ORCCA Dividing by a Monomial

Section 6.7 Dividing by a Monomial

Now that we know how to add, subtract, and multiply polynomials, we will learn how to divide a polynomial by a monomial.

Subsection 6.7.1 Dividing a Polynomial by a Monomial

One example of dividing a polynomial is something we already studied in Section 4.7, when we rewrote an equation in standard form in slope-intercept form. We'll briefly review this process.

Example 6.7.1.

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 divided each term on the right side of the equation by \(-2\text{.}\)

This is an example of polynomial division that we have already done. We'll extend it to more complicated examples, many of which involve dividing polynomials by variables (instead of just numbers).

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 by 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 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.

Example 6.7.2.

Simplify \(\dfrac{2x^3+4x^2-10x}{2}\text{.}\)

The first step will be to recognize that the \(2\) we're dividing by will be divided into every term of the numerator. Once we recognize that, we will simply perform 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*}

Example 6.7.3.

Simplify \(\dfrac{15x^4-9x^3+12x^2}{3x^2}\)

Explanation

The key to simplifying \(\frac{15x^4-9x^3+12x^2}{3x^2}\) is to recognize that each term in the numerator will be divided by \(3x^2\text{.}\) In doing this, each coefficient and exponent will change. Performing this division by distributing, we get:

\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 6.7.4.

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. Here's how Example 6.7.3 would be visualized:

\begin{equation*} \frac{15x^4-9x^3+12x^2}{3x^2}=\fbox{ }x^{\fbox{ }} - \fbox{ }x^{\fbox{ }} + \fbox{ }x^{\fbox{ }} \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*}

(Note that \(\frac{x^2}{x^2}\) is technically \(x^0\text{,}\) which is equivalent to \(1\text{.}\))

Example 6.7.5.

Simplify \(\dfrac{20x^3y^4+30x^2y^3-5x^2y^2}{-5xy^2}\)

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 6.7.6.

Example 6.7.7.

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.

Explanation

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.