Hi Everyone!

On this page you will find some material about Lesson 6. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

Lesson 6: Solving Equations by Using the Zero Product Rule


In this section you will find some important information about the specific resources related to this lesson:

  • the learning outcomes,
  • the section in the textbook,
  • the WeBWorK homework sets,
  • a link to the pdf of the lesson notes,
  • a link to a video lesson.

Learning Outcomes.

  • State the Zero Product Rule.
  • Solve an equation using the Zero Product Rule.

Topic. This lesson covers Section 4.8: Solving Equations by the Zero Product Rule.

WeBWorK. There is one WeBWorK assignment on today’s material:


Lesson Notes.

Video Lesson.

Video Lesson 6 (based on Lesson 6 Notes)



Warmup Questions

These are questions on fundamental concepts that you need to know before you can embark on this lesson. Don’t skip them! Take your time to do them, and check your answer by clicking on the “Show Answer” tab.

Warmup Question 1

Factor $3x^3-12x$.

Show Answer 1

\begin{align*} 3x^3-12x&=3x\underbrace{(x^2-4)}_{\text{difference of squares}}\\ &=3x(x-2)(x+2)\end{align*}

Warmup Question 2

Factor $3x^2+5x-2$.

Show Answer 2


Warmup Question 3

What must $x$ or $y$ be if $2xy=0$?

Show Answer 3

$x=0$ or $y=0$


If you are not comfortable with the Warmup Questions, don’t give up! Click on the indicated lesson for a quick catchup. A brief review will help you boost your confidence to start the new lesson, and that’s perfectly fine.

Need a review? Check Lesson 4 and Lesson 5.

Quick Intro

This is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.

A Quick Intro to Solving Equations by Using the Zero Product Rule

Key Words. Factoring, factor, solving an equation, the Zero Product Rule.

$\bigstar$ In the expression $(x+1)(x-4)$ the factors are the terms that are being multiplied: $x+1$ and $x-4$.

$\bigstar$ By the Warmup Question 3, we saw that the only way for $2xy=0$ is if $x$ or $y$ is zero. In general, a product can only be zero if one of the factors is zero. This is the Zero Product Rule.

$\bigstar$ In this lesson we solve equations by using the zero product rule. The first step is to set one of the sides equals zero. Then we factor the other side, and apply the Zero Product Rule by setting each factor equals zero.

Video Lesson

Many times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!

Video Lesson

A video lesson on Solving Equations by Using the Zero Product Rule [6:27]

A description of the video

In the video you will see how to solve

  • $(x-3)(x+4)=0$
  • $6x^3-29x^2+9x=0$

Try Questions

Now that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.

Try Question 1

Solve $(2x-3)(x+7)=0$ by using the Zero Product Rule.

Show Answer 1


By the Zero Product Rule,

\begin{align*}2x-3=0 \qquad &\text{ or }\qquad x+7=0\\2x=3\qquad &\text{ or }\qquad x=-7\\x=\dfrac{3}{2}\qquad &\text{ or }\qquad x=-7\end{align*}

The solution set is $\left\{-7,\dfrac{3}{2}\right\}$.

Try Question 2

Solve $6x^2-19x+10=0$ by using the Zero Product Rule.

Show Answer 2

\begin{align*} 6x^2-19x+10&=0 \\ (3x-2)(2x-5)&=0\end{align*}

\begin{align*} 3x-2=0\qquad & \text{ or }\qquad 2x-5=0\\3x=2\qquad & \text{ or }\qquad 2x=5\end{align*}

$$x=\dfrac{2}{3}\qquad\text{ or }\qquad x=\dfrac{5}{2}$$

The solution set is $\left\{\dfrac{2}{3},\dfrac{5}{2}\right\}$.

Try Question 3

Solve $4x^2+6x=0$ by using the Zero Product Rule.

Show Answer 3

\begin{align*} 4x^2+6x&=0\\ 2x(2x+3) &= 0 \\ x=0 \qquad & \text{ or }\qquad 2x+3 =0 \\x=0 \qquad & \text{ or }\qquad x=-\dfrac{3}{2} \end{align*}

The solution set is $\left\{-\dfrac{3}{2},0\right\}$.


You should now be ready to start working on the WeBWorK problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.


It is time to do the homework on WeBWork:


When you are done, come back to this page for the Exit Questions.

Exit Questions

After doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!

Exit Questions

  • What is the Zero Product Rule?
  • Is there an analogue of the Zero Product Rule if we replace 0 with 1?  Give an example supporting your answer. 
  • What is the goal of solving a quadratic equation? 
  • Must we check our answers even if we know we didn’t make a mistake? 

$\bigstar$ Solve the equation $x(x+1)=-x+35$ by using the Zero Product Rule.

Show Answer

\begin{align*}x(x+1)&=-x+35 \\ x^2+x&=-x+35\\x^2+2x-35&=0  \\(x+7)(x-5) &=0\\ \\ x+7 = 0 \qquad &\text{ or }\qquad x-5 = 0 \\ x=-7 \qquad &\text{ or }\qquad x=5 \end{align*}

The solution set is $\{-7,5\}$.

Need more help?

Don’t wait too long to do the following.

  • Watch the additional video resources.
Additional video resources on Solving Equations by Using the Zero Product Rule
  • Talk to your instructor.
  • Form a study group.
  • Visit a tutor. For more information, check the tutoring page.