6.5 Polynomial Equations
Topics covered in this section are:
- Use the Zero Product Property
- Solve quadratic equations by factoring
- Solve equations with polynomial functions
- 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 are 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. Listed below are some examples of quadratic equations:
| $x^{2}+5x+6=0$ | $3y^{2}+4y=10$ | $64u^{2}-81=0$ | $n(n+1)=42$ |
The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get $n^{2}+n$.
The general form of a quadratic equation is $ax^{2}+bx+c=0$, with $a≠0$. (If $a=0$, then $0 \cdot x^{2}=0$ and we are left with no quadratic term.)
QUADRATIC EQUATION
An equation of the form $ax^{2}+bx+c=0$ is called a quadratic equation.
$a$, $b$, and $c$, are real numbers and $a≠0$.
To solve quadratic equations we need methods different from the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter.
6.5.1 Use the Zero Product Property
We will first solve some quadratic equations by using 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 itself.
ZERO PRODUCT PROPERTY
If $a \cdot b=0$ then either $a=0$ or $b=0$ or both.
We will now use the Zero Product Property, to solve a quadratic equation.
Example 1
Solve: $(5n-2)(6n-1)=0$.
Solution
| Step 1. Set each factor equal to zero. | The product equals zero, so at least one factor must equal zero. | $(5n-2)(6n-1)=0$ $5n-2=0$ or $6n-1=0$ |
| Step 2. Solve the linear equations. | Solve each equation. | $n=\frac{2}{5}$ or $n=\frac{1}{6}$ |
| Step 3. Check. | Substitute each solution separately into the original equation. | $n=\textcolor{red}{\frac{2}{5}}\ \ \ $ $(5n-2)(6n-1)=0 \ \ \ \ $ $\left( 5 \cdot \textcolor{red}{\frac{2}{5}} – 2 \right) \left(6 \cdot \textcolor{red}{\frac{2}{5}} -1 \right) \overset{?}{=}0\ \ \ \ $ $(2-2) \left(\frac{12}{5}-1 \right) \overset{?}{=} 0\ \ \ \ $ $0 \cdot \frac{7}{5} \overset{?}{=}0\ \ \ \ $ $0=0 \ \checkmark$ $n=\textcolor{blue}{\frac{1}{6}}\ \ \ $ $\left( 5 \cdot \textcolor{blue}{\frac{1}{6}} – 2 \right) \left(6 \cdot \textcolor{blue}{\frac{1}{6}} -1 \right) \overset{?}{=}0\ \ \ \ $ $\left(\frac{5}{6}-\frac{12}{6} \right)(1-1) \overset{?}{=} 0\ \ \ \ $ $\left(-\frac{7}{6} \right)(0) \overset{?}{=} 0\ \ \ \ $ $0 =0 \ \checkmark$ |
HOW TO: Use the Zero Product Property.
- Set each factor equal to zero.
- Solve the linear equations.
- Check.
6.5.2 Solve Quadratic Equations by Factoring
The Zero Product Property works very nicely to solve quadratic equations. The quadratic equation must be factored, with zero isolated on one side. So we must be sure to start with the quadratic equation in standard form, $ax^{2}+bx+c=0$. Then we must factor the expression on the left.
Example 2
Solve $2y^{2}=13y+45$.
