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LESSON 6.5 Dividing Polynomials Name Class 6.5 HARDBOUND SE PAGE 281 BEGINS HERE Date Dividing Polynomials Essential Question: What are some ways to divide polynomials, and how do you know when the divisor is a factor of the dividend? Common Core Math Standards The student is expected to: A-APR.6 Evaluating a Polynomial Function Using Synthetic Substitution Explore Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system. Also A-APR.1, A-APR.3 Polynomials can be written in something called nested form. A polynomial in nested form is written in such a way that evaluating it involves an alternating sequence of additions and multiplications. For instance, the nested form of p(x) = 4x 3 + 3x 2 + 2x + 1 is p(x) = x(x(4x + 3) + 2) + 1, which you evaluate by starting with 4, multiplying by the value of x, adding 3, multiplying by x, adding 2, multiplying by x, and adding 1. Mathematical Practices A MP.2 Reasoning Given p(x) = 4x 3 + 3x 2 + 2x + 1, find p(-2). Rewrite p(x) as p(x) = x(x(4x + 3) + 2) + 1. Language Objective Multiply. Work in small groups to complete a compare and contrast chart for dividing polynomials. Add. Multiply. -2 · 4 = -8 -8 + 3 = -5 -5 · -2 = 10 10 + 2 = 12 Add. ENGAGE Multiply. Essential Question: What are some ways to divide polynomials, and how do you know when the divisor is a factor of the dividend? You can set up an array of numbers that captures the sequence of multiplications and additions needed to find p(a). Using this array to find p(a) is called synthetic substitution. PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss the photo and how the number of teams and the attendance are both functions of time. Then preview the Lesson Performance Task. Add. © Houghton Mifflin Harcourt Publishing Company Possible answer: You can divide polynomials using long division or, for a divisor of the form x - a, synthetic division. The divisor is a factor of the dividend when the remainder is 0. 12 · -2 = -24 -24 + 1 = -23 Given p(x) = 4x 3 + 3x 2 + 2x + 1, find p(-2) by using synthetic substitution. The dashed arrow indicates bringing down, the diagonal arrows represent multiplication by –2, and the solid down arrows indicate adding. The first two steps are to bring down the leading number, 4, then multiply by the value you are evaluating at, -2. -2 4 3 2 1 2 1 -8 4 B Add 3 and –8. -2 4 3 -8 -5 4 Module 6 be ges must EDIT--Chan DO NOT Key=NL-B;CA-B Correction Lesson 5 365 gh "File info" made throu Date Class Name Dividing s Polynomial when do you know s, and how polynomial ways to divide nd? A-APR.1, are some CA2. Also of the divide ion: What g on page r is a factor table startin the diviso rd, see the of this standa n the full text 6.5 Resource Locker Quest Essential A-APR.6 For A-APR.3 Explore IN3_MNLESE389885_U3M06L5 365 ial Functio a Polynom tion way in such a stitu Evaluating form is written thetic Sub form of mial in nested ce, the nested Using Syn form. A polyno by ns. For instan HARDCOVER PAGES 281290 lying licatio called nested 4, multip ns and multip in something starting with ce of additio evaluate by can be written ting sequen which you Polynomials es an alterna + 2) + 1, adding 1. ting it involv x(4x + 3) by x, and p(x) = x( that evalua 2 multiplying 2x + 1 is 3 adding 2, + 3x + lying by x, p(x) = 4x 3, multip of x, adding the value Turn to these pages to find this lesson in the hardcover student edition. p(-2). 2 2x + 1, find 3 + 3x + ) + 2) + 1. x(4x + 3 (x) = x( p(x) as p Rewrite -8 -2 · 4 = Multiply. -5 -8 + 3 = Add. = 10 -5 · -2 Multiply. 