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Lesson Plans and Worksheets for Algebra I
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Common Core For Algebra I
Examples, videos, and solutions to help Algebra I students learn how to solve equations involving factored expressions.
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 βorβ: 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
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 2β7
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₯ β2)(2π₯ β 3) = (π₯ β 2)(π₯ +5). Lulu chooses to multiply through by
1/π₯β2 and gets the answer π₯ = 8. But Poindexter points out that π₯ = 2 is also an answer, which Lulu missed.
a. Whatβs the problem with Luluβs approach?
b. Use factoring to solve the original equation for π₯.
Exercises 8β11
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) = π2 β25.
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₯.
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