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Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Lesson 2: The Multiplication of Polynomials Student Outcomes ο§ Students develop the distributive property for application to polynomial multiplication. Students connect multiplication of polynomials with multiplication of multi-digit integers. Lesson Notes This lesson begins to address standards A-SSE.A.2 and A-APR.C.4 directly and provides opportunities for students to practice MP.7 and MP.8. The work is scaffolded to allow students to discern patterns in repeated calculations, leading to some general polynomial identities that are explored further in the remaining lessons of this module. As in the last lesson, if students struggle with this lesson, they may need to review concepts covered in previous grades, such as: ο· The connection between area properties and the distributive property: Grade 7, Module 6, Lesson 21. ο· Introduction to the table method of multiplying polynomials: Algebra I, Module 1, Lesson 9. ο· Multiplying polynomials (in the context of quadratics): Algebra I, Module 4, Lessons 1 and 2. Since division is the inverse operation of multiplication, it is important to make sure that your students understand how to multiply polynomials before moving on to division of polynomials in Lesson 3 of this module. In Lesson 3, division is explored using the reverse tabular method, so it is important for students to work through the table diagrams in this lesson to prepare them for the upcoming work. There continues to be a sharp distinction in this curriculum between justification and proof, such as justifying the identity (π + π)2 = π2 + 2ππ + π using area properties and proving the identity using the distributive property. The key point is that the area of a figure is always a nonnegative quantity and so cannot be used to prove an algebraic identity where the letters can stand for negative numbers (there is no such thing as a geometric figure with negative area). This is one of many reasons that manipulatives such as Algebra Tiles need to be handled with extreme care: depictions of negative area actually teach incorrect mathematics. (A correct way to model expressions involving the subtraction of two positive quantities using an area model is depicted in the last problem of the Problem Set.) The tabular diagram described in this lesson is purposely designed to look like an area model without actually being an area model. It is a convenient way to keep track of the use of the distributive property, which is a basic property of the number system and is assumed to be true for all real numbersβregardless of whether they are positive or negative, fractional or irrational. Classwork Opening Exercise (5 minutes) The Opening Exercise is a simple use of an area model to justify why the distributive property works when multiplying 28 × 27. When drawing the area model, remember that it really matters that the length of the side of the big square is 1 2 about 2 times the length of the top side of the upper right rectangle (20 units versus 8 units) in the picture below and similarly for the lengths going down the side of the large rectangle. It should be an accurate representation of the area of a rectangular region that measures 28 units by 27 units. Lesson 2: The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 28 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Opening Exercise Show that ππ × ππ = (ππ + π)(ππ + π) using an area model. What do the numbers you placed inside the four rectangular regions you drew represent? ππ Scaffolding: π ππ × ππ = (ππ + π)(ππ + π) πππ ππ πππ = πππ + πππ + πππ + ππ = πππ πππ ππ π For students working above grade level, consider asking them to prove that (π + π)(π + π) = ππ + ππ + ππ + ππ, where π, π, π, and π are all positive real numbers. The numbers placed into the blanks represent the number of unit squares (or square units) in each sub-rectangle. Example 1 (9 minutes) MP.7 Explain that the goal today is to generalize the Opening Exercise to multiplying polynomials. Start by asking students how the expression (π₯ + 8)(π₯ + 7) is similar to the expression 28 × 27. Then suggest that students replace 20 with π₯ in the area model. Since π₯ in (π₯ + 8)(π₯ + 7) can stand for a negative number, but lengths and areas are always positive, an area model cannot be used to represent the polynomial expression (π₯ + 8)(π₯ + 7) without also saying that π₯ > 0. So it is not correct to say that the area model above (with 20 replaced by π₯) represents the polynomial expression (π₯ + 8)(π₯ + 7) for all values of π₯. The tabular method below is meant to remind students of the area model as a visual representation, but it is not an area model. Example 1 Use the tabular method to multiply (π + π)(π + π) and combine like terms. π + ππ π ππ + ππ ππ πππ ο§ π ππ (π + π)(π + π) = ππ + ππ π + ππ π ππ Explain how the result π₯ 2 + 15π₯ + 56 is related to 756 determined in the Opening Exercise. οΊ If π₯ is replaced with 20 in π₯ 2 + 15π₯ + 56, then the calculation becomes the same as the one shown in the Opening Exercise: (20)2 + 15(20) + 56 = 400 + 300 + 56 = 756. Lesson 2: The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 29 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II ο§ How can we multiply these binomials without using a table? οΊ Think of π₯ + 8 as a single number and distribute over π₯ + 7: π₯ + 8 (π₯ + 7) = π₯ + 8 β π₯ + π₯ + 8 β 7 Next, distribute the π₯ over π₯ + 8 and the 7 over π₯ + 8. Combining like terms shows that (π₯ + 8)(π₯ + 7) = π₯ 2 + 15π₯ + 56. (π₯ + 8)(π₯ + 7) = (π₯ + 8) β π₯ + (π₯ + 8) β 7 = π₯2 + 8π₯ + 7π₯ + 56 ο§ What property did we repeatedly use to multiply the binomials? οΊ ο§ MP.7 ο§ The table in the calculation above looks like the area model in the Opening Exercise. What are the similarities? What are the differences? οΊ The expressions placed in each table entry correspond to the expressions placed in each rectangle of the area model. The sum of the table entries represents the product, just as the sum of the areas of the sub-rectangles is the total area of the large rectangle. οΊ One difference is that we might have π₯ < 0 so that 7π₯ and 8π₯ are negative, which does not make sense in an area model. How would you have to change the table so that it represents an area model? οΊ ο§ The distributive property First, all numbers and variables would have to represent positive lengths. So, in the example above, we would have to assume that π₯ > 0. Second, the lengths should be commensurate with each other; that is, the side length for the rectangle represented by 7 should be slightly shorter than the side length represented by 8. How is the tabular method similar to the distributive property? οΊ ο§ The sum of the table entries is equal to the result of repeatedly applying the distributive property to (π₯ + 8)(π₯ + 7). The tabular method graphically organizes the results of using the distributive property. Scaffolding: Does the table work even when the binomials do not represent lengths? Why? οΊ Yes it does because the table is an easy way to summarize calculations done with the distributive propertyβa property that works for all polynomial expressions. Exercises 1β2 (6 minutes) If students need to work another problem, ask students to use an area model to find 16 × 19 and then use the tabular method to find (π₯ + 6)(π₯ + 9). Allow students to work in groups or pairs on these exercises. While Exercise 1 is analogous to the previous example, in Exercise 2, students may need time to think about how to handle the zero coefficient of π₯ in π₯ 2 β 2. Allow them to struggle and discuss possible solutions. Lesson 2: The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 30 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Exercises 1β2 1. Use the tabular method to multiply (ππ + ππ + π)(ππ β ππ + π) and combine like terms. Sample student work: (ππ + ππ + π)(ππ β ππ + π) = ππ β πππ β ππππ + π + π 2. Use the tabular method to multiply (ππ + ππ + π)(ππ β π) and combine like terms. Sample student work: (ππ + ππ + π)(ππ β π) = ππ + πππ β ππ β ππ β π Another solution method would be to omit the row for ππ in the table and to manually add all table entries instead