12 10 + 2 = Add. = -24 12 · -2 and Multiply. lications = -23 ce of multip etic substitution. -24 + 1 es the sequen synth Add. that captur find p(a) is called d to of numbers . The dashe up an array p(a). Using this array tic substitutionby –2, and the You can set find to d using synthe neede lication p(-2) by additions ent multip + 1, find 2 3 + 3x + 2x arrows repres diagonal the (x) = 4x p you Given ng down, by the value tes bringi multiply te adding. arrow indica er, 4, then arrows indica leading numb solid down down the are to bring 1 two steps 2 The first -2. 3 ting at, 4 are evalua ) = 4x Given p(x -2 -8 n Mifflin Harcour t Publishin y g Compan © Houghto 4 Add 3 and –8. -2 2 3 4 1 -8 4 -5 Lesson 5 365 Module 6 6L5 365 85_U3M0 ESE3898 IN3_MNL 365 Lesson 6.5 Resource Locker 19/07/14 2:20 AM 19/07/14 2:16 AM Multiply the previous answer by –2. -2 4 3 -8 Continue this sequence of steps until you reach the last addition. 1 -2 10 -5 4 2 4 4 3 2 1 -8 10 -24 -5 12 -23 Evaluating a Polynomial Function Using Synthetic Substitution p(-2) = -23 INTEGRATE TECHNOLOGY HARDBOUND SE Reflect 1. EXPLORE PAGE 282 BEGINS HERE Discussion After the final addition, what does this sum correspond to? The final sum represents the value of p(x) where x = -2. QUESTIONING STRATEGIES Dividing Polynomials Using Long Division Explain 1 What operation are you doing and what are you finding when you use synthetic substitution on the polynomial function p(x) to find p(a)? You are dividing the polynomial function p(x) by the quantity (x - a), and you are finding the value of p(a). Recall that arithmetic long division proceeds as follows. Divisor –––– 23 ← Quotient 12 ⟌ 277 ← Dividend 24 ― 37 36 ― 1 ← Remainder dividend remainder Notice that the long division leads to the result _______ = quotient + ________ . Using the divisor divisor 277 1 numbers from above, the arithmetic long division leads to ___ = 23 + __ . Multiplying through 12 12 by the divisor yields the result dividend = (divisor)(quotient) + remainder. (This can be used as EXPLAIN 1 © Houghton Mifflin Harcourt Publishing Company a means of checking your work.) Example 1 Given a polynomial divisor and dividend, use long division to find the quotient and remainder. Write the result in the form dividend = (divisor)(quotient) + remainder and then carry out the multiplication and addition as a check. (4x 3 + 2x 2 + 3x + 5) ÷ (x 2 + 3x + 1) Begin by writing the dividend in standard form, including terms with a coefficient of 0 (if any). 4x 3 + 2x 2 + 3x + 5 Dividing Polynomials Using Long Division AVOID COMMON ERRORS Write division in the same way as you would when dividing numbers. ––––––––––––––––– x 2 + 3x + 1 ⟌ 4x 3 + 2x 2 + 3x + 5 Module 6 Students have the option of completing the polynomial division activity either in the book or online. 366 Lesson 5 PROFESSIONAL DEVELOPMENT IN3_MNLESE389885_U3M06L5 366 Math Background Division of polynomials is related to division of whole numbers. Given P(x) R(x) polynomials P(x) and D(x), where D(x) ≠ 0, we can write _____ = Q(x) + _____ , D(x) D(x) where the remainder R(x) is a polynomial whose degree is less than that of D(x) . (If the degree of R(x) were not less than the degree of D(x), we would be able to continue dividing.) Equivalently, P(x) = Q(x)D(x) + R(x). This last expression can be used to justify the Remainder Theorem. Notice that when D(x) is a linear divisor of the form x - a, the expression becomes P(x) = Q(x)(x - a) + r, where the remainder r is a real number. 