ORCCA Exponents and Polynomials Chapter Review

Section 6.7 Exponents and Polynomials Chapter Review

Subsection 6.7.1 Exponent Rules

In Section 6.1 we covered the exponent rules and how to use them.

Example 6.7.1 Quotients and Exponents

Let $t$ and $q$ be real numbers, where $q \neq 0$ and $t \neq 0\text{.}$ Find another way to write $\left(\frac{q^9}{t\cdot q^3}\right)^2\text{.}$

Explanation

We first use the Quotient Rule, then the Quotient to a Power Rule, then the Power to a Power Rule.

\begin{align*} \left(\frac{q^9}{t\cdot q^3}\right)^2\amp=\left(\frac{q^{9-3}}{t}\right)^2\\ \amp=\left(\frac{q^{6}}{t}\right)^2\\ \amp=\frac{q^{6\cdot 2}}{t^2}\\ \amp=\frac{q^{12}}{t^2} \end{align*}

Example 6.7.2 The Zero Exponent

Recall that the Zero Exponent Rule says that any real number raised to the $0$-power is $1\text{.}$ Using this, and the other exponent rules, find another way to write $-9^0\text{.}$

Explanation

Remember that in expressions like $-9^0\text{,}$ the exponent only applies to what it is directly next to! In this case, the $0$ only applies to the $9$ and not the negative sign. So,

\begin{equation*} -9^0=-1 \end{equation*}

Example 6.7.3 Negative Exponents

Write $5x^{-3}$ without any negative exponents.

Explanation

Recall that the Negative Exponent Rule says that a factor in the numerator with a negative exponent can be flipped into the denominator. So

\begin{equation*} 5x^{-3}=\frac{5}{x^3} \end{equation*}

Note that the $5$ does not move to the denominator because the $-3$ exponent only applies to the $x$ to which it is directly attached.

Example 6.7.4 Summary of Exponent Rules

Use the exponent rules in List 6.1.15 to write the expressions in a different way. Reduce and simplify when possible. Always find a way to write your final simplification without any negative exponents.

$\dfrac{24p^3}{20p^{12}}$
$\left(\dfrac{2v^5}{4g^{-2}}\right)^4$
$12n^7\left(m^0\cdot n^2\right)^2$
$\dfrac{k^5}{k^{-4}}$

Explanation

$\begin{aligned}[t] \frac{24p^3}{20p^{12}}\amp=\frac{24}{20}\cdot\frac{p^3}{p^{12}}\\ \amp=\frac{6}{5}\cdot p^{3-12}\\ \amp=\frac{6}{5}\cdot p^{-9}\\ \amp=\frac{6}{5}\cdot\frac{1}{p^9}\\ \amp=\frac{6}{5p^9}\end{aligned}$
$\begin{aligned}[t] \left(\frac{2v^5}{4g^{-2}}\right)^4\amp=\left(\frac{v^5}{2g^{-2}}\right)^4\\ \amp=\left(\frac{v^5g^2}{2}\right)^4\\ \amp=\frac{v^{5\cdot4}g^{2\cdot4}}{2^4}\\ \amp=\frac{v^{20}g^{8}}{16}\end{aligned}$
$\begin{aligned}[t] \amp12n^7\left(m^0\cdot n^2\right)^2\\ \amp=12n^7\left(1\cdot n^2\right)^2\\ \amp=12n^7\left(n^2\right)^2\\ \amp=12n^7n^{2\cdot 2}\\ \amp=12n^7n^4\\ \amp=12n^{7+4}\\ \amp=12n^{11}\end{aligned}$
$\begin{aligned}[t] \frac{k^5}{k^{-4}}\amp=k^5\cdot k^{4}\\ \amp=k^{5+4}\\ \amp=k^9\end{aligned}$

Subsection 6.7.2 Scientific Notation

In Section 6.2 we covered the definition of scientific notation, how to convert to and from scientific notation, and how to do some calculations in scientific notation.

