Section 5.3 Dividing by a Monomial
We learned how to add and subtract polynomials in Section 1. Then in
[cross-reference to target(s) "section-introduction-to-exponent-properties" missing or not unique], we learned how to multiply monomials together (but not yet how to multiply general polynomials together). In this section we learn how to divide a general polynomial by a monomial.Subsection 5.3.1 Quotient of Powers Property
When we multiply the same base raised to powers, we add the exponents, as in \(2^{2}\cdot2^{3}=2^{5}\text{.}\) What happens when we divide the same base raised to powers?
Example 5.3.2.
Simplify \(\frac{x^5}{x^2}\) by first writing out what each power means.
Explanation.
Without knowing a property for simplifying this quotient of powers, we can write the expressions without exponents and simplify.
\begin{align*}
\frac{x^5}{x^2} \amp= \frac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x}\\
\amp= \frac{\cancel{x} \cdot \cancel{x} \cdot x \cdot x \cdot x}{\cancel{x} \cdot \cancel{x} \cdot 1}\\
\amp= \frac{x \cdot x \cdot x}{1}\\
\amp= x^3
\end{align*}
Notice that the difference of the exponents of the numerator and the denominator (\(5\) and \(2\text{,}\) respectively) is \(3\text{,}\) which is the exponent of the simplified expression.
When we divide as we’ve just done, we end up canceling factors from the numerator and denominator one-for-one. These common factors cancel to give us factors of \(1\text{.}\) The general rule for this is:
Fact 5.3.3. Quotient of Powers Property.
For any nonzero real number \(a\) and
[cross-reference to target(s) "item-integer-definition" missing or not unique] \(m\) and \(n\) where \(m\gt n\text{,}\)
\begin{equation*}
\frac{a^{m}}{a^{n}} = a^{m-n}
\end{equation*}
This rule says that when you’re dividing two expressions that have the same base, you can simplify the quotient by subtracting the exponents. In Example 2, this means that we can directly compute \(\frac{x^5}{x^2}\text{:}\)
\begin{align*}
\frac{x^5}{x^2} \amp= x^{5-2}\\
\amp=x^3
\end{align*}
Now we can update the list of exponent properties from
[cross-reference to target(s) "section-introduction-to-exponent-properties" missing or not unique].If \(a\) and \(b\) are real numbers, and \(m\) and \(n\) are positive integers, then we have the following properties:
- Product Property
- \(\displaystyle a^{m} \cdot a^{n} = a^{m+n}\)
- Power to a Power Property
- \(\displaystyle (a^{m})^{n} = a^{m\cdot n}\)
- Product to a Power Property
- \(\displaystyle (ab)^{m} = a^{m} \cdot b^{m}\)
- Quotient of Powers Property
- \(\dfrac{a^m}{a^n} = a^{m-n}\) (when \(m\gt n\))
Subsection 5.3.2 Dividing a Polynomial by a Monomial
Recall that dividing by a number \(c\) is the same as multiplying by the reciprocal \(\frac{1}{c}\text{.}\) For example, whether you divide \(8\) by \(2\) or multiply \(8\) by \(\frac{1}{2}\text{,}\) the result is \(4\) either way. In symbols,
\begin{equation*}
\frac{8}{2}=\frac{1}{2}\cdot8\qquad\text{(both work out to }4\text{)}
\end{equation*}
If we apply this idea to a polynomial being divided by a monomial, say with \(\frac{a+b}{c}\text{,}\) we can see that the distributive law works for this kind of division as well as with multiplication:
\begin{align*}
\frac{a+b}{c} \amp= \frac{1}{c}\cdot(a+b)\\
\amp= \frac{1}{c}\cdot a + \frac{1}{c}\cdot b\\
\amp= \frac{a}{c} + \frac{b}{c}
\end{align*}
In the end, the \(c\) has been “distributed” into the \(a\) and the \(b\text{.}\) Once we recognize that division by a monomial is distributive, we are left with individual monomial pairs that we can divide.
Example 5.3.5.
Simplify \(\dfrac{2x^3+4x^2-10x}{2}\text{.}\)
We recognize that the \(2\) we’re dividing by can be divided into each and every term of the numerator. Once we recognize that, we will simply perform those divisions.
\begin{align*}
\frac{2x^3+4x^2-10x}{2}\amp=\frac{2x^3}{2}+\frac{4x^2}{2}+\frac{-10x}{2}\\
\amp=x^3+2x^2-5x
\end{align*}
Example 5.3.6.