23/07/14 12:19 AM Students may have difficulty relating the familiar long-division process for whole numbers to identifying the process for polynomials using the algorithm for finding dividend = (divisor)(quotient) + remainder. Point out that polynomial long division remainder , dividend = quotient + _________ leads to this result: ________ divisor divisor which is equivalent to dividend = (divisor)(quotient) + remainder if you multiply each term by the divisor. Showing an example of arithmetic long division alongside an example of polynomial division may help students make the connection. Dividing Polynomials 366 Find the value you need to multiply the divisor by so that the first term matches with the first term of the dividend. In this case, in order to get 4x 2, we must multiply x 2 by 4x. This will be the first term of the quotient. 4x QUESTIONING STRATEGIES ––––––––––––––––– How can you tell if you are finished solving a polynomial division problem? The remainder has a degree less than the degree of the divisor, or has degree 0. What do you write as the final answer for a polynomial division problem? The answer should be written as the product of factors plus the remainder, where one factor is the divisor and the other factor is the quotient. x 2 + 3x + 1 ⟌ 4x 3 + 2x 2 + 3x + 5 Next, multiply the divisor through by the term of the quotient you just found and subtract that value from the dividend. (x 2 + 3x + 1)(4x) = 4x 3 + 12x 2 + 4x, so subtract 4x 3 + 12x 2 + 4x from 4x 3 + 2x 2 + 3x + 5. ––––––––––––––––– 4x x 2 + 3x + 1 ⟌ 4x 3 + 2x 2 + 3x + 5 -(4x 3 + 12x 2 + 4x) ――――――― -10x 2 - x + 5 HARDBOUND SE Taking this difference as the new dividend, continue in this fashion until the largest term of the remaining dividend is of lower degree than the divisor. PAGE 283 BEGINS HERE 4x - 10 ––––––––––––––––– x 2 + 3x + 1 ⟌ 4x 3 + 2x 2 + 3x + 5 -(4x 3 + 12x 2 + 4x) ――――――― -10x - x + 5 2 -(-10x 2 - 30x - 10) ―――――――― 29x + 15 Since 29x + 5 is of lower degree than x 2 + 3x + 1, stop. 29x + 15 is the remainder. Write the final answer. 4x 3 + 2x 2 + 3x + 5 = (x 2 + 3x + 1)(4x - 10) + 29x + 15 Check. 4x 3 + 2x 2 + 3x + 5 = (x 2 + 3x + 1)(4x - 10) + 29x + 15 © Houghton Mifflin Harcourt Publishing Company = 4x 3 + 12x 2 + 4x - 10x 2 - 30x - 10 + 29x + 15 = 4x 3 + 2x 2 + 3x + 5 B (6x 4 + 5x 3 + 2x + 8) ÷ (x 2 + 2x - 5) Write the dividend in standard form, including terms with a coefficient of 0. 6x 4 + 5x 3 + 0x 2 + 2x + 8 Write the division in the same way as you would when dividing numbers. –––––––––––––––––––––– x 2 + 2x - 5 ⟌ 6x 4 + 5x 3 + 0x 2 + 2x + 8 Module 6 367 Lesson 5 COLLABORATIVE LEARNING IN3_MNLESE389885_U3M06L5 367 Small Group Activity Help groups of students practice dividing polynomials using synthetic division. Provide each student with a different example of polynomial division, and ask them to show and explain the first step in dividing with synthetic division. Then have them pass the problem to another student, who writes the next step and explains it. They continue to pass the problem until each problem is completely solved and all steps are explained. The last student summarizes by giving the quotient and remainder in polynomial form. Encourage students to use these as examples of dividing polynomials when they write in their journals. 367 Lesson 6.5 19/07/14 2:44 AM Divide. 6x 2 - 7x + 44 –––––––––––––––––––––– x 2 + 2x - 5 ⟌ 6x 4 + 5x 3 + 0x 2 + 2x + 8 -(6x 4 + 12x 3 - 30x 2) ―――――――――― ( -7x 3 + 30x 2 + 2x 2 - -7x 3 -14x + 35x ) ――― 44x 2 - 33x + 8 - ( 44x + 88x - 220 ) 2 ――― -121x + 228 Write the final answer. 