In this section, you will develop a method to change a quadratic equation written as a sum into its product form (also called its factored form).


 Silvester Copeland
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2 CHAPTER 8 In Chapter 4, you used a web to organize the connections you found between each of the different representations of lines. These connections enabled you to use any representation (such as a graph, rule, situation, or table) to find any of the other representations. In this chapter, a quadratics web will challenge you to find connections between the different representations of a parabola. Through this endeavor, you will learn how to rewrite quadratic equations by using a process called factoring. You will also discover and use a very important property of zero. In this chapter, you will learn: Quadratics Think about these questions throughout this chapter:? How can I rewrite it? What s the connection? What s special about zero? What information do I need? Is there another method? How to factor a quadratic expression completely. How to find the roots of a quadratic equation, if they exist. How to move from all representations of a parabola (rule, graph, table, and situation) to each of the other representations directly. Section 8.1 In this section, you will develop a method to change a quadratic equation written as a sum into its product form (also called its factored form). Section 8.2 Through a fun application, you will find ways to generate each representation of a parabola from each of the others. You will also develop a method to solve quadratic equations using the Zero Product Property. Yo! Section 8.3 In this section, you will be introduced to another method to solve quadratic equations called the Quadratic Formula. 604 Algebra Connections
3 Chapter 8 Teacher Guide Section Lesson Days Lesson Objectives Materials Homework Introduction to Factoring Quadratics Factoring with Generic Rectangles Factoring with Special Cases Factoring Completely Investigating a Parabola Multiple Representations for Quadratics Algebra tiles 86 to 811 Algebra tiles 816 to 821 Overhead algebra tiles 827 to 832 Overhead algebra tiles (optional) Poster graph paper and markers OR overhead graph paper and pens Lesson Res. Pg. Poster paper and markers Colored pencils or pens 837 to to to Zero Product Property None 865 to Solving Quadratic Equations by Factoring None 876 to Completing the Quadratic Web Graphing calculators Overhead graphing calculator 885 to Introduction to the Quadratic Formula More Solving Quadratic Equations Overhead graphing calculator OR Lesson Res. Pg. on a transparency 896 to None to Choosing a Strategy Yoyo (optional) to Chapter Closure Total: Varied Format Options 12 days plus optional time for Chapter Closure Chapter 8: Quadratics 605
4 Overview of the Chapter In Chapter 8, students will broaden their understanding of multiple representations, systems of equations, and solving equations to include quadratic equations. This chapter has many learning objectives: Students will learn how to factor quadratic and simple cubic equations. Students will learn how to solve quadratic equations by factoring and using the Zero Product Property and will recognize that these solutions are the xintercepts of a corresponding parabola. Students will learn how to use the Quadratic Formula to solve quadratic equations. Students will examine when a parabola has zero, one, or two xintercepts. Students will apply quadratic equations to physical problems, such as the motion of an object under the force of gravity. Students begin this chapter by learning how to factor quadratics in Section 8.1. They will learn this very similarly to the way they learned how to multiply binomials: they will first factor with tiles and then will develop a shortcut method using a generic rectangle. Section 8.2 challenges students to find connections between the different representations of a parabola so that they can use any representation to get any of the others. Through this endeavor, students will need to find out how to graph a parabola directly from its rule, which requires a new algebraic process. Students then discover the Zero Product Property and learn how it can help them solve quadratic equations. Finally, students confront a quadratic equation that cannot be factored but does in fact have roots. This sets up the need for the Quadratic Formula, which is introduced and derived in Section 8.3. (See more notes about its derivation below.) Necessary Materials This text assumes that students have daily access to a scientific calculator, graph paper, and a ruler or straightedge. Furthermore, please note that most lessons in this chapter require the use of algebra tiles, which were previously used in Chapters 2, 3, 5, and 6. When algebra tiles is listed anywhere as a necessary material, please be prepared with: Algebra tiles, enough for each student or pair of students. Cornerpiece, one per student or pair of students. Overhead algebra tiles and cornerpiece for teacher use. Note: See the teacher notes in Chapter 2 to find out how to purchase algebra tiles or how to create them from resource pages. In addition, Lesson assumes that students have access to graphing technology, such as graphing calculators or computers with graphing software. The graphing technology allows 606 Algebra Connections
