Polynomial Equations

By the end of this section, you will be able to: use the Zero Product Property, solve quadratic equations by factoring, solve equations with polynomial functions, and solve applications modeled by polynomial equations.

We have spent considerable time learning how to factor polynomials. We will now look at polynomial equations and solve them using factoring, if possible.

A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

Polynomial equation. A polynomial equation is an equation that contains a polynomial expression. The degree of the polynomial equation is the degree of the polynomial.

We have already solved polynomial equations of degree one. Polynomial equations of degree one are linear equations of the form $ax+b=c$ .

We are now going to solve polynomial equations of degree two. A polynomial equation of degree two is called a quadratic equation. Some examples are

\begin{array}{l} x^2+5x+6=0 \\[4pt] 3y^2+4y=10 \\[4pt] 64u^2-81=0 \\[4pt] n(n+1)=42 \end{array}

The last equation does not appear to have the variable squared, but simplifying the left side gives $n^2+n$ .

The general form of a quadratic equation is $ax^2+bx+c=0$ , where $a\ne0$ . If $a=0$ , the squared term disappears and the equation is not quadratic.

Quadratic equation. An equation of the form

ax^2+bx+c=0

is called a quadratic equation, where

a

b

, and

c

are real numbers and

a\ne0

To solve quadratic equations we need methods different from the ones used for linear equations. We will use factoring here and study several other methods in a later chapter.

Use the Zero Product Property

The Zero Product Property says that if the product of two quantities is zero, then at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero.

Zero Product Property. If

a\cdot b=0

, then either

a=0

b=0

, or both.

Example. Solve $(5n-2)(6n-1)=0$ .

Set each factor equal to zero, then solve the two linear equations:

\begin{array}{rclcrcl} 5n-2&=&0 && 6n-1&=&0 \\[4pt] 5n&=&2 && 6n&=&1 \\[4pt] n&=&\tfrac{2}{5} && n&=&\tfrac{1}{6} \end{array}

Check each solution separately in the original equation:

\begin{array}{rcl} (5\cdot\tfrac{2}{5}-2)(6\cdot\tfrac{2}{5}-1)&=&0 \\[10pt] (2-2)(\tfrac{12}{5}-1)&=&0 \\[10pt] 0&=&0\ \checkmark \end{array}

\begin{array}{rcl} (5\cdot\tfrac{1}{6}-2)(6\cdot\tfrac{1}{6}-1)&=&0 \\[10pt] (\tfrac{5}{6}-\tfrac{12}{6})(1-1)&=&0 \\[10pt] 0&=&0\ \checkmark \end{array}

The solutions are $n=\tfrac{2}{5}$ and $n=\tfrac{1}{6}$ .

Solve $(3m-2)(2m+1)=0$ . Enter the two solutions separated by commas.

Solve $(4p+3)(4p-3)=0$ . Enter the two solutions separated by commas.

Use the Zero Product Property.

Set each factor equal to zero.
Solve the linear equations.
Check.

Solve quadratic equations by factoring

The Zero Product Property works well for solving quadratic equations. The quadratic equation must be factored, with zero isolated on one side. We first write the equation in standard form, $ax^2+bx+c=0$ , and then factor the expression on the left.

Example. Solve $2y^2=13y+45$ .

\begin{array}{lrcl} \text{Write the equation in standard form.} & 2y^2-13y-45&=&0 \\[4pt] \text{Factor the quadratic expression.} & (2y+5)(y-9)&=&0 \\[4pt] \text{Set each factor equal to zero.} & 2y+5&=&0 \quad\text{or}\quad y-9=0 \\[4pt] \text{Solve.} & y&=&-\tfrac{5}{2} \quad\text{or}\quad y=9 \end{array}

Substituting each solution into the original equation verifies both answers.

Solve $3c^2=10c-8$ . Enter the two solutions separated by commas.

Solve $2d^2-5d=3$ . Enter the two solutions separated by commas.

Solve a quadratic equation by factoring.

Write the quadratic equation in standard form, $ax^2+bx+c=0$ .
Factor the quadratic expression.
Use the Zero Product Property.
Solve the linear equations.
Check each solution separately in the original equation.

