Section6.5Dividing Polynomials
¶Now that we know how to add, subtract, and multiply polynomials, we will learn how to divide a polynomial by a monomial.
Subsection6.5.1Dividing a Polynomial by a Monomial
Example6.5.2
Rewrite \(4x-2y=10\) in slope-intercept form.
In being asked to rewrite this equation in slope-intercept form, we're really being asked to solve the equation \(4x-2y=10\) for \(y\text{.}\)
\begin{align*} 7x-2y \amp= 10 \\ 7x-2y \highlight{{}-7x} \amp= 10\highlight{{}-7x} \\ -2y \amp= -7x + 10 \\ \frac{-2y}{\highlight{-2}} \amp= \frac{-7x + 10}{\highlight{-2}} \\ y \amp= -\frac{7}{2}x -5 \end{align*}In the final step of work, we dividing each term on the right side of the equation by \(-2\text{.}\) In doing this, we used distribution and we divided a polynomial by a monomial!
The fact is that we have already been doing polynomial division.
Like polynomial multiplication, polynomial division will rely upon distribution.
It's important to remember that dividing by a number \(c\) is the same as multiplying the reciprocal \(\frac{1}{c}\text{:}\)
\begin{equation*} \frac{8}{2}=\frac{1}{2}\cdot8 ~~~\text{ and }~~~\frac{9}{3}=\frac{1}{3}\cdot 9 \end{equation*}If we apply this to idea to a situation involving polynomials, say \(\frac{a+b}{c}\text{,}\) we can show that distribution works for division as well:
\begin{align*} \frac{a+b}{c} \amp= \frac{1}{c}\cdot(a+b) \\ \amp= \frac{1}{c}\cdot a + \frac{1}{2}\cdot b\\ \amp= \frac{a}{c} + \frac{b}{c} \end{align*}Once we recognize that the division distributes just as multiplication distributed, we are left with individual monomial pairs that we will divide.
Example6.5.3
Simplify \(\frac{2x^3+4x^2-10x}{2}\text{.}\)
The first step will be to recognize that the \(2\) we're dividing by will be divided into every term of the numerator. Once you recognize that, you will simply do that division.
\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*}Once you become comfortable with this process, you will often leave out the step where we wrote out the distribution. You will do the distribution in your head and this will often become a one-step problem.
Example6.5.4
Simplify \(\frac{15x^4-9x^3+12x^2}{3x^2}\)
Solution
\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*}
Example6.5.5
Simplify \(\frac{20x^3y^4+30x^2y^3-5x^2y^2}{-5xy^2}\)
Solution
\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*}
Exercise6.5.6
Exercise6.5.7
Example6.5.8
A rectangular prism's volume can be calculated by the formula
\begin{equation*} V=Bh \end{equation*}where \(V\)stands for volume, \(B\) stands for base area, and \(h\) stands for height. A certain rectangular prism's volume can be modeled by \(4x^3-6x^2+8x\) cubic units. If its height is \(2x\) units, find the prism's base area.
Solution
Since \(V=Bh\text{,}\) we can use \(B=\frac{V}{h}\) to calculate the base area. After substitution, we have:
\begin{align*} B\amp=\frac{V}{h}\\ \amp=\frac{4x^3-6x^2+8x}{2x}\\ \amp=\frac{4x^3}{2x}-\frac{6x^2}{2x}+\frac{8x}{2x}\\ \amp=2x^2-3x+4 \end{align*}The prism's base area can be modeled by \(2x^2-3x+4\) square units.
We can also look at a variation of division, where we have to fill in the missing gaps from the division process:
Example6.5.9
Find the missing coefficients and exponents that are identified by “\(\mathord{?}\)”.
\begin{equation*} \frac{\mathord{?}x^6-8x^5+\mathord{?}x^3}{4x^\mathord{?}}=5x^\mathord{?}+\mathord{?}x^\mathord{?}+6x \end{equation*}Once these values are found, rewrite the entire equation without any “\(\mathord{?}\)”.
To find these missing values, we need to recognize that the division taking place has two components:
division of coefficients
division of the variable factors
We'll want to consider each component separately.