5 students to use a process of Guess and Check to find the equation of the parabola that best fits the information given in each complaint. If the technology is not available, students will need to use graphing shortcuts to check their answers to problem See the Technology Notes section in the front of this Teacher Guide for directions on using a TI83. For all other calculator models, consult the appropriate calculator user s guide or look online at The Derivation of the Quadratic Formula The Quadratic Formula is introduced at this stage to augment the methods students can use to solve quadratic equations. Using the algebraic skills they already have, students will derive the Quadratic Formula in this chapter. However, in Chapter 12, students will learn how to derive the Quadratic Formula another way: by completing the square. Accuracy of Solutions and Using a Scientific Calculator Before teaching this chapter, consider how you want students to display their answers to problems solved using the Quadratic Formula. For example, would you rather have them express the solutions to x 2! 3x! 5 = 0 as x = or as x! 4.2 or x! "1.2? 3± 29 2 Also, if you have your students use a scientific calculator to evaluate these expressions, what degree of accuracy do you expect? This text leaves these decisions to the teacher and only explicitly asks students to use their calculators to evaluate their answers when the application or context requires an interpretable value. You will need to instruct students exactly how you want them to leave answers when no special instructions are given in the text. This curriculum assumes that students have access to scientific calculators at school and at home. It is common, however, for students comfortable with using a calculator to have difficulty evaluating a complicated expression such as those that arise when using the Quadratic Formula.! !!3 Several problems that ask students to evaluate messy expressions, such as, are assigned in homework prior to Section 8.3. This will give you the opportunity to assess your students ability to use their calculators appropriately and to address common errors before Section 8.3. Where Is This Going? Quadratic equations will continue to be an area of focus throughout Chapters 9 through 12. For example, in Chapter 9, students will examine the solutions to quadratic inequalities. In Chapter 10, students will develop a new method to help solve quadratic equations called completing the square. In addition, students will rely on their understanding of factoring to simplify rational expressions in Chapter 10. Furthermore, in Chapter 12, students will derive the Quadratic Formula by completing the square. Chapter 8: Quadratics 607
6 Lesson How can I find the product? Introduction to Factoring Quadratics Lesson Objective: Students will review how to build rectangles with tiles and will learn shortcuts for finding the dimensions of a completed generic rectangle. Students will discover that the products of the terms in each diagonal of a generic rectangle are equal. Length of Activity: One day (approximately 50 minutes) Core Problems: Problems 81 through 85 Ways of Thinking: Materials: Suggested Lesson Activity: Reversing thinking, generalizing, justifying Algebra tiles Note: When algebra tiles is listed anywhere as a necessary material, please be prepared with: Algebra tiles, enough for each student or pair of students Cornerpiece, one per student or pair of students Overhead algebra tiles and cornerpiece for teacher use General Team Roles transparency (optional) Note: The General Team Roles transparency is the Lesson Resource Page ( General Team Roles ). It is listed as an optional resource throughout many lessons in the rest of the text. When you see General Team Roles transparency listed, it is always the Lesson Resource Page that is being referred to. Have a student volunteer read the lesson introduction. Then have teams start problem 81 as a warmup. This problem asks students to review what they learned in Chapter 5. After about five minutes, move teams on to problem 82, which asks students to factor expressions using algebra tiles to investigate whether a sum can always be written as a product. This will help students recognize that not all expressions are factorable. Also, the act of building composite rectangles with algebra tiles will strengthen students understanding of the factoring process that will be introduced in Lesson Specifically, students will need to understand the options for splitting the xtiles between the two corners of the rectangle. To ensure that there is enough time for problems 84 and 85, stop teams after 20 minutes. Then teams should move on to problem 83, which has students focus on finding the dimensions of a completed generic rectangle. As you circulate, emphasize that students should look for special strategies to find the dimensions. After 10 minutes, pull the class together and ask students to share any shortcuts they discovered. Students may notice that each 608 Algebra Connections
7 dimension is the greatest common factor (GCF) of its corresponding row or column, demonstrated in the diagram at right. Note that this is true when the quadratic expression does not have a common factor. Students will examine what happens when a quadratic has a common factor in Lesson As teams finish, have them post solutions on the board for others to check. Leave the solutions on the board so that they can be used during closure. Colorcoding work can help some students keep track of the different parts of the quadratic. For example, writing all x 2 terms in blue, all xterms in red, and all constants in green (in both the expression and in the generic rectangle) can help students see how the parts in the generic rectangle relate to the expression. See the example at right. Finally, problem 84 asks students to recognize that the product of the terms of a generic rectangle s diagonal equals the product of the terms in the other diagonal. This is an important pattern that will later help students develop an algorithm for factoring quadratics without algebra tiles. blue +1 3x 2x 5 6x 2 2x +5 6x 2 +17x + 5 6x 2 red 2x 5 15x 15x green Closure: (10 minutes) Team Roles: Either individually or as a class, have students respond to problem 85, which asks them to write a description of the diagonals pattern introduced in problem 84. Be sure that all students not only recognize the pattern, but also test the pattern on the other generic rectangles in problem 83. A Math Notes box proving that the pattern always works is not offered until Lesson so that students will not have the pattern in front of them while they try to discover it on their own. However, once students have found the pattern, there is no harm in proving this pattern for closure. If you have not changed your teams or roles recently, you may want to start this chapter with a new seating chart. Remind team members of their roles using the General Team Roles transparency from Chapter 7. Emphasize the need for Facilitators to keep teams together in problem 82. Some good phrases for Facilitators to use are, Did everyone get? and Are we all ready to move on? Resource Managers will not only need to get the algebra tiles for the team, but they will also be responsible for making sure the tiles are returned in complete sets. Homework: Problems 86 through 811 Chapter 8: Quadratics 609
