NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M1 ALGEBRA II Lesson 11: The Special Role of Zero in Factoring Student Outcomes . Students find solutions to polynomial equations where the polynomial expression is not factored into linear factors. Students construct a polynomial function that has a specified set of zeros with stated multiplicity. Lesson Notes This lesson focuses on the first part of standard A-APR.B.3, identifying zeros of polynomials presented in factored form. Although the terms root and zero are interchangeable, for consistency only the term zero is used throughout this lesson and in later lessons. The second part of the standard, using the zeros to construct a rough graph of a polynomial function, is delayed until Lesson 14. The ideas that begin in this lesson continue in Lesson 19, in which students will be able to associate a zero of a polynomial function to a factor in the factored form of the associated polynomial as a consequence of the remainder theorem, and culminate in Lesson 39, in which students apply the fundamental theorem of algebra to factor polynomial expressions completely over the complex numbers. Classwork Scaffolding: Here is an alternative opening Opening Exercise (12 minutes) activity that may better illuminate the special role of Opening Exercise zero. ퟐ ퟐ Find all solutions to the equation (풙 + ퟓ풙 + ퟔ)(풙 − ퟑ풙 − ퟒ) = ퟎ. For each equation, list some possible values for 푥 and 푦. The main point of this opening exercise is for students to recognize and then formalize that the statement “If 푎푏 = 0, then 푎 = 0 or 푏 = 0” applies not only when 푎 and 푏 are 푥푦 = 10, 푥푦 = 1, numbers or linear functions (which we used when solving a quadratic equation), but also 푥푦 = −1, 푥푦 = 0 applies to cases where 푎 and 푏 are polynomial functions of any degree. What do you notice? Does one equation tell you In small groups, let students discuss ways to solve this equation. Walk around the room more information than and offer advice such as, “Have you considered factoring each quadratic expression? others? What do you get?” As soon as one group factors both quadratic expressions, or when three minutes have passed, show, or let that group show, the factorization on the board. ⏟(푥 + 2 )( 푥 + 3 ) ⋅ ⏟(푥 − 4 )( 푥 + 1 ) = 0 푥2+5푥+6 푥2−3푥−4 . What are the solutions to this equation? −2, −3, 4, −1 Lesson 11: The Special Role of Zero in Factoring 120 This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M1 ALGEBRA II . Why? If 푥 is any number other than −2, −3, 4, or −1, then each factor is a nonzero number that is 푥 + 2 ≠ 0, 푥 + 3 ≠ 0, etc. However, the multiplication of four nonzero numbers is nonzero, so that value of 푥 cannot be a solution. Therefore, the only possible solutions are −2, −3, 4, and −1. It is easy to confirm that these are indeed solutions by substituting them each into the equation individually. Why are these numbers also solutions to the original equation? Because the expression (푥 + 2)(푥 + 3)(푥 − 4)(푥 + 1) is equivalent to (푥2 + 5푥 + 6)(푥2 − 3푥 − 4). Now let’s study the solutions to 푥2 + 5푥 + 6 = 0 and 푥2 − 3푥 − 4 = 0 separately. What are the solutions to 푥2 + 5푥 + 6 = 0? −2, −3 . What are the solutions to 푥2 − 3푥 − 4 = 0? 4, −1 . Relate the solutions of the equation (푥2 + 5푥 + 6)(푥2 − 3푥 − 4) = 0 to the solutions of the compound statement, “푥2 + 5푥 + 6 = 0 or 푥2 − 3푥 − 4 = 0.” They are the same. Given two polynomial functions 푝 and 푞 of any degree, the solution set of the equation 푝(푥)푞(푥) = 0 is the union of the solution set of 푝(푥) = 0 and the solution set of 푞(푥) = 0. Let’s think about why. Lead students in a discussion of the following proof: . Suppose 푎 is a solution to the equation 푝(푥)푞(푥) = 0; that is, it is a number that satisfies 푝(푎)푞(푎) = 0. Since 푝(푎) is a number and 푞(푎) is a number, one or both of them must be zero, by the zero product property that states, “If