6x 4 + 5x 3 + 2x + 8 = (x 2 + 2x - 5)(6x 2 - 7x + 44) - 121x + 228 HARDBOUND SE Check. PAGE 284 6x 4 + 5x 3 + 2x + 8 = (x 2 + 2x - 5)(6x 2 - 7x + 44) -121x + 228 BEGINS HERE = 6x 4 - 7x 3 + 44x 2 + 12x 3 -14x 2 + 88x - 30x 2 + 35x - 220 - 121x + 228 = 6x 4 + 5x 3 + 2x + 8 Reflect 2. How do you include the terms with coefficients of 0? You represent the term with 0 as the coefficient, e.g, 0x. 3. (15x 3 + 8x - 12) ÷ (3x 2 + 6x + 1) (3x 2 + 6x + 1)(5x - 10) + 63x - 2 5x - 10 –––––––––––––––––––– 3x 2 + 6x + 1 ⟌ 15x 3 + 0x 2 + 8x - 12 = 15x 3 - 30x 2 + 30x 2 - 60x + 5x - 10 + -(15x 3 + 30x 2 + 5x) 63x - 2 ―――――――― = 15x 3 + 8x - 12 -30x + 3x - 12 2 -(-30x 2 - 60x - 10) © Houghton Mifflin Harcourt Publishing Company Your Turn Use long division to find the quotient and remainder. Write the result in the form dividend = (divisor)(quotient) + remainder and then carry out a check. ――――――――― 63x - 2 Module 6 368 Lesson 5 DIFFERENTIATE INSTRUCTION IN3_MNLESE389885_U3M06L5 368 Graphic Organizers 19/07/14 3:04 AM Have groups of students create graphic organizers to help them divide polynomials using synthetic division. Have them show how to organize a problem into a form similar to the one shown. Then have them use organizers to write each of the steps, explain what goes into each of the cells, and then interpret the results. Dividing Polynomials 368 (9x 4 + x 3 + 11x 2 - 4) ÷ (x 2 + 16) 4. EXPLAIN 2 x2 Dividing p(x) by x - a Using Synthetic Division 9 -4 -6 7+(-4) 2 -2(2) 3 -2(3) 3 ) 16x + 2124 2 = 9x 4 + x 3 + 11x 2 - 4 x - 133x + 0x 3 2 -(x 3 + 0x 2 + 16x) ――――――― -133x 2 - 16x - 4 -(-133x 2 + 0x - 2128) ―――――――――― -16x + 2124 Explain 2 Dividing p(x) by x - a Using Synthetic Division Compare long division with synthetic substitution. There are two important things to notice. The first is that p(a) is equal to the remainder when p(x) is divided by x - a. The second is that the numbers to the left of p(a) in the bottom row of the synthetic substitution array give the coefficients of the quotient. For this reason, synthetic substitution is also called synthetic division. Long Division –––––––––––––––––– Synthetic Substitution 3x 2 + 10x + 20 x - 2 ⟌ 3x 3 + 4x 2 + 0x + 10 -(3x 3 - 6x 2) 10x 2 + 0x -(10x 2 - 20x) 20x + 10 -20x - 40 ―――― 50 ――――― 2 3 4 0 10 6 20 40 ―――――― 3 10 20 50 ―――――― 9+(-6) 3 Students should notice that the divisor is -2 because the form of the divisor is (x - a), or (x - (-2)); the rows are added; the remainder is the last digit in the last row, 3; and the last row includes the coefficients of the quotient, starting with the power of the variable decreased by 1. 3 (2x 2 + 7x + 9) ÷ (x + 2) = 2x + 3 + _____ x+2 = 9x 4 + x 3 - 133x 2 + 144x 2 + 16x - 2128 - 2 ―――――――― Example 2 HARDBOUND SE © Houghton Mifflin Harcourt Publishing Company bring down 2 7 3 -(9x + 0x + 144x patterns in synthetic division. Have students use arrows and expressions, if necessary, to help them understand the patterns. For example, (2x 2 + 7x + 9) ÷ (x + 2) may be shown as: 2 4 4 INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 Students should quickly see that there are -2 (x 2 + 16)(9x 2 + x - 133) - 16x + 2124 9x 2 + x - 133 + 16 ⟌–––––––––––––––––––––––– 9x + x + 11x + 0x - 4 Given a polynomial p(x), use synthetic division to divide by x - a and obtain the quotient and the (nonzero) remainder. Write the result in the form p(x) = (x - a) (quotient) + p(a) then carry out the multiplication and addition as a check. (7x 3 - 6x + 9) ÷ (x + 5) By inspection, a = -5. Write the coefficients and a in the synthetic division format. -5 Bring down the first coefficient. Then multiply and add for each column. -5 Write the result, using the non-remainder entries of the bottom row as the coefficients. PAGE 285 BEGINS HERE Check. 