# Equations Involving Factored Expressions

Examples, videos, and solutions to help Algebra I students learn how to solve equations involving factored expressions.

### New York State Common Core Math Algebra I, Module 1, Lesson 17

Lesson 17 Student Outcomes

Students learn that equations of the form (x - a)(x - b) = 0 have the same solution set as two equations joined by βorβ: x - a = 0 or x - b = 0.

Students solve factored or easily factorable equations.

Lesson 17 Summary
The zero-product property says that If ab = 0, then either a = 0 or b = 0 or a = b = 0.

Exercise 1

1. Solve each equation for π₯.
a. π₯ β 10 = 0
b. π₯2 + 20 = 0
c. Demanding Dwight insists that you give him two solutions to the following equation:
(π₯ β 10)(π₯2 + 20) = 0
Can you provide him with two solutions?
d. Demanding Dwight now wants FIVE solutions to the following equation:
(π₯ β 10)(2π₯ + 6)(π₯2 β36)(π₯2 + 10)(π₯/2 + 20) = 0
Can you provide him with five solutions?
Do you think there might be a sixth solution?
Consider the equation (π₯ β4)(π₯ + 3) = 0.
e. Rewrite the equation as a compound statement.
f. Find the two solutions to the equation.

Example 1
Solve 2π₯2 β 10π₯ = 0, for π₯.

Example 2
Solve π₯(π₯ β 3)+ 5(π₯ β 3) = 0, for π₯.

Exercises 2β7
2. (π₯ + 1)(π₯ + 2) = 0
3. (3π₯ β 2)(π₯ + 12) = 0
4. (π₯ β 3)(π₯ β 3) = 0
5. (π₯ + 4)(π₯ β 6)(π₯ β 10) = 0
6. π₯2 β 6π₯ = 0
7. π₯(π₯ β 5)+ 4(π₯ β 5) = 0

Example 3
Consider the equation (π₯ β2)(2π₯ β 3) = (π₯ β 2)(π₯ +5). Lulu chooses to multiply through by 1/π₯β2 and gets the answer π₯ = 8. But Poindexter points out that π₯ = 2 is also an answer, which Lulu missed.
a. Whatβs the problem with Luluβs approach?
b. Use factoring to solve the original equation for π₯.

Exercises 8β11
8. Use factoring to solve the equation for π₯: (π₯ β 2)(2π₯ β 3) = (π₯ β 2)(π₯ + 1).
9. Solve each of the following for π₯:
a. π₯ + 2 = 5
b. π₯2 + 2π₯ = 5π₯
c. π₯(5π₯ β 20)+ 2(5π₯ β 20) = 5(5π₯ β 20)
10. a. Verify: (π β 5)(π + 5) = π2 β25.
b. Verify: (π₯ β 88)(π₯ + 88) = π₯2 β 882
c. Verify: π΄2 β π΅2 = (π΄ β π΅)(π΄ + π΅).
d. Solve for π₯: π₯2 β 9 = 5(π₯ β 3).
e. Solve for π€: (π€ + 2)(π€ β 5) = π€2 β 4.
11. A string 60 inches long is to be laid out on a tabletop to make a rectangle of perimeter 60 inches. Write the width of the rectangle as 15 + π₯ inches. What is an expression for its length? What is an expression for its area? What value for π₯ gives an area of the largest possible value? Describe the shape of the rectangle for this special value of π₯.

Exit Ticket

1. Find the solution set to the equation 3x2 + 27x = 0
2. Determine if each statement is true or false. If the statement is false, explain why or show work proving that it is false.
a. If a = 5 then ac = 5c
b. If ac = 5c then a = 5

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.