Before we factor, we must make sure the quadratic equation is in standard form. Solving quadratic equations by factoring uses all the factoring techniques learned in this chapter.

Example. Solve $169x^2=49$ .

\begin{array}{lrcl} \text{Write the equation in standard form.} & 169x^2-49&=&0 \\[4pt] \text{Factor the difference of squares.} & (13x-7)(13x+7)&=&0 \\[4pt] \text{Use the Zero Product Property.} & 13x-7&=&0 \quad\text{or}\quad 13x+7=0 \\[4pt] \text{Solve.} & x&=&\tfrac{7}{13} \quad\text{or}\quad x=-\tfrac{7}{13} \end{array}

The check is left to you.

Solve $25p^2=49$ . Enter the two solutions separated by commas.

Solve $36x^2=121$ . Enter the two solutions separated by commas.

In the next example, the left side is factored but the right side is not zero. To use the Zero Product Property, one side must be zero. We multiply the factors and then write the equation in standard form.

Example. Solve $(3x-8)(x-1)=3x$ .

\begin{array}{lrcl} \text{Multiply the binomials.} & 3x^2-11x+8&=&3x \\[4pt] \text{Write in standard form.} & 3x^2-14x+8&=&0 \\[4pt] \text{Factor the trinomial.} & (3x-2)(x-4)&=&0 \\[4pt] \text{Use the Zero Product Property.} & 3x-2&=&0 \quad\text{or}\quad x-4=0 \\[4pt] \text{Solve.} & x&=&\tfrac{2}{3} \quad\text{or}\quad x=4 \end{array}

The check is left to you.

Solve $(2m+1)(m+3)=12m$ . Enter the two solutions separated by commas.

Solve $(k+1)(k-1)=8$ . Enter the two solutions separated by commas.

When a quadratic equation factors into three factors, the first factor may be a nonzero constant. Such a factor cannot equal zero.

Example. Solve $3x^2=12x+63$ .

\begin{array}{lrcl} \text{Write in standard form.} & 3x^2-12x-63&=&0 \\[4pt] \text{Factor the GCF.} & 3(x^2-4x-21)&=&0 \\[4pt] \text{Factor the trinomial.} & 3(x-7)(x+3)&=&0 \\[4pt] \text{Use the Zero Product Property.} & x-7&=&0 \quad\text{or}\quad x+3=0 \\[4pt] \text{Solve.} & x&=&7 \quad\text{or}\quad x=-3 \end{array}

The constant factor $3$ is not zero and produces no solution.

Solve $18a^2-30=-33a$ . Enter the two solutions separated by commas.

Solve $123b=-6-60b^2$ . Enter the two solutions separated by commas.

The Zero Product Property also applies to a product of three or more factors. If the product is zero, at least one factor must be zero. We can therefore solve some equations of degree greater than two in the same way.

Example. Solve $9m^3+100m=60m^2$ .

\begin{array}{lrcl} \text{Bring all terms to one side.} & 9m^3-60m^2+100m&=&0 \\[4pt] \text{Factor the GCF.} & m(9m^2-60m+100)&=&0 \\[4pt] \text{Factor the trinomial.} & m(3m-10)(3m-10)&=&0 \\[4pt] \text{Use the Zero Product Property.} & m&=&0 \quad\text{or}\quad 3m-10=0 \\[4pt] \text{Solve.} & m&=&0 \quad\text{or}\quad m=\tfrac{10}{3} \end{array}

Solve $8x^3=24x^2-18x$ . Enter the distinct solutions separated by commas.

Solve $16y^2=32y^3+2y$ . Enter the distinct solutions separated by commas.

Solve equations with polynomial functions

As our study of polynomial functions continues, it will often be important to know when a function has a certain value or which points lie on its graph. The Zero Product Property will help us find these answers.

Example. For $f(x)=x^2+2x-2$ :

(a) Find $x$ when $f(x)=6$ .