8 8.1.1 How can I find the product? Introduction to Factoring Quadratics Student pages for this lesson are In Chapter 5 you learned how to multiply algebraic expressions using algebra tiles and generic rectangles. This section will focus on reversing this process: How can you find a product when given a sum? 81. Review what you know about products and sums below. a. Write the area of the rectangle at right as a product and as a sum. Remember that the product represents the area found by multiplying the length by the width, while the sum is the result of adding the areas inside the rectangle. [ (x + 4)(y + x + 2) = xy + x 2 + 6x + 4 y + 8 ] x y x b. Use a generic rectangle to multiply (6x!1)(3x + 2). Write your solution as a sum. [ 18x 2 + 9x! 2 ] 82. The process of changing a sum to a product is called factoring. Can every expression be factored? That is, does every sum have a product that can be represented with tiles? Investigate this question by building rectangles with algebra tiles for the following expressions. For each one, write the area as a sum and as a product. If you cannot build a rectangle, be prepared to convince the class that no rectangle exists (and thus the expression cannot be factored). [ a: (2x + 3)(x + 2), b: (2x + 1)(3x + 2), c: no solution, d: (2x + y)(y + 3) ; Conclusion: Not every expression can be factored. ] a. 2x 2 + 7x + 6 b. 6x 2 + 7x + 2 c. x 2 + 4x +1 d. 2xy + 6x + y 2 + 3y 610 Algebra Connections
9 83. Work with your team to find the sum and the product for the following generic rectangles. Are there any special strategies you discovered that can help you determine the dimensions of the rectangle? Be sure to share these strategies with your teammates. [ a: (3x +1)(2x + 5) = 6x x + 5, b: (5x! 2)(y + 3) = 5xy +15x! 2y! 6, c: (4x! 3)(3x + 4) = 12x 2 + 7x! 12 ] a. b. c. 2x 5!2y!6!9 x!12 6x 2 15x 5xy 15x 12x 2 16x 84. While working on problem 83, Casey noticed a pattern with the diagonals of each generic rectangle. However, just before she shared her pattern with the rest of her team, she was called out of class! The drawing on her paper looked like the diagram below. Can you figure out what the two diagonals have in common? [ The product of each diagonal is equal: 6x 2! 5 = 30x 2 and 2x!15x = 30x 2. ] 2x 5 6x 2 15x 85. Does Casey s pattern always work? Verify that her pattern works for all of the 2by2 generic rectangles in problem 83. Then describe Casey s pattern for the diagonals of a 2by2 generic rectangle in your Learning Log. Be sure to include an example. Title this entry Diagonals of a Generic Rectangle and include today s date. [ Diagonals: part (a) are both 30x 2, part (b) are both!30xy, part (c) are both!144x 2. Students should state that this pattern always works and describe the pattern in their own words. Typical response: The product of one diagonal always equals the product of the other diagonal. ] Chapter 8: Quadratics 611
10 ETHODS AND MEANINGS MATH NOTES New Vocabulary to Describe Algebraic Expressions Since algebraic expressions come in many different forms, there are special words used to help describe these expressions. For example, if the expression can be written in the form ax 2 + bx + c and if a is not 0, it is called a quadratic expression. Study the examples of quadratic expressions below. Examples of quadratic expressions: x 2! 15x m 2! 25 12! 3k 2 + 5k The way an expression is written can also be named. When an expression is written in product form, it is described as being factored. When factored, each of the expressions being multiplied is called a factor. For example, the factored form of x 2! 15x + 26 is (x!13)(x! 2), so x!13 and x! 2 are each factors of the original expression. Finally, the number of terms in an expression can help you name the expression to others. If the expression has one term, it is called a monomial, while an expression with two terms is called a binomial. If the expression has three terms, it is called a trinomial. Study the examples below. Examples of monomials: 15xy 2 and!2m Examples of binomials: 16m 2! 25 and 7h h Examples of trinomials: 12! 3k 2 + 5k and x 2! 15x Write the area of the rectangle at right as a sum and as a product. [ (2x! 3)(x + 2y! 4) = 2x 2 + 4xy! 11x! 6y + 12 ]!3x!6y 2x 2 4xy 12!8x 612 Algebra Connections
11 87. Multiply the expressions below using a generic rectangle. Then verify Casey s pattern (that the product of one diagonal equals the product of the other diagonal). [ a: 12x x! 5, b: 4x 2! 28x + 49 ] a. (4x!1)(3x + 5) b. (2x! 7) Remember that a Diamond Problem is a pattern for which the product of two numbers is placed on top, while the sum of the same two numbers is placed on bottom. (This pattern is demonstrated in the diamond at right.) Copy and complete each Diamond Problem below. [ Find solutions in the diamonds below. ] x xy x+y y a. b. c. d e. f. 6x 2 2x 3x 5x!7x 2 7x x 6x 89. For each line below, name the slope and yintercept. [ a: m = 2, (0,!! 1 2) ; b: m =!3, (0,!!7) ; c: m =! 2 3, (0,!8) ; d: m = 0, (0,!!2) ] a. y =!1+4 x 2 b. 3x + y =!7 c. y =!2 3 x + 8 d. y =! On graph paper, graph y = x 2! 2x! 8. a. Name the yintercept. What is the connection between the yintercept and the rule y = x 2! 2x! 8? [ (0, 8); It is the constant in the equation. ] b. Name the xintercepts. [ ( 2, 0) and (4, 0); Students may notice that the product of the xintercepts equals the constant term. ] c. Find the lowest point of the graph, the vertex. [ (1, 9); Its xcoordinate is midway between the xintercepts. ] Calculate the value of each expression below. [ a:!1, b:! 7.24, c:! " 4.24 ] a. 5! 36 b c.!2! 5 Chapter 8: Quadratics 613