Simplify \(\dfrac{15x^4-9x^3+12x^2}{3x^2}\text{.}\)
Explanation.
We recognize that each term in the numerator can be divided by \(3x^2\text{.}\) To actually carry out that division we’ll need to use the Quotient of Powers Property. This is going to cause a change in each coefficient and exponent.
\begin{align*}
\frac{15x^4-9x^3+12x^2}{3x^2}\amp=\frac{15x^4}{3x^2}+\frac{-9x^3}{3x^2}+\frac{12x^2}{3x^2}\\
\amp=5x^2-3x+4
\end{align*}
Remark 5.3.7.
Once you become comfortable with this process, you might leave out the step where we wrote out the distribution. You will do the distribution in your head and this will become a one-step exercise. Here’s how Example 6 would be visualized:
\begin{equation*}
\frac{15x^4-9x^3+12x^2}{3x^2}=\boxed{\phantom{5}}x^{\boxed{\phantom{2}}} - \boxed{\phantom{3}}x^{\boxed{\phantom{1}}} + \boxed{\phantom{4}}x^{\boxed{\phantom{0}}}
\end{equation*}
And when calculated, we’d get:
\begin{equation*}
\frac{15x^4-9x^3+12x^2}{3x^2}=5x^2-3x+4
\end{equation*}
(With the last term, note that \(\frac{x^2}{x^2}\) reduces to \(1\text{.}\))
Example 5.3.8.
Simplify \(\dfrac{20x^3y^4+30x^2y^3-5x^2y^2}{-5xy^2}\text{.}\)
Explanation.
\begin{align*}
\frac{20x^3y^4+30x^2y^3-5x^2y^2}{-5xy^2}\amp=
\frac{20x^3y^4}{-5xy^2}+\frac{30x^2y^3}{-5xy^2}+\frac{-5x^2y^2}{-5xy^2}\\
\amp= -4x^2y^2 -6xy+x
\end{align*}
Checkpoint 5.3.9.
Simplify the following expression
\(\displaystyle\frac{{18r^{20}+18r^{16}-54r^{14}}}{{-6r^{2}}}=\)
Explanation.
We divide each term by \({-6r^{2}}\) as follows.
\begin{equation*}
\begin{aligned}
\displaystyle\frac{{18r^{20}+18r^{16}-54r^{14}}}{{-6r^{2}}} \amp = \frac{18r^{20}}{-6r^2}+\frac{18r^{16}}{-6r^2}+\frac{-54r^{14}}{-6r^2}\\
\amp ={-{\frac{18}{6}}r^{18} - {\frac{18}{6}}r^{14}+{\frac{54}{6}}r^{12}}\\
\amp ={-3r^{18}-3r^{14}+9r^{12}}\\
\end{aligned}
\end{equation*}
Example 5.3.10.
The density of an object, \(\rho\) (pronounced “rho”), can be calculated by the formula
\begin{equation*}
\rho=\frac{m}{V}
\end{equation*}
where \(m\) is the object’s mass, and \(V\) is its volume. The mass of a certain cancerous growth can be modeled by \(4t^3-6t^2+8t\) grams, where \(t\) is the number of days since the growth began. If its volume is \(2t\) cubic centimeters, find the growth’s density.
Explanation.
We have:
\begin{align*}
\rho\amp=\frac{m}{V}\\
\amp=\frac{4t^3-6t^2+8t}{2t}\,\frac{\text{g}}{\text{cm}^3}\\
\amp=\frac{4t^3}{2t}-\frac{6t^2}{2t}+\frac{8t}{2t}\quad\frac{\text{g}}{\text{cm}^3}\\
\amp=2t^2-3t+4\quad\frac{\text{g}}{\text{cm}^3}
\end{align*}
The growth’s density can be modeled by \(2t^2-3t+4\) g⁄cm3.
Reading Questions 5.3.3 Reading Questions
1.
How is dividing a polynomial by a monomial similar to distributing multiplication over a polynomial? For example, how is the process of simplifying \(\frac{15x^3+5x^2+10x}{5x}\) similar to simplifying \(5x\left(15x^3+5x^2+10x\right)\text{?}\)
Exercises 5.3.4 Exercises
Skills Practice
Quotient of Powers Property.
Use properties of exponents to simplify the expression.
1.
\(\displaystyle{\frac{m^{6}}{m^{4}}}\)
2.
