $3^{rd}$ power.

Remember what we said about the coefficient of a variable like $x$—if there is no visible coefficient, then the coefficient is an unwritten 1. You could write “$1x$” but it is not necessary. Similarly, if there is no exponent on a coefficient or variable, then you can think of it as having an unwritten exponent of 1. So 7 could be written as $7^1$. The constant 7 then, is to the $1^{st}$ power.

Also, the exponent is applied to the constant, variable, or quantity that is directly to its left. That value is called the base. In the monomial above, the base is $x$. The exponent, in this case, is not applied to the 7 because it is not directly to the left of the exponent.

What is the exponent? It’s a shortcut. It’s a way of writing many multiplications in a simpler way. In the monomial above, $7x^3$, the 3 indicates that the variable $x$ is multiplied by itself three times.

$7x^3=7 \cdot x \cdot x \cdot x$

You can see the amount of space that is saved by using the exponent. When we write all of the multiplications instead of using the exponent, it is called the expanded form. You can see that it is, indeed, expanded—it takes much more space to write. Imagine if the exponent were greater, like $7x^{27}$.

$7x^{27}=7 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x$

The exponent truly saves a lot of space! Let’s look at an example where we write an expression out into expanded form.

Example

$& (7x^3)(4x^5)\\&=(7 \cdot x \cdot x \cdot x)(4 \cdot x \cdot x \cdot x \cdot x \cdot x)$

Here we have written the expression out into expanded form.

Now, use the commutative property of multiplication to change the order of the factors so that similar factors are next to each other.

$&=7 \cdot 4 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x && \text{Parentheses disappear.} \\&=28x^8 && \text{Multiply coefficients 7 and 4.}\\& && \text{Use exponent as a shortcut for the variables.}$

This may seem like a cumbersome long way to work, but it is accurate. There is a simpler way though-let’s take a look at the next section.

II. Recognize and Apply the Power of a Product Property to Numeric and Variable Monomial Expressions

In the previous section, we said that an exponent is applied to the constant, variable, or quantity that is directly to its left. However, we only applied exponents to single variables. Exponents can also be applied to products using parentheses.

Take the following example: $(5x)^4$

If we apply the exponent 4 to whatever is directly to its left, we would apply it to the parentheses, not just the $x$. The parentheses are directly to the left of the 4. This indicates that the entire product in the parentheses is taken to the $4^{th}$ power. As in the previous section, we can write this in expanded form.

$& (5x)^4 \\&=(5x)(5x)(5x)(5x)$

Now we multiply the monomials as we have already learned—by placing like factors next to each other, multiplying the coefficients, and simplifying using exponents.

$&=5 \cdot 5 \cdot 5 \cdot 5 \cdot x \cdot x \cdot x \cdot x \\&=625x^4$

This is the Power of a Product Property which says, for any nonzero numbers $a$ and $b$ and any integer $n$

$(ab)^n=a^n b^n$

Example

$& (7h)^3 \\&=(7h)(7h)(7h) \\&=7 \cdot 7 \cdot 7 \cdot h \cdot h \cdot h \\&=343 h^3$

Example

$& (-2x^4)^5 \\&=(-2x^4)(-2x^4)(-2x^4)(-2x^4)(-2x^4) \\&=(-2 \cdot x \cdot x \cdot x \cdot x)(-2 \cdot x \cdot x \cdot x \cdot x)(-2 \cdot x \cdot x \cdot x \cdot x)(-2 \cdot x \cdot x \cdot x \cdot x) \\&=-2 \cdot -2 \cdot -2 \cdot -2 \cdot -2 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \\&=-32x^{20}$

You can see that whether we have positive or negative integers or both, we can still use the Power of a Product Property. You may have already noticed a pattern with the exponents and the final product. When you multiply like bases, there is another shortcut—you can add the exponents of like bases. Another way of saying it is:

$a^m \cdot a^n=a^{m+n}$

Example

$& (-2x^4)^5 \\&=(-2x^4)(-2x^4)(-2x^4)(-2x^4)(-2x^4) \\&=-2 \cdot -2 \cdot -2 \cdot -2 \cdot -2 \cdot x^4 \cdot x^4 \cdot x^4 \cdot x^4 \cdot x^4 \\&=-2 \cdot -2 \cdot -2 \cdot -2 \cdot -2 \cdot x^{4+4+4+4+4}\\&=-32x^{20}$

Write the definition of this property and one example down in your notebook.

III. Recognize and Appy the Power of a Quotient Property to Numeric and Variable Monomial Expressions

This may sound confusing, but in math, we can rewrite this as $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}$ or $\left(\frac{1}{2} \right)^4$. We can use exponents with fractions or quotients, too. In order to answer the question above, we would multiply the numerators and denominators across, like this: $\frac{1 \cdot 1 \cdot 1 \cdot 1}{2 \cdot 2 \cdot 2 \cdot 2}=\frac{1}{16}$. Half of a half of a half of a half is one sixteenth. Once again, we have repeating multiplication of the same number which we could write more easily as $\frac{1^4}{2^4}=\frac{1}{16}$.