12 Lesson Is there a shortcut? Factoring with Generic Rectangles Lesson Objective: Students will develop an algorithm to factor quadratic expressions without algebra tiles. Length of Activity: One day (approximately 50 minutes) Core Problems: Ways of Thinking: Materials: Suggested Lesson Activity: Problems 812 through 815 (parts (a) and (b)) Making connections, generalizing Algebra tiles Introduce today s lesson and use problem 812 as a quick warmup for students to review what they learned in Lesson Although problem 813 is written so that students can investigate factoring in teams, it is recommended that it instead be used as an outline for a wholeclass discussion. For part (a), have students build a composite rectangle with tiles in order to factor (as shown in the diagram at right). Then, as a class, figure out how to split the 10 xterms in the generic rectangle for part (b). At this point, expect students to guess and check how to place the xterms into the rectangle so that it is factorable. For example, they may first try to split the 10 xterms by placing 5x into each corner. However, when finding the dimensions, they will quickly learn that there are no dimensions that, when multiplied, will create that generic rectangle. Let students struggle for a bit, as this will help them appreciate the use of the Diamond 8 Problems later. Have overhead 3x algebra tiles ready to demonstrate 2 all of the different combinations for 10x and to verify the solution. At this point we expect that students will be glad to find a shortcut that will help them decide how to split the xterm in the generic rectangle. Part (c) hints that Casey s pattern can help factor another trinomial, 2x 2 + 7x + 6. Specifically, we know that the product of the two missing terms equals the product of the terms on the other diagonal (6 and 2x 2 ), based on the pattern introduced in problem 84. Also, the number of xtiles that are placed in opposite corners of a generic rectangle must add up to the total number of xtiles in the original expression (which, in this case, is 7x).? 2x 2 6? x x x 614 Algebra Connections
13 Part (d) of problem 813 points out that since students know the sum and product of the two missing terms, they can create and solve a Diamond Problem (see diagram at right) to find the missing terms in the generic rectangle. Once the generic rectangle is complete, students can find common factors in rows and columns to find the dimensions of the generic rectangle. To recap, this fourstep process will help students factor all factorable quadratic expressions: 1. Place the x 2 terms and constant terms of the quadratic expression in opposite corners of the generic rectangle. Determine the sum and product of the two remaining corners: The sum is simply the xterm of the quadratic expression, while the product is equal to the product of the x 2 terms and constant terms. 2. Place this sum and product into a Diamond Problem and solve. 3. Place the solutions from the Diamond Problem into the generic rectangle and find the dimensions of the generic rectangle. 4. Write the answer as a product. product 12x 2 Problem 814 offers teams a chance to recap the factoring process. Then students practice using this factoring method in problem Note that part (d) of problem 815 is not factorable. 7x sum Closure: (5 minutes) Pull the class together and point out that students have accomplished an important goal: They have developed an algorithm to factor quadratics without algebra tiles. Have a few student volunteers recap the factoring process. Ask questions that help students justify statements, such as, Why do you multiply the x 2 term with the units term and place that at the top of the Diamond Problem? Homework: Problems 816 through Is there a shortcut? Factoring with Generic Rectangles Student pages for this lesson are Since mathematics is often described as the study of patterns, it is not surprising that generic rectangles have many patterns. You saw one important pattern in Lesson (Casey s pattern from problem 84). Today you will continue to use patterns while you develop a method to factor trinomial expressions. Chapter 8: Quadratics 615
14 812. Examine the generic rectangle shown at right. a. Review what you learned in Lesson by writing the area of the rectangle at right as a sum and as a product. [ (5x! 2)(2x! 7) = 10x 2! 39x + 14 ]!35x 14 10x 2! 4x b. Does this generic rectangle fit Casey s pattern for diagonals? Demonstrate that the product of each diagonal is equal. [!35x "! 4x = 10x 2 "14 = 140x 2 ] FACTORING QUADRATICS To develop a method for factoring without algebra tiles, first study how to factor with algebra tiles, and then look for connections within a generic rectangle. a. Using algebra tiles, factor 2x 2 + 5x + 3 ; that is, use the tiles to build a rectangle, and then write its area as a product. [ (2x + 3)(x +1) ] b. To factor with tiles (like you did in part (a)), you need to determine how the tiles need to be arranged to form a rectangle. Using a generic rectangle to factor requires a different process. Miguel wants to use a generic rectangle to factor 3x x + 8. He knows that 3x 2 and 8 go into the rectangle in the locations shown at right. Finish the rectangle by deciding how to place the ten xterms. Then write the area as a product. [ One corner should contain 4x, while the other should contain 6x ; (3x + 4)(x + 2). ] 3x 2 8 c. Kelly wants to find a shortcut to factor 2x 2 + 7x + 6. She knows that 2x 2 and 6 go into the rectangle in the locations shown at right. She also remembers Casey s pattern for diagonals. Without actually factoring yet, what do you know about the missing two parts of the generic rectangle? [ Their sum is 7x, and their product is 12x 2. ] d. To complete Kelly s generic rectangle, you need two xterms that have a sum of 7x and a product of 12x 2. Create and solve a Diamond Problem that represents this situation. [ The product 12x 2 should be placed at the top of the diamond problem, 7x at the bottom, and terms 3x and 4x should be in the middle. ]? 6 2x 2? product sum e. Use your results from the Diamond Problem to complete the generic rectangle for 2x 2 + 7x + 6, and then write the area as a product of factors. [ (2x + 3)(x + 2) ] 616 Algebra Connections