Solution
| Step 1. Write the quadratic equation in standard form, $ax^{2}+bx+x=0$. | Write in standard form. | $2y^{2}=13y+45$ $2y^{2}-13y-45=0$ |
| Step 2. Factor the quadratic expression. | $(2y+5)(y-9)=0$ | |
| Step 3. Use the Zero Product Property. | Set each factor equal to zero. We have two linear equations. | $2y+5=0 \ \ y-9=0$ |
| Step 4. Solve the linear equations. | $y=-\frac{5}{2} \ \ \ \ y=9$ | |
| Step 5. Check. | Substitute each solution separately into the original equation. | $\begin{align*} y&=\textcolor{red}{-\frac{5}{2}} \\ 2y^{2}&=13y+45 \\ 2 \left( \textcolor{red}{-\frac{5}{2}} \right)^{2} & = 13 \left( \textcolor{red}{-\frac{5}{2}} \right) +45 \\ 2 \left( \frac{25}{4} \right)& = \left( -\frac{65}{2} \right) + \frac{90}{2} \\ \frac{25}{2} &= \frac{25}{2} \ \checkmark \end{align*}$ $ \begin{align*} y&= \textcolor{blue}{9} \\ 2(\textcolor{blue}{9})^{2} & = 13(\textcolor{blue}{9})+45 \\ 2(81) & = 117+45 \\ 162 &=162 \ \checkmark \end{align*}$ |
HOW TO: Solve a quadratic equation by factoring.
- Write the quadratic equation in standard form, ax2+bx+c=0.
- Factor the quadratic expression.
- Use the Zero Product Property.
- Solve the linear equations.
- Check. Substitute each solution separately into the original equation.
Before we factor, we must make sure the quadratic equation is in standard form.
Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example?
Example 3
Solve $169q^{2}=49$.
Solution
| $169q^{2}=49$ | |
| Write the quadratic in standard form. | $169q^{2}-49=0$ |
| Factor. It is a difference of squares. | $(13q-7)(13q+7)$ |
| Use the Zero Product Property to set each factor equal to $0$. Solve each equation. | $\begin{align*} 13q-7&=0 \\ 13q&=7 \\ q&= \frac{7}{13} \\ \\ 13q+7&=0 \\ 13q&=-7 \\ q&=-\frac{7}{13} \end{align*}$ |
Check: We leave the check to you.
In the next example, the left side of the equation is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form.
Example 4
Solve: $(3x-8)(x-1)=3x$.
Solution
| $(3x-8)(x-1)=3x$ | |
| Multiply the binomials. | $3x^{2}-11x+8=3x$ |
| Write the quadratic equation in standard form. | $3x^{2}-14x+8=0$ |
| Factor the trinomial. | $(3x-2)(x-4)=0$ |
| Use the Zero Product Property to set each factor to $0$. Solve each equation. | $\begin{align*} 3x-2&=0 \\ 3x&=2 \\ x&=\frac{2}{3} \\ \\ x-4&=0 \\ x&=4 \end{align*}$ |
| Check your answers. | The check is left to you. |
In the next example, when we factor the quadratic equation we will get three factors. However the first factor is a constant. We know that factor cannot equal 0.
Example 5
Solve: $3x^{2}=12x+63$.
Solution
| $3x^{2}=12x+63$ | |
| Write the quadratic equation in standard form. | $3x^{2}-12x-63=0$ |
| Factor the GCF first. | $3(x^{2}-4x-21)=0$ |
| Factor the trinomial. | $3(x-7)(x+3)=0$ |
| Use the Zero Product Property to set each factor to $0$. Solve each equation. | $3≠0 \ \ x-7 = 0 \ \ x+3=0$ $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=7 \ \ \ \ \ \ \ \ \ x=-3$ |
| Check your answers. | The check is left to you. |
The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree greater than two by using the Zero Product Property, just like we solved quadratic equations.
Example 6
Solve: $9m^{3}+100m=60m^{2}$.
Solution
| $9m^{3}+100m=60m^{2}$ | |
| Bring all the terms to one side so that the other side is zero. | $9m^{3}-60m^{2}+100m=0$ |
| Factor the GCF first. | $m(9m^{2}-60m+100)=0$ |
| Factor the trinomial. | $m(3m-10)(3m-10)=0$ |
| Use the Zero Product Property to set each factor to $0$. Solve each equation. | $m=0 \ \ 3m-10 = 0$ $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m=\frac{10}{3}$ |
| Check your answers. | The check is left to you. |
6.5.3 Solve Equations with Polynomial Functions
As our study of polynomial functions continues, it will often be important to know when the function will have a certain value or what points lie on the graph of the function. Our work with the Zero Product Property will be help us find these answers.
