# 2.8: Solving Polynomial Equations

- Page ID
- 92367

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We begin by introducing a property that will be used extensively in this and future sections.

If the product of two or more numbers equals zero, then at least one of the numbers must equal zero. That is, if

\(ab =0\)

then

\(a = 0\) or \(b =0\)

Let’s use the *zero factor property* to solve a few equations.

Solve for \(x\): \((x+3)(x-5)=0\)

**Solution **

The product of two factors equals zero.

\[(x+3)(x-5)=0 \nonumber \]

Hence, at least one of the factors must equal zero. Using the zero product property, set each factor equal to zero, then solve the resulting equations for \(x\).

\[\begin{aligned} x+3 &=0 \\ x &=-3 \end{aligned} \nonumber \]

or

\[\begin{aligned} x-5 &=0 \\ x &=5 \end{aligned} \nonumber \]

Hence, the solutions are \(x = −3\) and \(x =5\)

**Check: **

Check that each solution satisfies the original equation.

Substitute \(−3\) for \(x\):

\[\begin{aligned}(x+3)(x-5) &=0 \\(-3+3)(-3-5) &=0 \\(0)(-8) &=0 \\ 0 &=0 \end{aligned} \nonumber \]

Substitute \(5\) for \(x\):

\[\begin{aligned}(x+3)(x-5) &=0 \\(5+3)(5-5) &=0 \\(8)(0) &=0 \\ 0 &=0 \end{aligned} \nonumber \]

Because each check produces a true statement, both \(x = −3\) and \(x = 5\) are solutions of \((x + 3)(x−5) = 0\).

Solve for x: \((x-7)(x-2)=0\)

**Answer**-
\(7\), \(2\)

The zero product property also works equally well if more than two factors are present. For example, if \(abc = 0\), then either \(a = 0\) or \(b = 0\) or \(c = 0\). Let’s use this idea in the next example.

Solve for \(x\): \(x(2x + 9)(3x−5) = 0\)

**Solution**

The product of three factors equals zero.

\[x(2x + 9)(3x−5) = 0 \nonumber \]

Using the zero product property, set each factor equal to zero, then solve the resulting equations for \(x\).

\[x=0 \nonumber \]

or

\[\begin{align*} 2x + 9 &= 0\\ 2x &= -9\\ x &= -\dfrac{9}{2} \end{align*} \nonumber \]

or

\[\begin{align*} 3x - 5 &= 0\\ 3x &= 5\\ x &= \dfrac{5}{3} \end{align*} \nonumber \]

Hence, the solutions are \(x = 0\), \(x = −9/2\), and \(x =5 /3\). We encourage the reader to check the solution.

Solve for \(x\): \(6x(x + 4)(5x + 1) = 0\)

**Answer**-
\(0\), \(−4\), \(−1/5\)

## Linear versus Nonlinear Equations

Many equations that you have solved in the past were linear equations. Linear equations contain terms where the highest power of the variable we are solving for is one. Nonlinear equations have a power on the variable we are solving for that exceeds one. In this chapter we are learning how to solve nonlinear equations involving polynomials. L'ets briefly make sure we can recognize the difference between a linear and a nonlinear equation.

Use the following conditions to determine if an equation is linear or nonlinear.

- If the highest power of the variable we are solving for is one, then the equation is
*linear*. - If the highest power of the variable we are solving for is larger than one, then the equation is
*nonlinear*.

If the instruction is “solve for \(x\),” classify each of the following equations as linear or nonlinear.

- \(3x−5=4−7x\)
- \(x^2 =8x\)

**Solution**

Because the instruction is “solve for \(x\),” to determine whether the equation is linear or nonlinear, we identify the largest power of \(x\) present in the equation.

- The highest power of \(x\) present in the equation \(3x− 5=4− 7x\) is one. Hence, this equation is linear.
- The equation \(x^2 =8 x\) contains a power of \(x\) higher than one (it contains an \(x^2\)). Hence, this equation is nonlinear.