15 814. Factoring with a generic rectangle is especially convenient when algebra tiles are not available or when the number of necessary tiles becomes too large to manage. Using a Diamond Problem helps avoid guessing and checking, which can at times be challenging. Use the process from problem 813 to factor 6x x The questions below will guide your process. a. When given a trinomial, such as 6x x + 12, what two parts of a generic rectangle can you quickly complete? [ One corner contains 6x 2, and the opposite corner contains 12. ] b. How can you set up a Diamond Problem to help factor a trinomial such as 6x x + 12? What goes on the top? What goes on the bottom? [ The product of the x 2 and units terms (in this case, 72x 2 ) goes on top, while the xterm (17x ) goes on bottom. ] c. Solve the Diamond Problem for 6x x + 12 and complete its generic rectangle. d. Write the area of the rectangle as a product. [ (2x + 3)(3x + 4) ] product sum Use the process you developed in problem 813 to factor the following quadratics, if possible. If a quadratic cannot be factored, justify your conclusion. [ a: (x + 3)(x + 6), b: (4x! 3)(x + 5), c: (2x! 3)(2x! 1), d: not factorable because there are no integers that multiply to get!9x 2 (the diagonal of the generic rectangle) and add to get 5x. ] a. x 2 + 9x +18 b. 4x 2 +17x!15 c. 4x 2!8x + 3 d. 3x 2 + 5x!3 MATH NOTES ETHODS AND MEANINGS Diagonals of Generic Rectangles Why does Casey s pattern from problem 84 work? That is, why does the product of the d ad bd terms in one diagonal of a 2by2 generic rectangle always equal the c ac bc product of the terms in the other diagonal? a b Examine the generic rectangle at right for (a + b)(c + d). Notice Product = abcd Product = abcd that each of the resulting diagonals have a product of abcd. Thus, the product of the terms in the diagonals are equal. Chapter 8: Quadratics 617
16 816. Use the process you developed in problem 813 to factor the following quadratics, if possible. [ a: (x! 6)(x + 2), b: (2x + 1) 2, c: (x! 5)(2x +1), d: (x + 4)(3x! 2) ] a. x 2! 4x!12 b. 4x 2 + 4x +1 c. 2x 2! 9x! 5 d. 3x 2 +10x! For each rule represented below, state the x and yintercepts, if possible. [ a: xintercepts ( 1, 0) and (3, 0), yintercept: (0, 3); b: xintercept (2, 0), no yintercept; c: xintercepts ( 3, 0), ( 1, 0), and (1, 0), yintercept (0, 2); d: xintercept (8, 0), yintercept (0, 20) ] a. b. c. d. 5x! 2y = 40 x y Graph y = x 2! 9 on graph paper. a. Name the yintercept. What is the connection between the yintercept and the rule y = x 2! 9? [ (0, 9); It is the constant in the equation. ] b. Name the xintercepts. What is the connection between the xintercepts and the rule y = x 2! 9? [ (3, 0) and ( 3, 0); Students may notice that the product of the xintercepts equals the constant term. ] Find the point of intersection for each system. [ a: (6, 9), b: (0, 2) ] a. y = 2x! 3 x + y =15 b. 3x = y! 2 6x = 4! 2y 618 Algebra Connections
17 820. Solve each equation below for the given variable, if possible. [ a: x =! 10 23, b: all numbers, c: c = 0 ] a. 4 x 5 = x!2 7 b.!3(2b! 7) =!3b + 21! 3b c. 6! 2(c! 3) = Find the equation of the line that passes through the points ( 800, 200) and ( 400, 300). [ y = 1 4 x ] Lesson How can I factor this? Factoring with Special Cases Lesson Objective: Students will continue to practice their factoring skills while learning about special cases: quadratics with missing terms, quadratics that are not in standard form, and quadratics with more than one possible factored form. Length of Activity: One day (approximately 50 minutes) Core Problems: Problems 822 through 824 Ways of Thinking: Materials: Suggested Lesson Activity: Justifying, applying and extending, generalizing Overhead algebra tiles General Team Roles transparency (optional) Start class with an overview of the lesson objective. Then move teams into problem 822, which is a review of previous work and is an opportunity for you to assess where students are in their understanding. As you circulate, look for and correct common errors, and be available to answer teams questions. Be sure that students are completing both a Diamond Problem and a generic rectangle for each quadratic. Ask questions that help you know if students understand the factoring process. For example, when students state that 6x 2 must go at the top of a Diamond Problem, ask them to explain how they know it is 6x 2. If a student has part of a generic rectangle completed, ask him or her to justify one or two of the entries. Note that the quadratic in part (a) of problem 822 is a perfect square trinomial, so while debriefing this problem, you may want to point out that the answer can be written two ways: as (x + 3)(x + 3) or as (x + 3) 2. The quadratic in part (c) is not factorable. Problem 823 introduces students to quadratics that are not in standard form or that have terms missing. Remind teams that you expect to hear mathematical discussions for each case. Expect some questions from Chapter 8: Quadratics 619
18 teams that are unsure about putting 0 into their Diamond Problem. Note that the expression in part (d) of problem 823 (40!100m) is not quadratic; however, students can still use the same process to factor it. In this case, since the m 2 term is 0, the generic rectangle will have 0 and 40. Since the product of the missing terms is 0 and the sum is 100m, a Diamond Problem can help determine the other entries in the generic rectangle, as shown at right. Help students recognize that this generic rectangle can be abbreviated into a generic rectangle with only two parts, like the one shown at right. 100m Finally, problem 824 introduces students to a quadratic expression that has two factored forms, which will later lead to an investigation of what it means to be completely factored in Lesson After teams have finished problem 824, debrief as a whole class by having different volunteers demonstrate their solutions, so that students can see that there is more than one factored form. Have overhead algebra tiles ready to demonstrate that the rectangles have different arrangements indeed, as shown below m 0 100m m 40 If time allows, problem 825 offers a challenge for teams to use their factoring and multiplication skills. Closure: (8 minutes) Team Roles: Problem 826 asks students to explain how to factor a quadratic in their own words. Since this is an important entry, you may want to work together as a class to create a strong description that can be entered into each student s Learning Log. A Math Notes box in Lesson will also summarize this process for later reference. Today s work will depend on good communication between team members. Use the General Team Roles transparency to review the roles at the beginning of the lesson. Facilitators should keep their teams together as students apply their factoring skills. They should also make sure team members agree on an answer before moving to the next problem and should ask, Is another answer possible? Homework: Problems 827 through 832 Note: Problem 832 is a preview of the Zero Product Property. Furthermore, problem 828 is important to assign, as it introduces the idea of square root. This idea will be used in homework problems to prepare students to use the Quadratic Formula in Section 8.3. Plan on taking some time to go over problem 828 before you begin Lesson Algebra Connections