We can start by rewriting the division using distribution and the original equation becomes:
\begin{equation*} \frac{\mathord{?}x^6}{4x^\mathord{?}}+\frac{-8x^5}{4x^\mathord{?}}+\frac{\mathord{?}x^3}{4x^\mathord{?}}=5x^\mathord{?}+\mathord{?}x^\mathord{?}+6x \end{equation*}To determine the missing coefficients, we need to look at each division problem individually:
- \(\frac{\mathord{?}x^6}{4x^\mathord{?}}=5x^\mathord{?}\)
The only coefficient that would work is if we started with \(20x^6\text{.}\)
- \(\frac{-8x^5}{4x^\mathord{?}}=\mathord{?}x^\mathord{?}\)
The only coefficient that would work is if we ended with \(-2x^\mathord{?}\text{.}\)
- \(\frac{\mathord{?}x^3}{4x^\mathord{?}}=6x\)
The only coefficient that would work is if we started with \(24x^4\text{.}\)
Now that the coefficients are known, we'll turn our attention to the variable factors. The complication in trying to figure out the exponents is that we don't know what the exponent in the denominator is. Figuring this out must be our first step.
When we look at the first term, we have \(\frac{20x^6}{4x^\mathord{?}}=5x^\mathord{?}\text{.}\) We know we had six \(x\)'s initially, but we don't know how many were canceled by division or how many remained after the canceling. We have the same issue with the second term \(\frac{-8x^5}{4x^\mathord{?}}=-2x^\mathord{?}\text{.}\)
Only in the third term, \(\frac{24x^3}{4x^\mathord{?}}=6x\) do we know how many \(x\)'s we started with (three) and how many remained after the division (one). The only way this could happen is if we had canceled two \(x\)'s, so we must have divided by \(4x^2\text{.}\)
Now we know the exponent of the denominator, we can complete the other two terms:
\begin{equation*} \frac{20x^6}{4x^2}=5x^4 ~~~\text{and}~~~ \frac{-8x^5}{4x^2}=-2x^3 \end{equation*}We have now identified every missing coefficient and exponent and can write the complete equation:
\begin{equation*} \frac{20x^6-8x^5+24x^3}{4x^2}=5x^4-2x^3+6x \end{equation*}Example6.5.10
Find the missing coefficients and exponents that are identified by “\(\mathord{?}\)”.
\begin{equation*} \frac{60x^\mathord{?}+48x^\mathord{?}+\mathord{?}x^4}{\mathord{?}x^\mathord{?}}=\mathord{?}x^6+8x^4-3 \end{equation*}Once these values are found, rewrite the entire equation without any “\(\mathord{?}\)”.
Solution
We can start by rewriting the division using distribution and the original equation becomes:
\begin{equation*} \frac{60x^\mathord{?}}{\mathord{?}x^\mathord{?}}+\frac{48x^\mathord{?}}{\mathord{?}x^\mathord{?}}+\frac{\mathord{?}x^4}{\mathord{?}x^\mathord{?}}=\mathord{?}x^6+8x^4-3 \end{equation*}To detemine the missing coefficients, we need to first determine the coefficient of the denominator. Only the middle term, \(\frac{48x^\mathord{?}}{\mathord{?}x^\mathord{?}}=8x^4\text{,}\) provides enough information to find that the coefficient of the denominator is \(6\text{.}\)
Now that we know the denominator is \(6x^\mathord{?}\text{,}\) we can fill in the other missing coefficients:
\begin{equation*} \frac{60x^\mathord{?}}{6x^\mathord{?}}+\frac{48x^\mathord{?}}{6x^\mathord{?}}+\frac{-18x^4}{6x^\mathord{?}}=10x^6+8x^4-3 \end{equation*}Again we'll turn our attention to the variable factors. The most useful term to consider at this moment is the third term, \(\frac{-18x^4}{6x^\mathord{?}}=-3\text{.}\) We know the original third term contained four \(x\)'s. We also know there are none left after the division. This means all four were canceled, so our denominator must be \(6x^4\text{.}\)
Now that we know the denominator is \(6x^4\text{,}\) we can fill in the other missing exponents:
\begin{equation*} \frac{60x^{10}}{6x^4}+\frac{48x^8}{6x^4}+\frac{-18x^4}{6x^?}=10x^6+8x^4-3 \end{equation*}Finally, we will write the complete equation:
\begin{equation*} \frac{60x^{10}+48x^8-18x^4}{6x^4}=10x^6+8x^4-3 \end{equation*}Subsection6.5.2Exercises
Dividing Polynomials by Monomials
Application Problems