The Power of a Quotient Property says that for any nonzero numbers $a$ and $b$ and any integer $n$:

$\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

Example

$\left(\frac{5}{3}\right)^4=\frac{5^4}{3^4}=\frac{625}{81}$

You can see in this example that we have simplified the expression by figuring out what five to the fourth is and what three to the fourth is. The next step in this problem would be to divide.

Example

$\left(\frac{3k}{2j}\right)^4=\frac{(3k)^4}{(2j)^4}=\frac{(3k)(3k)(3k)(3k)}{(2j)(2j)(2j)(2j)}=\frac{81k^4}{16j^4}$

This problem has different variables, so this is as far as we can take this problem.

IV. Recognize and Appy the Power of a Power Property to Numeric and Variable Monomial Expressions

We have raised monomials to a power, products to a power, and quotients to a power. You can see that exponents are a useful tool in simplifying expressions. If you follow the rules of exponents, the patterns become clear. We have already seen powers taken to a power. For example, look at the quotient:

$\left(\frac{x^7}{y^9}\right)^4=\frac{(x^7)^4}{(y^9)^4}=\frac{(x^7)(x^7)(x^7)(x^7)}{(y^9)(y^9)(y^9)(y^9)}=\frac{x^{7+7+7+7}}{y^{9+9+9+9}}=\frac{x^{28}}{y^{36}}$

If you focus on just the numerator, you can see that $(x^7)^4=x^{28}$. You can get the exponent 28 by multiplying 7 and 4. This is an example of the Power of a Power Property which says for any nonzero numbers $a$ and $b$ and any integer $n$:

$(a^m)^n=a^{m \cdot n}$

Example

$(x^5)^3=x^{5.3}=x^{15}$

Example

$(x^6 y^3)^7=x^{6 \cdot 7} y^{3 \cdot 7}=x^{42} y^{21}$

Example

Combine the Power of a Quotient Property and the Power of a Power Property.

$\left(\frac{k^{10}}{f^6} \right)^4=\frac{k^{10 \cdot 4}}{f^{6 \cdot 4}}=\frac{k^{40}}{f^{24}}$

Now let’s go back and work on the problem from the introduction.

## Real-Life Example Completed

The Laboratory

Here is the original problem once again. Reread it and then solve for the number of grams in the sample.

One of the places that the students were able to visit when they went downtown was a laboratory at the city college. Downtown, the city college had some of its classrooms and one of the classrooms was a laboratory.

“This is a good friend of mine Professor Smith,” Mr. Travis said introducing the students to a woman with blonde hair and a wide smile.

“Welcome,” Professor Smith said. “Are you enjoying your trip downtown?”

Many students responded yes and then were drawn over to one of the laboratory tables where a lot of work was taking place.

“What is happening here?” Sam asked.

“Well, I started with a very small sample of cobalt. I actually had 10 grams of it and I took a third of a third of a third of a third of it,” She explained.

The students began figuring the math out in their heads. Can you figure it out? How many grams did the sample end up being?

Solution to Real – Life Example

To figure out the number of grams in the sample, we must use what we have learned about monomials and powers.

Professor Smith started off with 10 grams.

Then she took a third of a third of a third of a third of it. That is $\frac{1}{3}$ to the fourth power.

Here is how we can set up the problem.

$10 \left(\frac{1}{3}\right)^4=10 \left(\frac{1^4}{3^4}\right)=10 \left (\frac{1}{81}\right)=\frac{10}{81} \ grams$

We can convert that into a decimal by dividing the numerator by the denominator.

.12 grams is our answer as a decimal.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Monomial
a single term of variables, coefficients and powers.
Coefficient
the number part of a monomial or term.
Variable
the letter part of a term
Exponent
the little number, the power, that tells you how many times to multiply the base by itself.
Base
the number that is impacted by the exponent.
Expanded Form
write out all of the multiplication without an exponent.
Power of a Product Property
$(ab)^n=a^n(b^n)$
Power of a Quotient Property
the exponent is applied to both the top and bottom numbers in an expression.
Power of a Power Property
the exponent is applied to all the terms inside the parentheses by multiplying the powers together.

## Time to Practice

Directions: Simplify.

1. $(6x^5)(9x^2)$
2. $(-13d^5)(-5d^{15})$
3. $(3p^3 q^4)(7 pq^2 u^2)(2 pqu)$
4. $(10x)^4$
5. $(-4t^3)^5$
6. $(18 r^2 s^3)^2(2r^{11}s^3 t^2)^3$
7. $\left(\frac{2}{3}\right)^4$
8. $\left(\frac{7k}{-2m}\right)^3$
9. $\left(\frac{x^{11}}{y^9}\right)^5$
10. $(6a^4)^4$
11. $(w^{23}y^{21}z^{32})^3$
12. $\left(\frac{-5x^3}{3h^2 j^8}\right)^5$

Directions: Write out each expression in expanded form and simplify.

1. $(7x^2)(4x^9)$
2. $(12y^3)(4y^9)$
3. $(7x^6)(5x^5)$

Directions: Use what you have learned to solve each problem.

1. Write the equation for finding the area of this rectangle.
2. Find the area of the rectangle.
3. Find the volume of a cube whose side, $s$, is $15 g^4$.
4. A cheese-maker cuts a 54kg. block of cheese in half 5 times. How can you write this using exponents?
5. What is the mass of the smallest piece?

Jan 14, 2013

Aug 21, 2014