Equations Involving Factored Expressions


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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

Worksheets for Algebra 1

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€œor”: 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‘₯ βˆ’ 10 = 0
    b. π‘₯2 + 20 = 0
    c. Demanding Dwight insists that you give him two solutions to the following equation:
    (π‘₯ βˆ’ 10)(π‘₯2 + 20) = 0
    Can you provide him with two solutions?
    d. Demanding Dwight now wants FIVE solutions to the following equation:
    (π‘₯ βˆ’ 10)(2π‘₯ + 6)(π‘₯2 βˆ’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‘₯ βˆ’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 βˆ’ 10π‘₯ = 0, for π‘₯.

Example 2
Solve π‘₯(π‘₯ βˆ’ 3)+ 5(π‘₯ βˆ’ 3) = 0, for π‘₯.

Exercises 2–7
2. (π‘₯ + 1)(π‘₯ + 2) = 0
3. (3π‘₯ βˆ’ 2)(π‘₯ + 12) = 0
4. (π‘₯ βˆ’ 3)(π‘₯ βˆ’ 3) = 0
5. (π‘₯ + 4)(π‘₯ βˆ’ 6)(π‘₯ βˆ’ 10) = 0
6. π‘₯2 βˆ’ 6π‘₯ = 0
7. π‘₯(π‘₯ βˆ’ 5)+ 4(π‘₯ βˆ’ 5) = 0

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

Exercises 8–11
8. Use factoring to solve the equation for π‘₯: (π‘₯ βˆ’ 2)(2π‘₯ βˆ’ 3) = (π‘₯ βˆ’ 2)(π‘₯ + 1).
9. Solve each of the following for π‘₯:
a. π‘₯ + 2 = 5
b. π‘₯2 + 2π‘₯ = 5π‘₯
c. π‘₯(5π‘₯ βˆ’ 20)+ 2(5π‘₯ βˆ’ 20) = 5(5π‘₯ βˆ’ 20)
10. a. Verify: (π‘Ž βˆ’ 5)(π‘Ž + 5) = π‘Ž2 βˆ’25.
b. Verify: (π‘₯ βˆ’ 88)(π‘₯ + 88) = π‘₯2 βˆ’ 882
c. Verify: 𝐴2 βˆ’ 𝐡2 = (𝐴 βˆ’ 𝐡)(𝐴 + 𝐡).
d. Solve for π‘₯: π‘₯2 βˆ’ 9 = 5(π‘₯ βˆ’ 3).
e. Solve for 𝑀: (𝑀 + 2)(𝑀 βˆ’ 5) = 𝑀2 βˆ’ 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



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