of adding along the diagonals: (ππ + ππ + π)(ππ β π) = ππ + πππ + ππ β πππ β ππ β π = ππ + πππ β ππ β ππ β π Example 2 (6 minutes) Scaffolding: For further scaffolding, consider asking students to see the pattern using numerical expressions, such as: Prior to Example 2, consider asking students to find the products of each of these expressions. (π₯ β 1)(π₯ + 1) (π₯ β 1)(π₯ 2 + π₯ + 1) (π₯ β 1)(π₯ 3 + π₯ 2 + π₯ + 1) Students may work on this in mixed-ability groups and come to generalize the pattern. Lesson 2: The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 (2 β 1)(21 + 1) (2 β 1)(22 + 2 + 1) (2 β 1)(23 + 22 + 2 + 1). Can they describe in words or symbols the meaning of these quantities? 31 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Example 2 Multiply the polynomials (π β π)(ππ + ππ + ππ + π + π) using a table. Generalize the pattern that emerges by writing down an identity for (π β π)(ππ + ππβπ + β― + ππ + π + π) for π a positive integer. π ππ MP.8 ππ βπ π βππ ππ ππ βππ ππ ππ βππ ππ ππ βπ π π βπ π π π πππ πππ ππ (π β π)(ππ + ππ + ππ + π + π) = ππ β π βπ The pattern suggests (π β π)(ππ + ππβπ + β― + ππ + π + π) = ππ+π β π. ο§ What quadratic identity from Algebra I does the identity above generalize? οΊ This generalizes (π₯ β 1)(π₯ + 1) = π₯ 2 β 1, or more generally, the difference of squares formula (π₯ β π¦)(π₯ + π¦) = π₯ 2 β π¦ 2 with π¦ = 1. We will explore this last identity in more detail in Exercises 4 and 5. Exercises 3β4 (10 minutes) MP.1 Before moving on to Exercise 3, it may be helpful to scaffold the problem by asking students to multiply (π₯ β π¦)(π₯ + π¦) and (π₯ β π¦)(π₯ 2 + π₯π¦ + π¦ 2 ). Ask students to make conjectures about the form of the answer to Exercise 3. Exercises 3β4 3. Multiply (π β π)(ππ + ππ π + πππ + ππ ) using the distributive property and combine like terms. How is this calculation similar to Example 2? Distribute the expression ππ + ππ π + πππ + ππ through π β π to get (π β π)(ππ + ππ π + πππ + ππ ) = ππ + ππ π + ππ ππ + πππ β ππ π β ππ ππ β πππ β ππ = ππ β ππ Substitute π in for π to get the identity for π = π in Example 2. This calculation is similar to Example 2 because it has the same structure. Substituting π for π results in the same expression as Example 2. Exercise 3 shows why the mnemonic FOIL is not very helpfulβand in this case does not make sense. By now, students should have had enough practice multiplying to no longer require such mnemonics to help them. They understand that the multiplications they are doing are really repeated use of the distributive property, an idea that started when they learned the multiplication algorithm in Grade 4. However, it may still be necessary to summarize the process with a mnemonic. If this is the case, try Each-With-Each, or EWE, which is short for the process of multiplying each term of one polynomial with each term of a second polynomial and combining like terms. Lesson 2: The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 32 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II To introduce Exercise 4, consider starting with a group activity to help illuminate the generalization. For example, students could work in groups again to investigate the pattern found in expanding these expressions. (π₯ 2 + π¦ 2 )(π₯ 2 β π¦ 2 ) 3 3 3 Scaffolding: 3 For further scaffolding, consider asking students to see the pattern using numerical expressions, such as: (π₯ + π¦ )(π₯ β π¦ ) (π₯ 4 + π¦ 4 )(π₯ 4 β π¦ 4 ) (π₯ 5 + π¦ 5 )(π₯ 5 β π¦ 5 ) (32 + 22 )(32 β 22 ) (33 + 23 )(33 β 23 ) (34 + 24 )(34 β 24 ). Multiply (ππ β ππ )(ππ + ππ ) using the distributive property and combine like terms. Generalize the pattern that emerges to write down an identity for (ππ β ππ )(ππ + ππ ) for positive integers π. 4. (ππ π )(ππ βπ π) +π Generalization: (ππ = (ππ βπ π )(ππ βπ π) π (ππ π) ππ β π + +π =π π) βπ π π π π π π π π π β π =π βπ π +π π βπ =π βπ . ππ βπ . How do 13 β 5 and 34 β 24 relate to the first line? How do 35 β 19 and 36 β 26 relate to the second line? Etc. Sample student work: (ππ β ππ )(ππ + ππ ) = πππ β πππ ο§ The generalized identity π₯ 2π β π¦ 2π = (π₯ π β π¦ π )(π₯ π + π¦ π ) is used several times in this module. For example, it helps to recognize that 2130 β 1 is not a prime number because it can be written as (265 β 1)(265 + 1). Some of the problems in the Problem Set rely on this type of thinking. Closing (4 minutes) Ask students to share two important ideas from the dayβs lesson with their neighbor. You can also use this opportunity to informally assess their understanding. ο§ Multiplication of two polynomials is performed by repeatedly applying the distributive property and combining like terms. ο§ There are several useful identities: - (π + π)(π + π) = ππ + ππ + ππ + ππ (an example of each-with-each) - (π + π)2 = π2 + 2ππ + π 2 - (π₯ π β π¦ π )(π₯ π + π¦ π ) = π₯ 2π β π¦ 2π , including (π₯ β π¦)(π₯ + π¦) = π₯ 2 β π¦ 2 and (π₯ 2 β π¦ 2 )(π₯ 2 + π¦ 2 ) = π₯ 4 β π¦ 4 - (π₯ β 1)(π₯ π + π₯ πβ1 + β― π₯ 2 + π₯ + 1) = π₯ π+1 β 1 Lesson 2: The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 33 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II ο§ (Optional) Consider a quick white board activity in which students build fluency with applying these identities. The vocabulary used in this lesson was introduced and taught in Algebra I. The definitions included in this lesson are for reference. To support students, consider creating a poster with these vocabulary words for the classroom wall. Relevant Vocabulary EQUIVALENT POLYNOMIAL EXPRESSIONS: Two polynomial expressions in one variable are equivalent if, whenever a number is substituted into all instances of the variable symbol in both expressions, the numerical expressions created are equal. POLYNOMIAL IDENTITY: A polynomial identity is a statement that two polynomial expressions are equivalent. For example, (π + π)π = ππ + ππ + π for any real number π is a polynomial identity. COEFFICIENT OF A MONOMIAL: The coefficient of a monomial is the value of the numerical expression found by substituting the number π into all the variable symbols in the monomial. The coefficient of πππ is π, and the coefficient of the monomial (ππππ) β π is ππ. TERMS OF A POLYNOMIAL: When a polynomial is expressed as a monomial or a sum of monomials, each monomial in the sum is called a term of the polynomial. LIKE TERMS OF A POLYNOMIAL: Two terms of a polynomial that have the same variable symbols each raised to the same power are called like terms. STANDARD FORM OF A POLYNOMIAL IN ONE VARIABLE: A polynomial expression with one variable symbol, π, is in standard form if it is expressed as ππ ππ + ππβπ ππβπ + β― + ππ π + ππ , where π is a non-negative integer, and ππ , ππ , ππ , β¦ , ππ are constant coefficients with ππ β π. A polynomial expression in π that is in standard form is often just called a polynomial in π or a polynomial. The degree of the polynomial in standard form is the highest degree of the terms in the polynomial, namely π. The term ππ ππ is called the leading term and ππ (thought of as a specific number) is called the leading coefficient. The constant term is the value of the numerical expression found by substituting π into all the variable symbols of the polynomial, namely ππ . Exit Ticket (5 minutes) Lesson 2: The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 34 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Name Date Lesson 2: The Multiplication of Polynomials Exit Ticket Multiply (π₯ β 1)(π₯ 3 + 4π₯ 2 + 4π₯ β 1) and combine like terms. Explain how you reached your answer. Lesson 2: The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 35 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Exit Ticket Sample Solutions Multiply (π β π)(ππ + πππ + ππ β π) and combine like terms. Explain how you reached your answer. Tabular method: ππ πππ ππ π βπ ππ βππ ππ πππ βπππ πππ πππ βππ ππ βπ π βπ (π β π)(ππ + πππ + ππ β π) = ππ + πππ β ππ + π π βππ π Using the distributive property (Each-With-Each): (π β π)(ππ + πππ + ππ β π) = ππ + πππ + πππ β π β ππ β πππ β ππ + π = ππ + πππ β ππ + π. Problem Set Sample Solutions 1. Complete the following statements by filling in the blanks. a. (π + π)(π + π + π) = ππ + ππ + ππ + ____ + ____ + ____ b. (π β π)π = ( c. (ππ + ππ)π = (ππ)π + π(ππ)(ππ) + ( d. (π β π)(π + π + ππ ) = e. ππ β ππ = (π + f. (ππ + ππ)(ππ β ππ) = g. (πππ β π)(πππ + π) = h. [(π β π) β π][(π β π) + π] = ( Lesson 2: π ) β( )ππ + ππ π, π π ) βπ )(π β ππ, ππ , ππ ππ ππ ) π, π β πππ , ππππ β π πππ π ) β π The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 πβπ 36 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II 2. Use the tabular method to multiply and combine like terms. a. (ππ β ππ + π)(π + π) Sample student work: (ππ β ππ + π)(π + π) = ππ β ππ β ππ + ππ b. (ππ β πππ β πππ )(ππ β ππππ ) Sample student work: (ππ β πππ β πππ )(ππ β ππππ ) = πππππ + πππππ β πππππ β ππππ + πππ c. (πππ + ππ β ππ β π)(ππ β ππ β π) Sample student work: (πππ + ππ β ππ β π)(ππ β ππ β π) = πππ β ππππ β ππππ β ππ + πππ + ππ d. (ππ β ππ + π)(ππ + ππ + π) Sample student work: (ππ β ππ + π)(ππ + ππ + π) = ππ + πππ + ππ Lesson 2: The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 37 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II 3. Multiply and combine like terms to write as the sum or difference of monomials. a. ππ(π + ππ) b. π ππ + πππ c. ππ + πππ + π π (πππ + ππππ ) π d. (π β π)(π + π) f. π (πππ β π)(πππ + π) h. ππ³ π β ππ i. ππππππ (ππππππ + π. ππππ) j. (π β π)(π + π)(ππ + π) l. n. π π(π + π)(π + π)(π + π) π π p. π (π + π)(ππ β ππ + π β π) r. ππ β π s. (π + π)(ππ β ππ + ππ β ππ + π β π) ππ β π (π + π)(ππ β ππ + ππ β ππ + ππ β ππ + π β π) t. ππ β π 4. π(π + π)(π + π)(π + π)(π + π) ππ + ππππ + ππππ + ππππ + πππ π + ππ + πππ + ππ q. π(π + π)(π + π) ππ + πππ + ππ π + ππ + πππ + π o. (π β π)(ππ + ππ + ππ + ππ + π + π) ππ β π (π + π)(π + π)(π + π) π (ππ + π)(πππ + π) πππ + πππ + ππ + π ππ β π m. (βππ β π)ππ βπππ β πππ πππππππ + πππππ k. (πππ β π)(πππ + π) πππππ β π π + π β ππ g. πππ (πππ β πππ ) ππππ β ππππ πππ + πππ e. ππ (π + π) + π (ππ β ππ + π)(ππ β π + π) ππ β ππ + πππ β ππ + π Polynomial expressions can be thought of as a generalization of place value. a. Multiply πππ × πππ using the standard paper-and-pencil algorithm. b. Multiply (πππ + π + π)(ππ + π + π) using the tabular method and combine like terms. (πππ + π + π)(ππ + π + π) = πππ + πππ + πππ + ππ + π Lesson 2: The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 38 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 2 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II c. Substitute π = ππ into your answer from part (b). ππ, πππ d. Is the answer to part (c) equal to the answer from part (a)? Compare the digits you computed in the algorithm to the coefficients of the entries you computed in the table. How do the place-value units of the digits compare to the powers of the variables in the entries? Yes. The digits computed in the algorithm are the same as the coefficients computed in the table entries. The zero-degree term in the table corresponds to the ones unit, the first-degree terms in the table correspond to the tens unit, the second-degree terms in the table correspond to the hundreds unit, and so on. 5. Jeremy says (π β π)(ππ + ππ + ππ + ππ + ππ + ππ + π + π) must equal ππ + ππ + ππ + ππ + ππ + ππ + π + π because when π = ππ, multiplying by π β π is the same as multiplying by π. a. Multiply (π β π)(ππ + ππ + ππ + ππ + ππ + ππ + π + π). ππ β πππ β πππ β πππ β πππ β πππ β πππ β ππ β π b. Substitute π = ππ into your answer. πππβπππβπππ β ππβπππβπππ β πβπππβπππ β πππβπππ β πππππ β ππππ β πππ β ππ β π πππβπππβπππ β ππβπππβπππ = ππβπππβπππ c. Is the answer to part (b) the same as the value of ππ + ππ + ππ + ππ + ππ + ππ + π + π when π = ππ? Yes d. Was Jeremy right? No, just because it is true when π is ππ does not make it true for all real π. The two expressions are not algebraically equivalent. 6. In the diagram, the side of the larger square is π units, and the side of the smaller square is π units. The area of the shaded region is (ππ β ππ ) square units. Show how the shaded area might be cut and rearranged to illustrate that the area is (π β π)(π + π) square units. π π Solution: π+π π πβπ Lesson 2: π πβπ The Multiplication of Polynomials This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 39 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.