Equations Involving Factored Expressions

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1 Exploratory Exercise 1. A. Jenna said the product of two numbers is 20. Would the factors have to be 4 and 5? Why? B. Julie said the product of two numbers is 20. Would both factors have to be less than 20? Why? C. Justin said the product of two numbers is 20. Would both factors have to be positive? Why? 2. Jeremy said the product of two numbers is 0. What do you know must be true about the factors? Why? 3. A. Demanding Dwight insists that you give him two solutions to the following equation. What are the two solutions? (xx 10) xx = 0 Equations written in this way are in factored form. B. Demanding Dwight now wants FIVE solutions to the following equation: What are the five solutions? (xx 10)(2xx + 6)(xx 2 36)(xx ) xx = 0 Do you think there might be a sixth solution? Explain. S.135

2 4. How could we summarize what we ve observed? If aaaa = 0, then either aa = or aa = or aa = aa =. This is known as the zero-product property. 5. Consider the equation (xx 4)(xx + 3) = 0. A. Rewrite the equation as a compound statement. B. Find the two solutions to the equation. Sometimes the equation is not written in factored form. In these cases you need to think about the reverse of the distributive property. 6. Solve 2xx 2 10xx = 0, for xx. 7. Solve xx(xx 3) + 5(xx 3) = 0, for xx. S.136

3 Solve. 8. (xx + 1)(xx + 2) = 0 9. (3xx 2)(xx + 12) = (xx 3)(xx 3) = (xx + 4)(xx 6)(xx 10) = xx 2 6xx = xx(xx 5) + 4(xx 5) = Consider the equation (xx 2)(2xx 3) = (xx 2)(xx + 5). Lulu chooses to multiply both sides by xx 2 and gets the answer xx = 8. But Marius points out that xx = 2 is also an answer, which Lulu missed. A. What s the problem with Lulu s approach? 1 B. Marius used factoring to solve the original equation for xx. He wrote (x 2)(2x 3) (x 2)(x + 5) = 0. How did he get this equation? C. Marius said there is a common factor of (x 2) in each term. Finish his work below. ( x 2) (( ) ( )) = 0 ( x 2) ( ) = Use Marius factoring method to solve the equation for xx: (xx 2)(2xx 3) = (xx 2)(xx + 1). S.137

4 16. 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 + xx inches. What is an expression for its length? What is an expression for its area? What value for xx gives an area of the largest possible value? Describe the shape of the rectangle for this special value of xx. Lesson Summary The zero-product property says that if aaaa = 0, then either aa = or aa = or aa = aa =. S.138

5 Homework Problem Set 1. Find the solution set of each equation: a. (xx 1)(xx 2)(xx 3) = 0 b. (xx 16.5)(xx 109) = 0 c. xx(xx + 7) + 5(xx + 7) = 0 2. Solve xx 2 11xx = 0, for xx. 3. A. Solve (pp + 3)(pp 5) = 2(pp + 3), for pp. B. What solution do you lose if you simply divide by pp + 3 to get pp 5 = 2? 4. Using what you learned in this lesson, create an equation that has 53 and 22 as its only solutions. 5. Solve each of the following for xx: a. xx + 2 = 5 b. xx 2 + 2xx = 5xx c. CHALLENGE PROBLEM: xx(5xx 20) + 2(5xx 20) = 5(5xx 20) S.139

6 6. In Lesson 9 you learned how to multiply polynomials using the distributive property. Use the products below to help you look for patterns when doing the reverse of the distributive property factoring. A. Verify: (aa 5)(aa + 5) = aa B. Verify: (xx 88)(xx + 88) = xx C. Verify: AA 2 BB 2 = (AA BB)(AA + BB). 7. Use the patterns from Problem Set #6 to find the solutions to each equation. A. Solve for x: x 2 9 = 0 B. Solve for w: w 2 4 = 0 C. Solve for xx: xx 2 9 = 5(xx 3). D. Solve for ww: (ww + 2)(ww 5) = ww 2 4. S.140

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