Example 7
For the function $f(x)=x^{2}+2x-2$,
- find $x$ when $f(x)=6$
- find two points that lie on the graph of the function
Solution
Part 1
| $f(x)=x^{2}+2x-2$ | |
| Substitute $6$ for $f(x)$. | $6=x^{2}+2x-2$ |
| Put the quadratic in standard form. | $x^{2}+2x-8=0$ |
| Factor the trinomial. | $(x+4)(x-2)=0$ |
| Use the Zero Product Property. Solve. | $x+4=0 \ \ \ \ x-2=0$ $\ \ \ \ \ \ \ x=-4 \ \ \ \ \ \ \ \ x=2$ |
| Check: | |
| $\begin{align*} f(x)&=x^{2}+2x-2 \\ f(-4)&=(-4)^{2}+2(-4)-2 \\ f(-4)&=16-8-2 \\ f(-4)&=6 \ \checkmark \end{align*}$ | $\begin{align*} f(x)&=x^{2}+2x-2 \\ f(2)&=(2)^{2}+2(2)-2 \\ f(2)&=4+4-2 \\ f(2)&=6 \ \checkmark \end{align*}$ |
Part 2
Since $f(-4)=6$ and $f(2)=6$, the points $(-4, 6)$ and $(2, 6)$ lie on the graph of the function.
The Zero Product Property also helps us determine where the function is zero. A value of $x$ where the function is $0$, 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 $(x, 0)$ is a point on the graph. This point is an $x$-intercept of the graph. It is often important to know where the graph of a function crosses the axes. We will see some examples later.
Example 8
For the function $f(x)=3x^{2}+10x-8$, find
- the zeros of the function
- any $x$-intercepts of the graph of the function
- any $y$-intercepts of the graph of the function
Solution
Part 1
| $f(x)=3x^{2}+10x-8$ | |
| Substitute $0$ for $f(x)$. | $0=3x^{2}+10x-8$ |
| Factor the trinomial. | $(x+4)(3x-2)=0$ |
| Use the Zero Product Property. Solve. | $x+4=0 \ \ \ \ \ 3x-2=0$ $\ \ \ \ \ \ \ x=-4 \ \ \ \ \ \ \ \ \ \ \ x=\frac{2}{3}$ |
Part 2
An $x$-intercept occur when $y=0$. Since $f(-4)=0$ and $f \left( \frac{2}{3} \right) =0$, the points $(-4, 0)$ and $(\frac{2}{3}, 0)$ lie on the graph. These points are $x$-intercepts of the function.
Part 3
A $y$-intercept occurs when $x=0$. To find the $y$-intercepts, we need to find $f(0)$.
| $f(x)=3x^{2}+10x-8$ | |
| Find $f(0)$ by substituting $0$ for $x$. | $f(0)=3(0)^{2}+10(0)-8$ |
| Simplify. | $f(0)=-8$ |
Since $f(0)=-8$, the point $(0, -8)$ lies on the graph. This point is the $y$-intercept of the function.
6.5.4 Solve Applications Modeled by Polynomial Equations
The problem-solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to polynomial equations. We will copy the problem-solving strategy here so we can use it for reference.
HOW TO: Use a problem solving strategy to solve word problems.
- Read the problem. Make sure all the words and ideas are understood.
- Identify what we are looking for.
- Name what we are looking for. Choose a variable to represent that quantity.
- Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
- Solve the equation using appropriate algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
We will start with a number problem to get practice translating words into a polynomial equation.
Example 9
The product of two consecutive odd integers is $323$. Find the integers.