the product of two numbers is zero, then at least one of the numbers is zero.” Therefore, 푝(푎) = 0 or 푞(푎) = 0, which means 푎 is a solution to the compound statement, “푝(푥) = 0 or 푞(푥) = 0.” . Now let’s prove the other direction and show that if 푎 is a solution to the compound statement, then it is a solution to the equation 푝(푥)푞(푥) = 0. This direction is also easy: Suppose 푎 is a number such that either 푝(푎) = 0 or 푞(푎) = 0. In the first case, 푝(푎)푞(푎) = 0 ∙ 푞(푎) = 0. In the second case, 푝(푎)푞(푎) = 푝(푎) ∙ 0 = 0. Hence, in either case, 푎 is a solution to the equation 푝(푥)푞(푥) = 0. Students may have difficulty understanding the distinction between the equations 푝(푥)푞(푥) = 0 and 푝(푎)푞(푎) = 0. Help students understand that 푝(푥)푞(푥) = 0 is an equation in a variable 푥, while 푝(푎) is the value of the function 푝 when it is evaluated at the number 푎. Thus, 푝(푎)푞(푎) is a number. For example, if 푝 and 푞 are the quadratic polynomials 푝(푥) = 푥2 + 5푥 + 6 and 푞(푥) = 푥2 − 3푥 − 4 from the Opening Exercise, and students are considering the case when 푎 is 5, then 푝(5) = 56 and 푞(5) = 6. Therefore, 5 cannot be a solution to the equation 푝(푥)푞(푥) = 0. Communicate to students that they can use the statement below to break problems into simpler parts: MP.1 Given any two polynomial functions 푝 and 푞, the set of solutions to the equation 푝(푥)푞(푥) = 0 can be found by solving 푝(푥) = 0, solving 푞(푥) = 0, and combining the solutions into one set. Ask students to try the following exercise on their own. Lesson 11: The Special Role of Zero in Factoring 121 This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M1 ALGEBRA II Exercise 1 (2 minutes) Exercise 1 1. Find the solutions of (풙ퟐ − ퟗ)(풙ퟐ − ퟏퟔ) = ퟎ. The solutions to (풙ퟐ − ퟗ)(풙ퟐ − ퟏퟔ) are the solutions of 풙ퟐ − ퟗ = ퟎ combined with the solutions of 풙ퟐ − ퟏퟔ = ퟎ. These solutions are −ퟑ, ퟑ, −ퟒ, and ퟒ. The next example looks at a polynomial equation for which the solution is already known. The goal of this example and the discussion that follows is to use a solution to the equation 푓(푥) = 0 to further factor the polynomial 푓. In doing so, the class ends with a description of the zeros of a function, a concept first introduced in Algebra I, Module 4. Example 1 (8 minutes) Example 1 Suppose we know that the polynomial equation ퟒ풙ퟑ − ퟏퟐ풙ퟐ + ퟑ풙 + ퟓ = ퟎ has three real solutions and that one of the factors of ퟒ풙ퟑ − ퟏퟐ풙ퟐ + ퟑ풙 + ퟓ is (풙 − ퟏ). How can we find all three solutions to the given equation? Steer the discussion to help students conjecture that the polynomial 4푥3 − 12푥2 + 3푥 + 5 must be the product of (푥 − 1) and some quadratic polynomial. Since (푥 − 1) is a factor, and we know how to divide polynomials, we can find the quadratic polynomial by dividing: 4푥3 − 12푥2 + 3푥 + 5 = 4푥2 − 8푥 − 5. 푥 − 1 Now we know that 4푥3 − 12푥2 + 3푥 + 5 = (푥 − 1)(4푥2 − 8푥 − 5), and we also know that 4푥2 − 8푥 − 5 is a quadratic polynomial that has linear factors (2푥 + 1) and Scaffolding: (2푥 − 5). Allow students to generate ideas about how the linear Therefore, 4푥3 − 12푥2 + 3푥 + 5 = 0 has the same solutions factors affect the behavior of as (푥 − 1)(4푥2 − 8푥 − 5) = 0, which has the same solutions as the graph so that they can use (푥 − 1)(2푥 + 1)(2푥 − 5) = 0. the graph of a function to identify zeros. 1 5 In this factored form, the solutions of 푓(푥) = 0 are readily apparent: − , 1, and . 2 2 Lesson 11: The Special Role of Zero in Factoring 122 This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 11 M1 ALGEBRA II Discussion (8 minutes) . In Example 1 above, we saw that factoring the Zeros of 푓 are polynomial into linear factors helped us to find solutions 1 5 푦 = 푓(푥) − , 1, and . to the original polynomial equation 2 2 4푥3 − 12푥2 + 3푥 + 5 = 0. There is a corresponding notion for the zeros of a function. Let 푓 be a function whose domain is a subset of the real numbers and whose range is a subset of the real numbers.