\(\displaystyle{\frac{p^{9}}{p^{8}}}\)
3.
\(\displaystyle{\frac{t^{7}}{t}}\)
4.
\(\displaystyle{\frac{v^{4}}{v}}\)
5.
\(\displaystyle{\frac{24y^{9}}{3y^{3}}}\)
6.
\(\displaystyle{\frac{18b^{7}}{3b^{6}}}\)
7.
\(\displaystyle{\frac{-12d^{4}}{-4d^{2}}}\)
8.
\(\displaystyle{\frac{-27g^{9}}{3g^{5}}}\)
9.
\(\displaystyle{\frac{2i^{9}}{14i^{6}}}\)
10.
\(\displaystyle{\frac{2m^{5}}{10m^{3}}}\)
11.
\(\displaystyle{\frac{48t^{3}}{54t^{2}}}\)
12.
\(\displaystyle{\frac{35c^{8}}{63c^{2}}}\)
13.
\(\displaystyle{\frac{182^{470}}{182^{254}}}\)
14.
\(\displaystyle{\frac{194^{578}}{194^{296}}}\)
15.
\(\displaystyle{\frac{42^{119}\cdot 68^{129}}{42^{51}\cdot 68^{76}}}\)
16.
\(\displaystyle{\frac{49^{175}\cdot 47^{150}}{49^{86}\cdot 47^{99}}}\)
17.
\(\displaystyle{\frac{g^{16}v^{14}}{g^{11}v^{12}}}\)
18.
\(\displaystyle{\frac{i^{18}k^{22}}{i^{5}k^{20}}}\)
19.
\(\displaystyle{\frac{10a^{13}y^{17}}{35a^{11}y^{10}}}\)
20.
\(\displaystyle{\frac{42k^{39}r^{23}}{48k^{19}r^{16}}}\)
Dividing Polynomials by Monomials.
Simplify the expression.
21.
\(\displaystyle{\frac{35i^{9}+63i^{5}}{7}}\)
22.
\(\displaystyle{\frac{16c^{9}+40c^{8}}{8}}\)
23.
\(\displaystyle{\frac{63v^{15}+54v^{11}}{9v^{6}}}\)
24.
\(\displaystyle{\frac{8n^{16}+4n^{15}}{2n^{7}}}\)
25.
\(\displaystyle{\frac{-7f^{17}+9f^{13}+2f^{15}}{f^{8}}}\)
26.
\(\displaystyle{\frac{-5z^{13}+4z^{12}+9z^{11}}{z^{4}}}\)
27.
\(\displaystyle{\frac{-20x^{8}-12x^{12}+20x^{9}}{4x^{3}}}\)
28.
\(\displaystyle{\frac{45x^{14}+25x^{7}-20x^{8}}{5x^{5}}}\)
29.
\(\displaystyle{\frac{36d^{9}-18d^{6}}{6d}}\)
30.
\(\displaystyle{\frac{28x^{9}-49x^{10}}{7x}}\)
31.
\(\displaystyle{\frac{72p^{9}w^{9}+16p^{8}w^{13}}{8p^{4}w^{4}}}\)
32.
\(\displaystyle{\frac{54h^{16}w^{16}+45h^{11}w^{13}}{9h^{7}w^{9}}}\)
Applications
Exercise Group.
A rectangular prism’s base area can be calculated by the formula \(B=\frac{V}{h}\text{,}\) where \(V\) is the volume and \(h\) is the height.
33.
A certain rectangular prism’s volume can be modeled by \({6x^{5}+30x^{3}+18x^{2}}\) cubic units. If its height is \({3x}\) units, find the prism’s base area.
34.
A certain rectangular prism’s volume can be modeled by \({24x^{4}-16x^{3}+12x^{2}}\) cubic units. If its height is \({4x}\) units, find the prism’s base area.
Exercise Group.
A cylinder’s height can be calculated by the formula \(h=\frac{V}{B}\text{,}\) where \(V\) is the volume and \(B\) is the base area.
35.
A certain cylinder’s volume can be modeled by \({18\pi x^{7}-24\pi x^{5}-27\pi x^{3}}\) cubic units. If its base area is \({3\pi x^{2}}\) square units, find the cylinder’s height.
36.
A certain cylinder’s volume can be modeled by \({9\pi x^{7}-9\pi x^{4}+27\pi x^{3}}\) cubic units. If its base area is \({3\pi x^{2}}\) square units, find the cylinder’s height.
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