7 0 -6 9 ―――――― 0 -6 9 -35 175 -845 ――――――――― 7 -35 169 -836 7 (7x 3 - 6x + 9) = (x + 5)(7x 2 - 35x + 169) - 836 (7x 3 - 6x + 9) = (x + 5)(7x 2 - 35x + 169) - 836 = 7x 3 - 35x 2 - 35x 2 - 175x + 169x + 845 - 836 = 7x 3 - 6x + 9 Module 6 369 Lesson 5 LANGUAGE SUPPORT IN3_MNLESE389885_U3M06L5 369 Connect Vocabulary Help students understand how the method synthetic division is used as a symbolic representation of a polynomial division problem. Point out that the English word synthetic means not genuine, unnatural, artificial, or contrived. That implies they will have to understand how to interpret the numerical results in the last row of the synthetic division problem. To help students remember the value of a for the divisor (x - a), show them how to form the equation x - a = 0, and then solve that equation for a. This procedure will remind students to use the correct sign for the divisor. 369 Lesson 6.5 23/07/14 12:27 AM B (4x 4 - 3x 2 + 7x + 2) ÷ (x - _21 ) QUESTIONING STRATEGIES Find a. Then write the coefficients and a in the synthetic division format. How can you recognize the quotient and remainder when using the synthetic division method? The bottom row of the synthetic division problem gives the coefficients of the quotient along with the remainder of the division problem. 1 Find a = _ 2 1 _ 2 4 0 -3 7 2 ――――――― Bring down the first coefficient. Then multiply and add for each column. 1 _ 2 4 0 -3 7 2 2 1 -1 3 ――――――― 4 2 -2 6 5 Write the result. (x - _12 )(4x + 2x - 2x + 6) + 5 (4x 4 - 3x 2 + 7x + 2)= 2 3 Check. (4x 4 - 3x 2 + 7x + 2)= (x - _12 )(4x + 2x - 2x + 6) + 5 3 2 = 4x 4 + 2x 3 - 2x 2 + 6x - 2x 3 - x 2 + x - 3 + 5 = 4x 4 + 3x 2 + 7x + 2 Reflect 5. Can you use synthetic division to divide a polynomial by x 2 + 3? Explain. No, the divisor must be a linear binomial in the form x - a; x 2 + 3 is a quadratic binomial. Your Turn 6. (2x 3 + 5x 2 - x + 7) ÷ (x - 2) 2 2 5 -1 4 (x - 2)(2x 2 + 9x + 17) + 41 7 = 2x 3 - 4x 2 + 9x 2 - 18x + 17x - 34 + 41 18 34 ―――――― 2 9 17 41 7. (6x 4 - 25x 3 - 3x + 5) ÷ 1 -_ 3 6 -25 0 -3 = 2x 3 + 5x 2 - x + 7 (x + _31 ) (x + _13 )(6x - 27x + 9x - 6) + 7 3 5 2 © Houghton Mifflin Harcourt Publishing Company Given a polynomial p(x), use synthetic division to divide by x - a and obtain the quotient and the (nonzero) remainder. Write the result in the form p(x) = (x - a)(quotient) + p(a). You may wish to perform a check. = 6x + 2x - 27x - 9x 2 + 9x 2 + 3x - 6x -2 9 -3 2 ――――――― 6 -27 9 -6 7 4 3 3 -2+7 = 6x 4 - 25x 3 - 3x + 5 Module 6 IN3_MNLESE389885_U3M06L5 370 370 Lesson 5 19/07/14 3:07 AM Dividing Polynomials 370 HARDBOUND SE EXPLAIN 3 Using the Remainder Theorem and Factor Theorem Explain 3 PAGE 286 BEGINS HERE When p(x) is divided by x - a, the result can be written in the form p(x) = (x - a)q(x) + r where q(x) is the quotient and r is a number. Substituting a for x in this equation gives p(a) = (a - a)q(a) + r. Since a - a = 0, this simplifies to p(a) = r. This is known as the Remainder Theorem. Using the Remainder Theorem and Factor Theorem If the remainder p(a) in p(x) = (x - a)q(x) + p(a) is 0, then p(x) = (x - a)q(x), which tells you that x - a is a factor of p(x). Conversely, if x - a is a factor of p(x), then you can write p(x) as p(x) = (x - a)q(x), and when you divide p(x) by x - a, you get the quotient q(x) with a remainder of 0. These facts are known as the Factor Theorem. QUESTIONING