Example 6.7.5 Scientific Notation for Large Numbers

The distance to the star Betelgeuse is about $3{,}780{,}000{,}000{,}000{,}000$ miles. Write this number in scientific notation.
The gross domestic product (GDP) of California in the year 2017 was about $\$2.746\times10^{13}\text{.}$ Write this number in standard notation.

Explanation

$3{,}780{,}000{,}000{,}000{,}000=3.78\times10^{15}\text{.}$
$\$2.746\times10^{13}=\$2{,}746{,}000{,}000{,}000\text{.}$

Example 6.7.6 Scientific Notation for Small Numbers

Human DNA forms a double helix with diameter $2\times10^{-9}$ meters. Write this number in standard notation.
A single grain of Forget-me-not (Myosotis) pollen is about $0.00024$ inches in diameter. Write this number in scientific notation.

Explanation

$2\times10^{-9}=0.000000002\text{.}$
$0.00024=2.4\times10^{-4}\text{.}$

Example 6.7.7 Multiplying and Dividing Using Scientific Notation

The fastest spacecraft so far have traveled about $5\times10^6$ miles per day.

If that spacecraft traveled at that same speed for $2\times10^4$ days (which is about $55$ years), how far would it have gone? Write your answer in scientific notation.
The nearest star to Earth, besides the Sun, is Proxima Centauri, about $2.5\times10^{13}$ miles from Earth. How many days would you have to fly in that spacecraft at top speed to reach Proxima Centauri

Explanation

Remember that you can find the distance traveled by multiplying the rate of travel times the time traveled: $d=r\cdot t\text{.}$ So this problem turns into

\begin{align*} d\amp=r\cdot t\\ d\amp=\left(\highlight{5}\times10^6\right)\cdot \left(\highlight{2}\times10^4\right)\\ \end{align*}
Multiply coefficient with coefficient and power of $10$ with power of $10\text{.}$
\begin{align*} \amp=\left(\highlight{5}\cdot\highlight{2}\right) \left(10^6\times10^4\right)\\ \amp=\highlight{10}\times10^{10}\\ \end{align*}
Remember that this still isn't in scientific notation. So we convert like this:
\begin{align*} \amp=\highlight{1.0\times10^1}\times10^{10}\\ \amp=1.0\times10^{11} \end{align*}

So, after traveling for $2\times10^4$ days (55 years), we will have traveled about $1.0\times10^{11}$ miles. That's one-hundred million miles. I hope someone remembered the snacks.
Since we are looking for time, let's solve the equation $d=r\cdot t$ for $t$ by dividing by $r$ on both sides: $t=\frac{d}{r}\text{.}$ So we have:

\begin{align*} t\amp=\frac{d}{r}\\ t\amp=\frac{2.5\times10^{13}}{5\times10^6}\\ \end{align*}
Now we can divide coefficient by coefficient and power of $10$ with power of $10\text{.}$
\begin{align*} t\amp=\frac{2.5}{5}\times\frac{10^{13}}{10^6}\\ t\amp=\highlight{0.5}\times10^7\\ t\amp=\highlight{5\times10^{-1}}\times10^7\\ t\amp=5\times10^6 \end{align*}

This means that to get to Proxima Centauri, even in our fastest spacecraft, would take $5\times10^6$ years. Converting to standard form, this is $5{,}000{,}000$ years. I think we're going to need a faster ship.

Subsection 6.7.3 Adding and Subtracting Polynomials

In Section 6.3 we covered the definitions of a polynomial, a term of a polynomial, a coefficient of a term, the degree of a term, the degree of a polynomial, theleading term of a polynomial, a constant term, monomials, binomials, and trinomials, and how to write a polynomial in standard form.

Example 6.7.8 Polynomial Vocabulary

Decide if the following statements are true or false.