Solution
| Step 1. Read the problem. | |
| Step 2. Identify what we are looking for. | We are looking for two consecutive odd integers. |
| Step 3. Name what we are looking for. | Let $n=$ the first integer. $n+2=$ the next consecutive odd integer. |
| Step 4. Translate into an equation. Restate the problem in a sentence. | The product of two consecutive odd integers is $323$. |
| $n(n+2)=323$ | |
| Step 5. Solve the equation. | $n^{2}+2n=323$ |
| Bring all terms to one side. | $n^{2}+2n-323=0$ |
| Factor the trinomial. | $(n-17)(n+19)=0$ |
| Use the Zero Product Property. Solve the equations. | $n-17=0 \ \ \ \ n+19=0$ $ \ \ \ \ \ \ \ \ \ \ n=17 \ \ \ \ \ \ \ \ \ \ n=-19$ |
| There are two values for $n$ that are solutions to this problems. | So there are two sets of consecutive odd integers that will work. |
| If the first integer is $n=17$ | If the first odd integer is $-19$ |
| then the next odd integer is | then the next odd integer is |
| $n+2$ | $n+2$ |
| $17+2$ | $-19+2$ |
| $19$ | $-17$ |
| $17, 19$ | $ -17, -19$ |
| Step 6. Check the answer. | |
| The results are consecutive odd integers $17, 19$ and $-19, -17$ | $17 \cdot 19 = 323 \ \checkmark$ $(-19)(-17)=323 \ \checkmark$ |
| Step 7. Answer the question. | The consecutive integers are $17, 19$ and $-19, -17$. |
Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give positive results.
In some applications, negative solutions will result from the algebra, but will not be realistic for the situation.
Example 10
A rectangular bedroom has an area of $117$ square feet. Then length of the bedroom is four feet more than the width. Find the length and width of the bedroom.
Solution
| Step 1. Read the problem. In problems involving geometric figures, a sketch can help you visualize the situation. |  |
| Step 2. Identify what you are looking for. | We are looking for the length and width. |
| Step 3. Name what you are looking for. | Let $w=$ the width of the bedroom. |
| The length is four feet more than the width. | $w+4=$ the length of the garden. |
| Step 4. Translate into an equation. | |
| Restate the important information in a sentence. | The area of the bedroom is $117$ square feet. |
| Use the formula for the area of a rectangle. | $A = l \cdot w$ |
| Substitute in the variables. | $117=(w+4)w$ |
| Step 5. Solve the equation. | $117=w^{2}+4w$ |
| Get zero on one side. | $w^{2}+4w-117=0$ |
| Factor the trinomial. | $(w+13)(w-9)=0$ |
| Use the Zero Product Property. | $w+13=0 \ \ \ \ \ w-9=0$ |
| Solve each equation. | $w=-13 \ \ \ \ \ \ \ \ \ w=9$ |
| Since $w$ is the width of the bedroom, it does not make sense for it to be negative. We eliminate that value for $w$. | $\cancel{w=-13} \ \ \ \ \ \ \ \ w=9$ |
| The width is $9$ feet. | |
| Find the value of the length. | $w+4$ $9+4$ $13$ |
| Step 6. Check the answer. Does the answer make sense? |  Yes, this makes sense. |
| Step 7. Answer the question. | The width of the bedroom is $9$ feet and the length is $13$ feet. |
In the next example, we will use the Pythagorean Theorem $a^{2}+b^{2}=c^{2}$. This formula give the relation between the legs and the hypotenuse of a right triangle.
We will use this formula in the next example.
Example 11
A boat’s sail is in the shape of a right triangle as shown. The hypotenuse will be $17$ feet long. The length of one side will be $7$ feet less than the length of the other side. Find the lengths of the sides of the sail.