STRATEGIES How are the Remainder Theorem and the Factor Theorem related? The Remainder Theorem implies that if a polynomial p(x) is divided by x - a, and the remainder p(a) = 0, then (x - a) is a factor of the polynomial. The Factor Theorem says that if (x - a) is a factor of p(x), you can rewrite the polynomial as the quotient q(x) times this factor, or p(x) = (x - a)q(x). Determine whether the given binomial is a factor of the polynomial p(x). If so, find the remaining factors of p(x). Example 3 p(x) = x 3 + 3x 2 - 4x - 12; (x + 3) Use synthetic division. -3 3 -4 -12 -3 0 12 ―――――――― 1 0 -4 0 1 Since the remainder is 0, x + 3 is a factor. Write q(x) and then factor it. q(x) = x 2 - 4 = (x + 2)(x - 2) So, p(x) = x 3 + 3x 2 - 4x - 12 = (x + 2)(x - 2)(x + 3). p(x) = x 4 - 4x 3 - 6x 2 + 4x + 5; (x + 1) Use synthetic division. © Houghton Mifflin Harcourt Publishing Company -1 -4 -6 4 5 -1 5 1-5 ――――――― -5 -1 5 0 1 1 Since the remainder is 0 , (x + 1) is a factor. Write q(x). q(x) = x - 5x - x +5 3 2 Now factor q(x) by grouping. q(x) = x 3 - 5x 2 - x +5 = x 2(x - 5) - (x - 5) = (x 2 - 1)(x - 5) = (x + 1)(x - 1)(x - 5) So, p(x) = x 4 - 4x 3 - 6x 2 + 4x + 5 = Module 6 IN3_MNLESE389885_U3M06L5 371 371 Lesson 6.5 (x + 1)(x + 1)(x - 1)(x - 5) 371 . Lesson 5 19/07/14 3:08 AM HARDBOUND SE Your Turn PAGE 287 Determine whether the given binomial is a factor of the polynomial p(x). If it is, find the remaining factors of p(x). 8. p(x) = 2x 4 + 8x 3 + 2x + 8; (x + 4) -4 9. 2 8 0 2 8 -8 0 0 -8 ―――――――― 2 0 0 2 0 3 0 -2 5 3 3 1 ―――――― ――――― 3 3 1 6 ELABORATE QUESTIONING STRATEGIES Since the remainder is 0, (x + 4) is a factor. When do you use synthetic substitution, and when do you use synthetic division? You use synthetic substitution when you want to find the function value of a polynomial for a certain number. You use synthetic division when you want to find the quotient of polynomial division. q(x) = 2x 3 + 2 = 2(x 3 + 1) = 2(x + 1)(x 2 - x + 1) So, p(x) = 2x 4 + 8x 3 + 2x + 8 = 2(x + 1)(x 2 - x + 1)(x + 4) p(x) = 3x 3 - 2x + 5; (x - 1) 1 BEGINS HERE Since the remainder is 6, x - 1 is not a factor. INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Have students work in pairs to complete a Elaborate 10. Compare long division and synthetic division of polynomials. The numbers generated by synthetic division are equal to the coefficients of the terms of the polynomial quotient, including the remainder. They are essentially the same process. chart like the following, showing similarities and differences: 11. How does knowing one linear factor of a polynomial help find the other factors? If one linear factor of the polynomial is known, synthetic division can be used to find the Alike Long Division product of the other factors, which may be easily factorable. written in standard form with 0 representing any missing terms. 13. Essential Question Check-In How do you know when the divisor is a factor of the dividend? The divisor is a factor of the dividend when the remainder is 0. IN3_MNLESE389885_U3M06L5 372 372 Synthetic Substitution or Division © Houghton Mifflin Harcourt Publishing Company 12. What conditions must be met in order to use synthetic division? The divisor must be a linear binomial with a leading coefficient of 1. The dividend must be Module 6 Different SUMMARIZE THE LESSON How do you explain the process of synthetic division, and when and why it is useful? What are possible sources of error in the process? Synthetic division is a division process that uses only the coefficients of a polynomial. If the