The expression $\frac{3}{5}x^2-\frac{1}{5}x^7+\frac{x}{2}-4$ is a polynomial.
The expression $4x^6-3x^{-2}-x+1$ is a polynomial.
The degree of the polynomial $\frac{3}{5}x^2-\frac{1}{5}x^7+\frac{x}{2}-4$ is $10\text{.}$
The degree of the term $5x^2y^4$ is $6\text{.}$
The leading coefficient of $\frac{3}{5}x^2-\frac{1}{5}x^7+\frac{x}{2}-4$ is $\frac{3}{5}\text{.}$
There are 4 terms in the polynomial $\frac{3}{5}x^2-\frac{1}{5}x^7+\frac{x}{2}-4\text{.}$
The polynomial $\frac{3}{5}x^2-\frac{1}{5}x^7+\frac{x}{2}-4$ is in standard form.

Explanation

True. The expression $\frac{3}{5}x^2-\frac{1}{5}x^7+\frac{x}{2}-4$ is a polynomial.
False. The expression $4x^6-3x^{-2}-x+1$ is not a polynomial. Variables are only allowed to have whole number exponents in polynomials and the second term has a $-2$ exponent.
False. The degree of the polynomial $\frac{3}{5}x^2-\frac{1}{5}x^7+\frac{x}{2}-4$ is not $10\text{.}$ It is $7\text{,}$ which is the highest power of any variable in the expression.
True. The degree of the term $5x^2y^4$ is $6\text{.}$
False. The leading coefficient of $\frac{3}{5}x^2-\frac{1}{5}x^7+\frac{x}{2}-4$ is not $\frac{3}{5}\text{.}$The leading coefficient comes from the degree $7$ term which is $-\frac{1}{5}\text{.}$
True. There are 4 terms in the polynomial $\frac{3}{5}x^2-\frac{1}{5}x^7+\frac{x}{2}-4\text{.}$
False. The polynomial $\frac{3}{5}x^2-\frac{1}{5}x^7+\frac{x}{2}-4$ is not in standard form. The exponents have to be written from highest to lowest, i.e. $-\frac{1}{5}x^7+\frac{3}{5}x^2+\frac{x}{2}-4\text{.}$

Example 6.7.9 Adding and Subtracting Polynomials

Simplify the expression $\left(\frac{2}{9}x-4x^2-5\right)+\left(6x^2-\frac{1}{6}x-3\right)\text{.}$

Explanation

First identify like terms and group them either physically or mentally. Then we will look for common denominators for these like terms and combine appropriately.

\begin{align*} \amp\left(\frac{2}{9}x-4x^2-5\right)+\left(6x^2-\frac{1}{6}x-3\right)\\ \amp=\highlight{\frac{2}{9}x}\mathbin{\lighthigh{-}}\mathbin{\lighthigh{4x^2}}-5\mathbin{\lighthigh{+}}\mathbin{\lighthigh{6x^2}}\mathbin{\highlight{-}}\mathbin{\highlight{\frac{1}{6}x}}-3\\ \amp=\lighthigh{\left(-4x^2+6x^2\right)}+\highlight{\left(\frac{2}{9}x-\frac{1}{6}x\right)}+\left(-3-5\right)\\ \amp=\lighthigh{2x^2}+\highlight{\left(\frac{4}{18}x-\frac{3}{18}x\right)}-8\\ \amp=\lighthigh{2x^2}+\highlight{\frac{1}{18}x}-8 \end{align*}

Subsection 6.7.4 Multiplying Polynomials

In Section 6.4 we covered how to multiply two polynomials together using distribution, FOIL, and generic rectangles.

Example 6.7.10 Multiplying Binomials

Expand the expression $(5x-6)(3+2x)$ using the binomial multiplication method of your choice: distribution, FOIL, or generic rectangles.