Solve: \(3 x−5=4−7x\)

**Solution**

Note that the largest power of the variable in the equation,\(x\), is one so the equation \(3x−5=4−7x\) is linear. Hence, the strategy is to move all terms containing the variable, \(x\), to one side of the equation, then move all the remaining terms to the other side of the equation.

\[\begin{array}{rlrl}{3 x-5} & {=4-7 x} & {\color {Red} \text { Original equation. }} \\ {3 x-5+7 x} & {=4} & {\color {Red} \text { Add } 7 x \text { to both sides. }} \\ {3 x+7 x} & {=4+5} & {\color {Red} \text { Add } 5 \text { to both sides. }}\end{array} \nonumber \]

Note how we have succeeded in moving all terms containing \(x\) to one side of the equation and all terms that do not contain \(x\) to the other side of the equation.

\[\begin{array}{rlrl}{10 x} & {=9} & {} & {\color {Red} \text { Simplify both sides. }} \\ {x} & {=\dfrac{9}{10}} & {} & {\color {Red} \text { Divide both sides by } 10 .}\end{array} \nonumber \]

Hence, the solution of \(3x−5=4−7x\) is \(x =9 /10\). Readers are encouraged to check this solution.

The situation is much different when the equation is nonlinear.

If an equation is *nonlinear*, first move everything to one side of the equation, making one side of the equation equal to zero. Continue the solution process by factoring and applying the zero *factor property*.

Solve: \(x^2 = 8x\)

**Solution**

Because the instruction is “solve for \(x\),” and the highest power of \(x\) is larger than one, the equation \(x^2 =8x\) is nonlinear. Hence, the strategy requires that we move all terms to one side of the equation, making one side zero.

\[\begin{array}{rlrl}{x^{2}} & {=8 x} \quad {\color {Red} \text { Original equation. }} \\ {x^{2}-8 x} & {=0} \quad {\color {Red} \text { Get zero on one side. }}\end{array} \nonumber \]

Next, we factor out the \(\mathrm{GCF}\) on the left-hand side.

\[x(x-8) = 0 \quad \color {Red} \text {Factor out the GCF.} \nonumber \]

By the zero factor property, either the first factor is zero or the second factor is zero.

\[\begin{array}{r}{x=0 \quad \text { or } \quad x-8=0} \\ {x=8}\end{array} \nonumber \]

Hence, the solutions are \(x = 0\) and \(x = 8\).

**Check: **

Check that each solution satisfies the original equation.

\[\begin{array}{l}{\text { Substitute } 0 \text { for } x :} \\ {\qquad \begin{aligned} x^{2} &=8 x \\(0)^{2} &=8(0) \\ 0 &=0 \end{aligned}}\end{array} \nonumber \]

\[\begin{array}{l}{\text { Subtitute } 8 \text { for } x :} \\ {\qquad \begin{aligned} x^{2} &=8 x \\(8)^{2} &=8(8) \\ 64 &=64 \end{aligned}}\end{array} \nonumber\]

Note that both results are true statements, guaranteeing that both \(x = 0\) and \(x = 8\) are solutions of \(x^2 =8x\)

Solve: \(t^2 =−5t\)

**Answer**-
\(0\), \(-5\)

Consider what would happen if we divided both sides of the equation \(x^2 =8x\) in Example \(\PageIndex{5}\) by \(x\):

\[\begin{aligned} x^{2} &=8 x \\ \dfrac{x^{2}}{x} &=\dfrac{8 x}{x} \\ x &=8 \end{aligned} \nonumber \]

We have lost one of the answers found in Example \(\PageIndex{5}\), \(x = 0\).

You should never divide by the variable you are solving for because you could be dividing by zero which is not mathematically possible. You also will lose answers to your equation.

Let’s try solving a nonlinear equation that requires factoring by grouping.

Solve: \(6x^2 +9x−8x−12 = 0\)

**Solution**

This equation is nonlinear. Begin by moving everything to one side of the equation, making one side equal to zero. Then factor.