19 8.1.3 How can I factor this? Factoring with Special Cases Student pages for this lesson are Practice your new method for factoring quadratic expressions without tiles as you consider special types of quadratic expressions Factor each quadratic below, if possible. Use a Diamond Problem and generic rectangle for each one. [ a: (x + 3) 2, b: (2x + 3)(x +1), c: not factorable, d: (3m + 7)(m! 2) ] a. x 2 + 6x + 9 b. 2x 2 + 5x + 3 c. x 2 + 5x! 7 d. 3m 2 + m! SPECIAL CASES Most quadratics are written in the form ax 2 + bx + c. But what if a term is missing? Or what if the terms are in a different order? Consider these questions while you factor the expressions below. Share your ideas with your teammates and be prepared to demonstrate your process for the class. [ a: (3x! 2)(3x + 2), b: 4x(3x! 4), c: (4k! 3)(2k! 1), d: 20(2! 5m) ] a. 9x 2! 4 b. 12x 2! 16x c k 2! 10k d. 40!100m Now turn your attention to the quadratic below. Use a generic rectangle and Diamond Problem to factor this expression. Compare your answer with your teammates answers. Is there more than one possible answer? [ Students should get either (2x! 6)(2x + 1) or (x! 3)(4x + 2). Note that some may get the error (2x! 6)(4x + 2) by looking only for common factors. Be sure to correct this by asking the students to verify their answer by multiplying the factors. ] 4x 2!10x! 6 Chapter 8: Quadratics 621
20 825. The multiplication table below has factors along the top row and left column. Their product is where the row and column intersect. With your team, complete the table with all of the factors and products. Multiply x! 2 2x + 1 x + 7 x 2 + 5x! 14 2x x + 7 3x + 1 3x 2! 5x! 2 6x 2 + 5x In your Learning Log, explain how to factor a quadratic expression. Be sure to offer examples to demonstrate your understanding. Include an explanation of how to deal with special cases, such as when a term is missing or when the terms are not in standard order. Title this entry Factoring Quadratics and include today s date. ETHODS AND MEANINGS MATH NOTES Standard Form of a Quadratic A quadratic expression in the form ax 2 + bx + c is said to be in standard form. Notice that the terms are in order from greatest exponent to least. Examples of quadratic expressions in standard form: 3m 2 + m!1, x 2! 9, and 3x 2 + 5x. Notice that in the second example, b = 0, while in the third example, c = The perimeter of a triangle is 51 cm. The longest side is twice the length of the shortest side. The third side is 3 cm longer than the shortest side. How long is each side? Write an equation that represents the problem and then solve it. [ s + 2s + s + 3 = 51 ; 12, 24, and 15 cm ] 622 Algebra Connections
21 828. Remember that a square is a rectangle with four equal sides. a. If a square has an area of 81 square units, how long is each side? [ 9 units ] b. Find the length of the side of a square with area 225 square units. [ 15 units ] c. Find the length of the side of a square with area 10 square units. [ 10 units ] d. Find the area of a square with side 11 units. [ 121 square units ] Factor the following quadratics, if possible. [ a: (k! 2)(k! 10), b: (2x + 7)(3x! 2), c: (x! 4) 2, d: (3m + 1)(3m!1) ] a. k 2!12k + 20 b. 6x 2 +17x!14 c. x 2! 8x +16 d. 9m 2! Examine the two equations below. Where do they intersect? [ (2, 5) ] y = 4x! 3 y = 9x! Find the equation of a line perpendicular to the one graphed at right that passes through the point (6, 2). [ y =!x + 8 ] Solve each equation below for x. Check each solution. [ a: 5, b: 6, c: 5 or 6, d:! 1 4, e: 8, f:! 1 4 or 8 ] a. 2x!10 = 0 b. x + 6 = 0 c. (2x!10)(x + 6) = 0 d. 4x +1 = 0 e. x! 8 = 0 f. (4 x +1)(x! 8) = 0 Chapter 8: Quadratics 623
22 Lesson Can it still be factored? Factoring Completely Lesson Objective: Students will complete their focus on factoring by considering expressions that can be factored first with a common factor and then again using the quadratic factoring method. Length of Activity: One day (approximately 50 minutes) Core Problems: Ways of Thinking: Materials: Suggested Lesson Activity: Problems 833 through 836 (parts (a) and (b)) Applying and extending, making connections Overhead algebra tiles (optional) Problem 832, from homework, may have raised some questions for your students because they are not used to equations having more than one possible answer. Be prepared to address any questions regarding these problems at the beginning of the lesson. You may want to ask students who found possible solutions to explain their thinking to the class. Choose a volunteer to read the lesson introduction, which connects today s lesson with what students learned about factors of numbers in previous courses. Then start teams with a factoring warmup, problem This collection of expressions includes perfect square trinomials (which give students a nice shorthand answer), missing terms, a quadratic that is not in standard form, and a quadratic with more than one possible factored form. After about 15 minutes, ask volunteers to post their solutions on the board or overhead. Look for students who got different answers to part (d) of problem 833 and ask them to post their answers. This will enable the class to look at the similarities and differences between both results. Leave this work on the board or save transparencies of the work for a discussion during closure of this lesson. Then have teams move on to problem 834, which asks them to analyze why some quadratics have unique factored forms while others have more than one possible factored form. As you circulate, ask questions that help teams focus on the terms of each quadratic, such as, What do you notice about all of the terms of 12t 2! 10t + 2? Is that true for the other quadratics as well? Students should recognize that each term is a multiple of 2. Problem 834 introduces students to the idea of a common factor. If teams get frustrated, be ready to pull the class together to review common factors. Problem 835 gives students an opportunity to practice factoring completely and gives you the opportunity to work with any teams that might be struggling. 624 Algebra Connections