Explanation

We will show work using the FOIL method.

\begin{align*} (5x-6)(3-2x)\amp=(5x\cdot3)+\left(5x\cdot(-2x)\right)+(-6\cdot3)+\left(-6\cdot(-2x)\right)\\ \amp=15x-10x^2-18+12x\\ \amp=-10x^2+27x-18 \end{align*}

Example 6.7.11 Multiplying Polynomials Larger than Binomials

Expand the expression $(3x-2)\left(4x^2-2x+5\right)$ by multiplying every term in the first factor with every term in the second factor.

Explanation

$\begin{aligned}[t]\amp(\highlight{3x}\mathbin{\lighthigh{-}}\mathbin{\lighthigh{2}})\left(4x^2-2x+5\right)\\ \amp=\highlight{3x}\cdot4x^2+\highlight{3x}\cdot(-2x)+\highlight{3x}\cdot5+(\mathbin{\lighthigh{-}}\mathbin{\lighthigh{2}})\cdot4x^2+(\mathbin{\lighthigh{-}}\mathbin{\lighthigh{2}})\cdot(-2x)+(\mathbin{\lighthigh{-}}\mathbin{\lighthigh{2}})\cdot5\\ \amp=12x^3-6x^2+15x-8x^2+4x-10\\ \amp=12x^3-14x^2+19x-10\end{aligned}$

Subsection 6.7.5 Special Cases of Multiplying Polynomials

In Section 6.5 we covered how to square a binomial and how to find the product of the sum or difference of two terms.

Example 6.7.12 Squaring a Binomial

Recall that Fact 6.5.4 gives formulas that help square a binomial.

Simplify the expression $(2x+3)^2\text{.}$

Explanation

Remember that you can use FOIL to do these problems, but in the interest of understanding concepts at a higher level for use in later chapters, we will use the relevant formula from Fact 6.5.4. In this case, since we have a sum of two terms being squared, we will use $(a+b)^2 = a^2+2ab+b^2\text{.}$

First identify $\highlight{a}$ and $\lighthigh{b}\text{.}$ In this case, $\highlight{a=2x}$ and $\lighthigh{b=3}\text{.}$ So, we have:

\begin{alignat*}{3} (\mathbin{\highlight{a}}+\mathbin{\lighthigh{b}})^2 \amp= (\mathbin{\highlight{a}})^2\amp\amp+2(\mathbin{\highlight{a}})(\mathbin{\lighthigh{b}})\amp\amp+(\mathbin{\lighthigh{b}})^2\\ (\mathbin{\highlight{2x}}+\mathbin{\lighthigh{3}})^2 \amp= (\mathbin{\highlight{2x}})^2\amp\amp+2(\mathbin{\highlight{2x}})(\mathbin{\lighthigh{3}})\amp\amp+(\mathbin{\lighthigh{3}})^2\\ \amp=4x^2\amp\amp+12x\amp\amp+9 \end{alignat*}

Example 6.7.13 The Product of the Sum and Difference of Two Terms

Recall that Fact 6.5.12 gives a formula to help multiply things that look like $(a+b)(a-b)\text{.}$

Simplify the expression $(7x+4)(7x-4)\text{.}$

Explanation

Remember that you can use FOIL to do these problems, but in the interest of understanding concepts at a higher level for use in later chapters, we will use the formula from Fact 6.5.12. In this case, that means we will use $(a+b)(a-b) = a^2-b^2\text{.}$

First identify $\highlight{a}$ and $\lighthigh{b}\text{.}$ In this case, $\highlight{a=7x}$ and $\lighthigh{b=4}\text{.}$ So, we have:

\begin{alignat*}{2} (\mathbin{\highlight{a}}+\mathbin{\lighthigh{b}})(\mathbin{\highlight{a}}-\mathbin{\lighthigh{b}}) \amp= (\mathbin{\highlight{a}})^2\amp\amp-(\mathbin{\lighthigh{b}})^2\\ (\mathbin{\highlight{7x}}+\mathbin{\lighthigh{4}})(\mathbin{\highlight{7x}}-\mathbin{\lighthigh{4}}) \amp= (\mathbin{\highlight{7x}})^2\amp\amp-(\mathbin{\lighthigh{4}})^2\\ \amp=49x^2\amp\amp-16 \end{alignat*}

Example 6.7.14 Binomials Raised to Other Powers

To raise binomials to powers higher than $2\text{,}$ we start by expanding the expression and multiplying all factors together from left to right.