Since zero is one side already, let’s factor the left-hand side by grouping since there are four terms,

\[\begin{align*} 6x^2+9x-8x-12 &=0 \\ (6x^2+9x)-(8x+12)&=0\\ 3x {\color{Red}(2x+3)}-4 {\color{Red}(2x+3)}&=0\\ (3x-4){\color {Red}(2x + 3)}&=0 \end{align*}\]

We now have a product of two factors that equals zero. Use the zero product property to write:

\[\begin{aligned} 3x-4 &=0 \\ 3x &= 4 \\ x &=\dfrac{4}{3} \end{aligned} \nonumber \]

or

\[\begin{aligned} 2x+3 &=0 \\ 2x &= -3 \\ x &= -\dfrac{3}{2} \end{aligned}\]

Hence, the solutions are \(x =4 /3\) and \(x =−3/2\). We leave it to the reader to check that the solutions in the original equation.

Solve: \(5w^2 −20w−4w + 16 = 0\)

**Answer**-
\(4/5\), \(4\)

## Using the Graphing Calculator

In this section we will employ two different calculator routines to find the solution of a nonlinear equation. Before picking up the calculator, let’s first use an algebraic method to solve the equation \(x^2 = −5x\). The equation is nonlinear, so the first step is to move everything to one side of the equation, making one side equal to zero.

\[\begin{aligned} x^{2} &= -5x \quad \color {Red} \text { Nonlinear. Make one side zero. } \\ x^{2}+5 x &= 0 \quad \color {Red} \text { Add } 5x \text { to both sides. } \\ x(x+5) &= 0 \quad \color {Red} \text { Factor out the GCF. } \end{aligned} \nonumber \]

Use the zero product property, setting each factor equal to zero, then solving the resulting equations for \(x\).

\[x=0 \nonumber \]

or

\[\begin{aligned} x+5&=0 \\ x&=-5 \end{aligned} \nonumber \]

Hence, the solutions are \(x = 0\) and \(x = −5\).

We’ll now use the calculator to find the solutions of \(x^2 = −5x\). The first technique employs the **5:intersect** routine on the calculator’s **CALC** menu.

Use the **5:intersect** utility on the graphing calculator to solve the equation \(x^2 = −5x\) for \(x\).

**Solution**

Load the left-hand side of \(x^2 = −5x\) in \(\mathbb{Y1}\) and the right-hand side in \(\mathbb{Y2}\) (see Figure \(\PageIndex{1}\)). Selecting **6:ZStandard** from the **ZOOM** menu produces the graphs shown in the image on the right in Figure \(\PageIndex{1}\).

Note that the graph of \(y = x^2\) is a parabola that opens upward, with vertex (turning point) at the origin. This graph reveals why the equation \(x^2 = −5x\) is called a nonlinear equation (not all the graphs involved are lines). Next, the graph of \(y = −5x\) is a line with slope \(−5\) and \(y\)-intercept at the origin.

The two graphs obviously intersect at the origin, but it also appears that there may be another point of intersection that is off the screen. Let’s increase \(\mathbb{Ymax}\) in an attempt to reveal the second point of intersection. After some experimentation, the settings shown in the first image in Figure \(\PageIndex{2}\) reveal both points of intersection. Pushing the **GRAPH** button produces the image on the right in Figure \(\PageIndex{2}\).

To find the solutions of the equation \(x^2 =−5x\), we must find the coordinates of the points where the graphs of \(y = x^2\) and \(y = −5x\) intersect. The \(x\)-coordinate of each point of intersection will be a solution of the equation \(x^2 = −5x\).