23 Closure: (10 minutes) During the introduction of the lesson, students were told that while 12 can be rewritten as 3 4, as 2 6, as 1 12, or as 2 2 3, only is considered to be factored completely, since the factors are prime and cannot be factored themselves. For closure, connect this with the two solutions found for part (d) of problem 833: (3n + 3)(n + 2) and (n +1)(3n + 6). Neither of these expressions is completely factored, because each contains a binomial that can be factored further. It is also a good idea to point out that each of the answers for part (d) of problem 833 can be factored again to find the most factored form. For example, if 3n + 3 from (3n + 3)(n + 2) is factored, the end result is 3(n + 1)(n + 2). Likewise, if 3n + 6 from (n +1)(3n + 6) is factored, the result is the same. Although factoring out a common factor first often makes factoring easier by lowering the coefficients of the expression, it does not need to happen first. If students forget to factor out a common factor first, they can always factor it out later. Homework: Problems 837 through Can it still be factored? Factoring Completely Student pages for this lesson are There are many ways to write the number 12 as a product of factors. For example, 12 can be rewritten as 3 4, as 2 6, as 1 12, or as While each of these products is accurate, only is considered to be factored completely, since the factors are prime and cannot be factored themselves. During this lesson you will learn more about what it means for a quadratic expression to be factored completely Review what you have learned by factoring the following expressions, if possible. [ a: (3x! 2) 2, b: (9m + 1)(9m! 1), c: (x! 4)(x! 7), d: Students will probably get one of two possible answers: (3n + 3)(n + 2) or (n + 1)(3n + 6). ] a. 9x 2! 12x + 4 b. 81m 2! 1 c x 2! 11x d. 3n 2 + 9n + 6 Chapter 8: Quadratics 625
24 834. Compare your solutions for problem 833 with the rest of your class. a. Is there more than one factored form of 3n 2 + 9n + 6? Why or why not? [ Yes, because there are two different arrangements of tiles that build a rectangle. ] b. Why does 3n 2 + 9n + 6 have more than one factored form while the other quadratics in problem 833 only have one possible answer? Look for clues in the original expression ( 3n 2 + 9n + 6 ) and in the different factored forms. [ Because there is a common factor of 3 in each of the terms of the original expression and in one of the two binomials in either of the two partially factored forms. ] c. Without factoring, predict which quadratic expressions below may have more than one factored form. Be prepared to defend your choice to the rest of the class. [ (i) and (iii) both have common factors, so they could have more than one factored form. ] i. 12t 2! 10t + 2 ii. 5 p 2! 23p! 10 iii. 10x x! 15 iv. 3k 2 + 7k! FACTORING COMPLETELY In part (c) of problem 834, you should have noticed that each term in 12t 2! 10t + 2 is divisible by 2. That is, it has a common factor of 2. a. What is the common factor for 10x x! 15? [ 5 ] b. For an expression to be completely factored, each factor must have all common factors separated out. Sometimes it is easiest to do this first. Since 5 is a common factor of 10x x! 15, you can factor 10x x! 15 using a special generic rectangle, which is shown below. Find the length of this generic rectangle and write its area as a product of its length and width. [ 5(2x 2 + 5x! 3) ] 5 10x x! 15 c. Can the result be factored even more? That is, can either factor from the result from part (b) above also be factored? Factor any possible expressions and write your solution as a product of all three factors. [ Yes; 2x 2 + 5x! 3 can be factored as (2x!1)(x + 3). Therefore, the final factored form of 10x x! 15 is 5(2x! 1)(x + 3). ] 626 Algebra Connections
25 836. Factor each of the following expressions as completely as possible. [ a: 5(x + 4)(x!1), b: 3x(x + 3)(x! 5), c: 2(x + 5)(x! 5), d: y(x! 5)(x + 2) ] a. 5x 2 +15x! 20 b. 3x 3! 6x 2! 45x c. 2x 2! 50 d. x 2 y! 3xy!10y ETHODS AND MEANINGS MATH NOTES Review the process of factoring quadratics developed in problem 813 and outlined below. This example demonstrates how to factor 3x 2 +10x Place the x 2  and constant terms of the quadratic expression in opposite corners of the generic rectangle. Determine the sum and product of the two remaining corners: The sum is simply the xterm of the quadratic expression, while the product is equal to the product of the x 2  and constant terms. 2. Place this sum and product into a Diamond Problem and solve it. Factoring Quadratic Expressions 3x 2 6x 24x 2 10x 8 4x 3. Place the solutions from the Diamond Problem into the generic rectangle and find the dimensions of the generic rectangle. 4. Write your answer as a product: (3x + 4)(x + 2). 2 x 6x 3x 2 3x 8 4x Factor the quadratic expressions below. If the quadratic is not factorable, explain why not. [ a: (2x + 5)(x!1), b: (x! 3)(x + 2), c: (3x +1)(x + 4), d: It is not factorable because no integers have a product of 14 and a sum of 5. ] a. 2x 2 + 3x! 5 b. x 2! x! 6 c. 3x 2 +13x + 4 d. 2x 2 + 5x + 7 Chapter 8: Quadratics 627