Substitute $6$ for $f(x)$ , put the quadratic in standard form, and factor:

\begin{array}{lrcl} \text{Substitute.} & 6&=&x^2+2x-2 \\[4pt] \text{Write in standard form.} & x^2+2x-8&=&0 \\[4pt] \text{Factor.} & (x+4)(x-2)&=&0 \\[4pt] \text{Solve.} & x&=&-4 \quad\text{or}\quad x=2 \end{array}

(b) Find two points on the graph of the function.

Since $f(-4)=6$ and $f(2)=6$ , the points $(-4,6)$ and $(2,6)$ lie on the graph.

For $f(x)=x^2-2x-8$ , find $x$ when $f(x)=7$ . Enter the two values separated by commas.

For $f(x)=x^2-8x+3$ , find $x$ when $f(x)=-4$ . Enter the two values separated by commas.

The Zero Product Property also helps us determine where a function is zero. A value of $x$ where the function is zero is called a zero of the function.

Zero of a function. For any function

f

, if

f(x)=0

, then

x

is a zero of the function.

When $f(x)=0$ , the point $(x,0)$ lies on the graph. This point is an $x$ -intercept. We find the $y$ -intercept by setting $x=0$ .

Example. For $f(x)=3x^2+10x-8$ , find (a) the zeros of the function, (b) the $x$ -intercepts, and (c) the $y$ -intercept.

To find the zeros, solve $f(x)=0$ :

\begin{array}{lrcl} \text{Substitute 0 for }f(x). & 0&=&3x^2+10x-8 \\[4pt] \text{Factor.} & (x+4)(3x-2)&=&0 \\[4pt] \text{Solve.} & x&=&-4 \quad\text{or}\quad x=\tfrac{2}{3} \end{array}

Thus the zeros are $-4$ and $\tfrac{2}{3}$ , and the $x$ -intercepts are $(-4,0)$ and $(\tfrac{2}{3},0)$ . For the $y$ -intercept, find $f(0)$ :

f(0)=3\cdot0^2+10\cdot0-8=-8.

The $y$ -intercept is $(0,-8)$ .

For $f(x)=2x^2-7x+5$ , find the zeros of the function. Enter the two values separated by commas.

For $f(x)=6x^2+13x-15$ , find the zeros of the function. Enter the two values separated by commas.

Solve applications modeled by polynomial equations

The problem-solving strategy used earlier for applications that translate to linear equations works just as well for applications that translate to polynomial equations.

Use a problem-solving strategy to solve word problems.

Read the problem. Make sure all words and ideas are understood.
Identify what you are looking for.
Name what you are looking for. Choose a variable to represent it.
Translate into an equation. Restate the problem in one sentence with the important information, then translate that sentence.
Solve the equation using appropriate algebraic techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

Consecutive integer applications

Example. The product of two consecutive odd integers is $323$ . Find the integers.

Let $n$ be the first odd integer. Then $n+2$ is the next consecutive odd integer. Translate and solve:

\begin{array}{lrcl} \text{Translate.} & n(n+2)&=&323 \\[4pt] \text{Write in standard form.} & n^2+2n-323&=&0 \\[4pt] \text{Factor.} & (n-17)(n+19)&=&0 \\[4pt] \text{Solve.} & n&=&17 \quad\text{or}\quad n=-19 \end{array}

If $n=17$ , the next integer is $19$ . If $n=-19$ , the next integer is $-17$ . Both pairs check because $17\cdot19=323$ and $(-19)(-17)=323$ . The consecutive odd integers are $17,19$ and $-19,-17$ .

The product of two consecutive odd integers is 255. Enter all four integers in increasing order, separated by commas.

The product of two consecutive odd integers is 483. Enter all four integers in increasing order, separated by commas.

The product of two positive integers and the product of two negative integers both give positive results. In some applications, however, negative solutions from the algebra are not realistic for the situation.

Rectangle applications

Example. A rectangular bedroom has an area of $117$ square feet. The length is four feet more than the width. Find the dimensions.

Let $w$ be the width, so $w+4$ is the length.