Solution
| Step 1. Read the problem. | |
| Step 2 Identify what you are looking for. | We are looking for the lengths of the sides of the sail. |
| Step 3. Name what you are looking for. One side is $7$ less than the other. | Let $x=$ the length of a side of the sail. $x-7=$ length of the other side. |
| Step 4. Translate into an equation. Since this is a right triangle, we can use the Pythagorean Theorem. | $a^{2}+b^{2}=c^{2}$ |
| Substitute the variables. | $x^{2}+(x-7)^{2}=17^{2}$ |
| Step 5. Solve the equation. | $x^{2}+x^{2}-14x+49=289$ |
| Simplify. | $2x^{2}-14x+49=289$ |
| It is a quadratic equation, so get zero on one side. | $2x^{2}-14x-240=0$ |
| Factor the GCF. | $2(x^{2}-7x-120)=0$ |
| Factor the trinomial. | $2(x-15)(x+8)=0$ |
| Use the Zero Product Property. | $2≠0 \ \ \ x-15=0 \ \ \ x+8=0$ |
| Solve. | $2≠0 \ \ \ \ \ \ x=15 \ \ \ \ \ \ \ \ \ x=-8$ |
| Since $x$ is a side of a triangle, $x=-8$ does not make sense. | $2≠0 \ \ \ \ \ \ x=15 \ \ \ \ \ \ \ \ \ \cancel{x=-8}$ |
| Find the length of the other side. | The length of one side is $15$. The length of the other side is $x-7$ $\textcolor{red}{15}-7$ $8$ |
| Step 6. Check the answer in the problem. Do these numbers make sense? |  Yes, these numbers make sense. |
| Step 7. Answer the question. | The sides of the sail are $8$, $15$, and $17$ feet. |
The next example uses the function that gives the height of an object as a function of time when it is thrown from $80$ feet above the ground.
Example 12
Dennis is going to throw his rubber band ball upward from the top of a campus building. When he throws the rubber band ball from $80$ feet above the ground, the function $h(t)=-16t^{2}+64t+80$ models the height, $h$, of the ball above the ground as a function of time, $t$. Find:
- the zeroes of this function which tell us when the ball hits the ground
- when the ball will be $80$ feet above the ground
- the height of the ball at $t=2$ seconds.
Solution
Part 1
The zeros of this function are found by solving $h(t)=0$. This will tell us when the ball will hit the ground.
| $h(t)=-16t^{2}+64t+80$ | |
| Set $h(t)=0$. | $-16t^{2}+64t+80=0$ |
| Factor out the GCF, $-16$. | $-16(t^{2}-4t-5)=0$ |
| Factor the trinomial. | $-16(t-5)(t+1)=0$ |
| Use the Zero Product Property. Solve. | $t-5=0 \ \ \ \ \ \ t+1=0$ $\ \ \ \ \ \ \ t=5 \ \ \ \ \ \ \ \ \ \ \ \ \ t=-1$ |
The result $t=5$ tells us the ball will hit the ground $5$ seconds after it is thrown. Since time cannot be negative, the result $t=-1$ is discarded.
Part 2
The ball will be $80$ feet above the ground when $h(t)=80$.
| $h(t)=-16t^{2}+64t+80$ | |
| Set $h(t)=80$. | $-16t^{2}+64t+80=80$ |
| Subtract $80$ from both sides. | $-16t^{2}+64t=0$ |
| Factor out the GCF, $-16t$. | $-16t(t-4)=0$ |
| Use the Zero Product Property. Solve. | $-16t=0 \ \ \ \ \ \ t-4=0$ $\ \ \ \ \ \ \ t=0 \ \ \ \ \ \ \ \ \ \ \ \ \ t=4$ |
The ball will be at $80$ feet the moment Dennis tosses the ball and then $4$ seconds later when the ball is falling.
Part 3
To find the height of the ball at $t=2$ seconds we need to find $h(2)$.
| $h(t)=-16t^{2}-64t+80$ | |
| To find $h(2)$, substitute $2$ for $t$. | $h(2)=-16(2)^{2}+64(2)+80$ |
| Simplify. | $h(2)=144$ |
| After $2$ seconds, the ball will be at $144$ feet. |
Licenses & Attributions
CC Licensed Content, Original
- Revision and Adaption. Provided by: Minute Math. License: CC BY 4.0
CC Licensed Content, Shared Previously
- Marecek, L., & Mathis, A. H. (2020). Polynomial Equations. In Intermediate Algebra 2e. OpenStax. https://openstax.org/books/intermediate-algebra-2e/pages/6-5-polynomial-equations. License: CC BY 4.0. Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