26 838. A line has intercepts (4, 0) and (0, 3). Find the equation of the line. [ y = 3 4 x! 3 ] As Jhalil and Joman practice for the SAT, their scores on practice tests rise. Jhalil s current score is 850, and it is rising by 10 points per week. On the other hand, Joman s current score is 570 and is growing by 50 points per week. a. When will Joman s score catch up to Jhalil s? [ in 7 weeks ] b. If the SAT test is in 12 weeks, who will score highest? [ Joman will score more with 1170 points, while Jhalil will have 970. ] Mary says that you can find an xintercept by substituting 0 for x, while Michelle says that you need to substitute 0 for y. a. Who, if anyone, is correct and why? [ Michelle is correct. One way to view this is graphically: The xintercept always has a ycoordinate of 0 because it lies on the xaxis. ] b. Use the correct approach to find the xintercept of! 4x + 5y = 16. [ ( 4, 0) ] Find three consecutive numbers whose sum is 138 by writing and solving an equation. [ 45, 46, 47; x + (x + 1) + (x + 2) = 138 ] Match each rule below with its corresponding graph. Can you do this without making any tables? Explain your selections. [ a: 2, b: 3, c: 1 ] a. y =!x 2! 2 b. y = x 2! 2 c. y =!x Algebra Connections
27 Lesson What do I know about a parabola? Investigating a Parabola Lesson Objective: Students will describe a parabola, using its intercepts, vertex, symmetry, and whether it is pointing up or down. Length of Activity: One day (approximately 50 minutes) Core Problems: Problem 843 Ways of Thinking: Materials: Suggested Lesson Activity: Generalizing, making connections Poster graph paper and markers OR overhead graph paper and overhead pens Have a student volunteer read the problem statement for problem 843, Functions of America. Be sure that all students understand the goals of this investigation. Describe your expectations for the team posters (such as being readable and full of mathematics). Emphasize that completed posters should address all four bullets in the student text. If you use team roles, you may want to discuss the importance of each student fulfilling his or her responsibility in their team. See suggestions in the Team Roles section below. Assign each team a parabola from the list below and get teams started on their investigation. Each parabola will offer new insights during presentations. See the notes below: y = x 2! 2x! 8 This parabola has xintercepts at x = 4 and 2. Observations can be that the yintercept is the constant term in the equation and that the product of the roots is the yintercept. y =!x y = x 2! 4x + 5 This parabola points downward, and its yintercept is the vertex. Its xintercepts are at (2, 0) and ( 2, 0), and the yaxis is its line of symmetry. This parabola has no xintercepts. y = x 2! 2x + 1 This parabola has only one xintercept: (1, 0). y = x 2! 6x + 5 This parabola has xintercepts at x = 1 and x = 5. If students graph it using a narrow domain (such as! 4 " x " 4 ), they will not get a complete graph with both intercepts. Chapter 8: Quadratics 629
28 y =!x 2 + 3x + 4 This parabola has xintercepts at x = 4 and x = 1. The vertex of the parabola occurs at x = 1.5, which will not show up in a table with integer inputs. Therefore, students need to use the idea of symmetry to find the vertex. y =!x 2 + 2x! 1 y = x 2 + 5x + 1 This parabola has only one xintercept at (1, 0) and is the same as y = x 2! 2x + 1 but flipped over. (Students may connect this with flipping tiles to find the opposite expression in Chapter 2.) This is a parabola with no integer xintercepts, so the xintercepts need to be estimated. As you circulate, be prepared to help students with the calculations for their rules, especially teams with rules that include!x 2. Be sure to ask students to justify their conclusions. Here are some questions that can help students support their reasons and use mathematical reasoning: Do you predict that all parabolas behave that way? How do you know? How do you know if that point is the highest (or lowest) point on the parabola? Why does your parabola point downward? What part of the rule causes the parabola to behave that way? Are there any patterns in the table? Do all parabolas have this (these) pattern(s)? Why or why not? If any team is stuck on the meaning of the word symmetry, you may want to point out the Math Notes box at the end of the lesson. This box not only defines the word, but it also offers examples of some lines of symmetry. Remind students to put their findings on their team poster or transparency. Leave enough time for quick presentations. Closure: (15 minutes) Have teams quickly share their findings. If there is not much time remaining to have complete presentations, then encourage one team member (such as the Recorder/Reporter) to make a short statement on one or two important observations that no other team has yet mentioned. As teams share, create a class list of all of the observations from the investigation. Record each statement using the language the student uses, as long as the rest of the class understands the statement. Expect some simplistic statements along with some unique and interesting observations. 630 Algebra Connections
29 The list should contain all of the key elements that we want students to use for the rest of this chapter when describing parabolas: xintercepts, yintercepts, symmetry, vertex, and whether the parabola is pointing up or down. The posters of student work will be referred to in Chapter 11 when students do similar investigations of other types of functions, such as y = x and y = x 3. Team Roles: If you use team roles, emphasize that the Recorder/Reporter has the important task of organizing the creation of the poster or presentation or team report. If you are going to require individual reports from each student, then the Recorder/Reporter is responsible to help teammates describe the team s findings. On the other hand, the team will need the Facilitator to make sure that all team members understand the task and agree with the findings. The Resource Manager will be responsible for quickly asking for assistance from the teacher when the entire team is stuck. Finally, the team will need the Task Manager to be very mindful of the time remaining for the investigation. This student needs to remind the other team members respectfully to stay on task and to make sure that the roles are being fulfilled. Homework: Problems 845 through What do I know about a parabola? Investigating a Parabola Student pages for this lesson are In previous chapters, you have investigated linear equations. In Section 8.2, you will study parabolas. You will learn all you can about their shape, study different equations used to graph them, and see how they can be used in reallife situations. Chapter 8: Quadratics 631
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