- Start by selecting
**5:intersect**from the**CALC**menu. When prompted for the “First curve?”, press**ENTER**. When prompted for the “Second curve?”, press**ENTER**. When prompted for a “Guess,” press**ENTER**. The result is the point \((0,0)\) shown in the image on the left in Figure \(\PageIndex{3}\). - Repeat the process a second time. Select
**5:intersect**from the**CALC**menu. When prompted for the “First curve?”, press**ENTER**. When prompted for the “Second curve?”, press**ENTER**. When prompted for a “Guess,” use the left-arrow key to move the cursor closer to the leftmost point of intersection, then press**ENTER**. The result is the point \((−5,25)\) shown in the image on the right in Figure \(\PageIndex{3}\).

Hence, the solutions of \(x^2 = −5x\) are \(x = −5\) and \(x = 0\).

Use the **5:intersect** utility on the graphing calculator to solve the equation \(x^2 =4x\) for \(x\).

**Answer**

Before demonstrating a second graphing calculator technique for solving nonlinear equations, let’s take a moment to metnion the relationship between zeros and x-intercepts.

The points where the graph of \(f\) crosses the \(x\)-axis are called the **\(x\)-intercepts** of the graph of \(f\). The \(x\)-value of each \(x\)-intercept is called a **zero** of the function \(f\).

We recently solved the equation \(x^2 = −5x\) by using the **5:intersect** feature on the graphing calculator. We’ll now employ the **2:zero** utility from the **CALC** menu to find the solutions of the equation \(x^2 = −5x\).

Use the **2:zero** utility on the graphing calculator to solve the equation \(x^2 = −5x\) for \(x\).

**Solution**

First, make one side of the equation equal to zero.

\[\begin{aligned}x^{2} &=-5 x \quad \color {Red} \text { Make one side zero. } \\ x^{2}+5 x &=0 \quad \color {Red} \text { Add } 5 x \text { to both sides. }\end{aligned} \nonumber \]

To determine the values of \(x\) that make \(x^2 +5x = 0\), we must locate the points where the graph of \(f(x)=x^2 +5x\) crosses the \(x\)-axis. These points are the \(x\)-intercepts of the graph of \(f\) and the \(x\)-values of these points are the zeros of the function \(f\).

Load the function \(f(x)=x^2 +5 x\) in \(\mathbb{Y1}\), then select **6:ZStandard** to produce the image in Figure \(\PageIndex{4}\). Note that the graph of \(f\) has two \(x\)-intercepts, and the \(x\)-values of each of these points are the zeros of the function \(f\).

Select **2:zero** from the **CALC** menu (see Figure \(\PageIndex{5}\)).

- The calculator responds by asking for a “Left Bound?” Use the left-arrow key to move the cursor so that it lies to the left of the \(x\)-intercept near \((−5,0)\) (see the second image in Figure \(\PageIndex{5}\)), then press the
**ENTER**key. - The calculator responds by asking for a “Right Bound?” Move the cursor so that is slightly to the right of the x-intercept near \((−5,0)\) (see the third image Figure \(\PageIndex{6}\)), then press the
**ENTER**key. - The calculator responds by asking for a “Guess?” Note the two triangular marks near the top of the viewing window in the first image in Figure \(\PageIndex{6}\) that mark the left- and right-bounds. As long as you place the cursor so that the x-value of the cursor location lies between these two marks, you’ve made a valid guess. Because the cursor already lies between these two marks, we usually leave it where it is and press the
**ENTER**key.

After making your guess and pressing the **ENTER** key, the calculator proceeds to find an approximation of the \(x\)-intercept that lies between the left- and right-bounds previously marked (see the second image in Figure \(\PageIndex{6}\). Hence, this \(x\)-intercept is \((−5,0)\), making \(−5\) a zero of \(f(x)=x^2 +5x\) and a solution of the equation \(x^2 +5x = 0\).

We’ll leave it to our readers to repeat the **2:zero** process to find the second zero at the origin.

Hence, the solutions of \(x^2 = −5x\) are \(x = −5\) and \(x = 0\).

It’s often easier to find the solutions of a nonlinear equation by making one side zero and identifying where the graph of the resulting function crosses the \(x\)-axis.