Expand the expression $(2x-5)^3\text{.}$

Explanation

\begin{align*} \amp(2x-5)^3\\ \amp=\highlight{(2x-5)(2x-5)}(2x-5)\\ \amp=\highlight{\left[(2x)^2-2(2x)(5)+5^2\right]}(2x-5)\\ \amp=\highlight{\left[4x^2-20x+25\right]}(2x-5)\\ \amp=\highlight{\left[4x^2\right]}(2x)+\highlight{\left[4x^2\right]}(-5)+\highlight{\left[-20x\right]}(2x)+\highlight{\left[-20x\right]}(-5)+\highlight{\left[25\right]}(2x)+\highlight{\left[25\right]}(-5)\\ \amp=8x^3-20x^2-40x^2+100x+50x-125\\ \amp=8x^3-60x^2+150x-125 \end{align*}

Subsection 6.7.6 Dividing by a Monomial

In Section 6.6 we covered how you can split a fraction up into multiple terms if there is a sum or difference in the numerator. Mathematically, this happens using the rule $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}\text{.}$ This formula can be used for any number of terms in the numerator, and for both sums and differences.

Example 6.7.15

Expand the expression $\frac{12x^5+2x^3-4x^2}{4x^2}\text{.}$

Explanation

\begin{align*} \frac{12x^5+2x^3-4x^2}{4x^2}\amp=\frac{12x^5}{4x^2}+\frac{2x^3}{4x^2}-\frac{4x^2}{4x^2}\\ \amp=3x^3+\frac{x}{2}-1 \end{align*}

Subsection 6.7.7 Exercises

Exponent Rules

1

Use the properties of exponents to simplify the expression.

$\left(4r^{12}\right)^3$

2

Use the properties of exponents to simplify the expression.

$\left(2y^{2}\right)^2$

3

Use the properties of exponents to simplify the expression.

$\displaystyle{({6y^{4}})\cdot({9y^{17}})}$

4

Use the properties of exponents to simplify the expression.

$\displaystyle{({10y^{7}})\cdot({-8y^{11}})}$

5

Use the properties of exponents to simplify the expression.

$\displaystyle{\left({-\frac{y^{9}}{4}}\right) \cdot \left({\frac{y^{4}}{9}}\right)}$

6

Use the properties of exponents to simplify the expression.

$\displaystyle{\left({-\frac{r^{11}}{7}}\right) \cdot \left({-\frac{r^{16}}{8}}\right)}$

7

Use the properties of exponents to simplify the expression.

$\left(-21\right)^0=$

8

Use the properties of exponents to simplify the expression.

$\left(-16\right)^0=$

9

Use the properties of exponents to simplify the expression.

$-42^0=$

10

Use the properties of exponents to simplify the expression.

$-47^0=$

11

Use the properties of exponents to simplify the expression.

$\left(\displaystyle\frac{-3}{2x^{9}}\right)^{3}=$

12

Use the properties of exponents to simplify the expression.

$\left(\displaystyle\frac{-3}{8x^{4}}\right)^{3}=$

13

Use the properties of exponents to simplify the expression.

$\displaystyle\frac{{5r^{14}}}{{15r^{3}}}=$

14

Use the properties of exponents to simplify the expression.

$\displaystyle\frac{{7r^{4}}}{{14r^{3}}}=$

15

Use the properties of exponents to simplify the expression.

$\left(\displaystyle\frac{x^{7}}{2y^{6}z^{8}}\right)^{3}=$