Using $A=l\cdot w$ gives

\begin{array}{lrcl} \text{Translate.} & 117&=&(w+4)w \\[4pt] \text{Write in standard form.} & 0&=&w^2+4w-117 \\[4pt] \text{Factor.} & 0&=&(w+13)(w-9) \\[4pt] \text{Solve.} & w&=&-13 \quad\text{or}\quad w=9 \end{array}

A width cannot be negative, so $w=9$ . The length is $9+4=13$ . The bedroom is $9$ feet wide and $13$ feet long, and $9\cdot13=117$ .

A rectangular sign has area 30 square feet. Its length is one foot more than its width. Enter the width and length, separated by a comma.

A rectangular patio has area 180 square feet. Its width is three feet less than its length. Enter the width and length, separated by a comma.

Right-triangle applications

The Pythagorean Theorem, $a^2+b^2=c^2$ , gives the relation between the legs and hypotenuse of a right triangle.

Example. A boat’s sail is in the shape of a right triangle. The hypotenuse is $17$ feet. One side is $7$ feet less than the other. Find the side lengths.

Let $x$ be one leg and $x-7$ the other leg.

\begin{array}{lrcl} \text{Use the Pythagorean Theorem.} & x^2+(x-7)^2&=&17^2 \\[4pt] \text{Simplify.} & 2x^2-14x-240&=&0 \\[4pt] \text{Factor the GCF and trinomial.} & 2(x-15)(x+8)&=&0 \\[4pt] \text{Solve.} & x&=&15 \quad\text{or}\quad x=-8 \end{array}

The negative length is impossible, so $x=15$ and $x-7=8$ . The sides of the sail are $8$ , $15$ , and $17$ feet. The check is $15^2+8^2=225+64=289=17^2$ .

A right-triangle deck has one side 7 feet longer than the other and hypotenuse 13 feet. Enter the two leg lengths, separated by a comma.

A right-triangle meditation garden has one leg 7 feet and a hypotenuse one foot longer than the other leg. Enter the other leg and hypotenuse, separated by a comma.

Projectile applications

Example. Dennis throws a rubber-band ball upward from the top of an $80$ -foot building. Its height is modeled by $h(t)=-16t^2+64t+80$ . Find (a) when the ball hits the ground, (b) when it is $80$ feet above the ground, and (c) its height at $t=2$ seconds.

(a) The zeros tell us when the ball hits the ground:

\begin{array}{lrcl} \text{Set }h(t)=0. & -16t^2+64t+80&=&0 \\[4pt] \text{Factor.} & -16(t-5)(t+1)&=&0 \\[4pt] \text{Solve.} & t&=&5 \quad\text{or}\quad t=-1 \end{array}

Time cannot be negative, so the ball hits the ground after $5$ seconds.

(b) Set $h(t)=80$ :

\begin{array}{lrcl} \text{Substitute.} & -16t^2+64t+80&=&80 \\[4pt] \text{Subtract 80 and factor.} & -16t(t-4)&=&0 \\[4pt] \text{Solve.} & t&=&0 \quad\text{or}\quad t=4 \end{array}

The ball is $80$ feet high when Dennis releases it and again after $4$ seconds.

h(2)=-16(2)^2+64\cdot2+80=144.

After $2$ seconds, the ball is at $144$ feet.

A rock's height is $h(t)=-16t^2+48t+160$ . How many seconds after release does it hit the ocean?

For the rock with $h(t)=-16t^2+48t+160$ , find its height at $t=1.5$ seconds.

A penny's height is $h(t)=-16t^2+32t+128$ . How many seconds after release does it hit the ocean?

Key terms

polynomial equation — an equation that contains a polynomial expression. degree of the polynomial equation — the degree of the polynomial in the equation. quadratic equation — a polynomial equation of degree two. Zero Product Property — if a product is zero, then at least one of its factors is zero. zero of a function — a value of $x$ for which $f(x)=0$ .

This section is adapted from Intermediate Algebra 2e, Section 6.5: Polynomial Equations by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: reformatted the worked-example tables as accessible aligned math, omitted the Be Prepared quiz, media links, and end-of-section exercises, recreated geometric figures accessibly, and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

General Strategy for Factoring Polynomials Knowledge Check: Chapters 1–6