last sum is 0, then the binomial is a factor of the polynomial; possible errors are not bringing down the first coefficient, forgetting to add instead of subtract, and forgetting to include coefficients that are 0. Lesson 5 19/07/14 3:09 AM Dividing Polynomials 372 Evaluate: Homework and Practice EVALUATE • Online Homework • Hints and Help • Extra Practice Given p(x), find p(-3) by using synthetic substitution. p(x) = 8x 3 + 7x 2 + 2x + 4 1. -3 8 Concepts and Skills Practice Explore Evaluating a Polynomial Function Using Synthetic Substitution Exercises 1–4 Example 1 Dividing Polynomials Using Long Division Exercises 5–8 Example 2 Dividing p(x) by x - a Using Synthetic Division Exercises 9–11 Example 3 Using the Remainder Theorem and Factor Theorem Exercises 12–15 2 -3 4 -24 51 -159 -17 53 -155 p(-3) = -155 p(x) = 2x 3 + 5x 2 - 3x 3. -3 2 p(-3) = 0 Students might make errors in signs when doing synthetic division and synthetic substitution because values are added rather than subtracted as in long division. Remind them that terms are always added for synthetic substitution and synthetic division. 5. 3 0 -1 0 0 -3 -9 -25 1 3 6 -2 -19 p(-3) = -19 p(x) = -x 4 + 5x 3 - 8x + 45 -3 -1 5 0 -8 45 3 -24 72 -192 ―――――――――――― -1 8 -24 64 -147 p(-3) = -147 (18x 3 - 3x 2 + x -1) ÷ (x 2 - 4) 18x - 3 2 Check. - 4 ⟌––––––––––––––––––– 18x - 3x + x - 1 3 (x 2 - 4)(18x - 3) + 73x - 13 2 -(18x 3 + 0x 2 - 72x) = 18x 3 - 72x - 3x 2 + 12 + 73x - 13 ――――――――― -3x 2 + 73x - 1 = 18x 3 - 3x 2 + x - 1 -(-3x 2 + 0x + 12) ―――――――― 73x - 13 6. (6x 4 + x 3 - 9x + 13) ÷ (x 2 + 8) 6x 2 + x - 48 x 2 + 8 ⟌––––––––––––––––––––––––– 6x + x + 0x - 9x + 13 4 3 2 -(6x 4 + 0x 3 + 48x 2) ―――――――――――― x 3 - 48x 2 - 9x AVOID COMMON ERRORS Students may be confused about when to use synthetic division and when to use long division. Point out that for a divisor other than a linear binomial with leading coefficient 1, long division is the best method. It may, however, be possible to divide the dividend and divisor by a constant to make the leading coefficient 1. -6 0 7 Given a polynomial divisor and dividend, use long division to find the quotient and remainder. Write the result in the form dividend = (divisor)(quotient) + remainder. You may wish to carry out a check. © Houghton Mifflin Harcourt Publishing Company AVOID COMMON ERRORS -3 6 ――――――― 4. 5 2 1 ―――――――― x Lesson 6.5 7 p(x) = x 3 + 6x 2 + 7x - 25 ―――――――― ASSIGNMENT GUIDE 373 8 2. -(x + 0x + 8x) Check. (x 2 + 8)(6x 2 + x - 48) - 17x + 397 = 6x 4 + x 3 - 48x 2 + 48x 2 + 8x - 384 - 17x + 397 = 6x 4 + x 3 - 9x + 13 ――――――――― 3 2 -48x 2 - 17x + 13 -(-48x 2 + 0x - 384) ――――――――― -17x + 397 Module 6 Exercise IN3_MNLESE389885_U3M06L5.indd 373 Lesson 5 373 Depth of Knowledge (D.O.K.) Mathematical Practices 1–15 1 Recall of Information MP.2 Reasoning 16–19 2 Skills/Concepts MP.4 Modeling 20 2 Skills/Concepts MP.2 Reasoning 21 3 Strategic Thinking MP.2 Reasoning 22 3 Strategic Thinking MP.6 Precision 23 3 Strategic Thinking MP.2 Reasoning 4/7/14 3:18 PM 7. (x 4 + 6x - 2.5) ÷ (x 2 + 3x + 0.5) x 2 - 3x + 8.5 –––––––––––––––––––––––– ( x 2 + 3x + 0.5)(x 2 - 3x + 8.5) - 18x - 6.75 x 2 + 3x + 0.5 ⟌ x 4 + 0x 3 + 0x 2 + 6x - 2.5 -(x 4 + 3x 3 + 0.5x 2) A common error when doing synthetic division is to subtract the second row rather than adding it. Show students a long division problem alongside its solution, using synthetic division to emphasize that the results will be different if they make this error. = x 4 - 3x 3 + 8.5x 2 + 3x 3 - 9x 2 + 25.5x + 0.5x 2 ――――――――――― -3x 3 - 0.5x 2 + 6x - 1.5x + 4.25 - 18x - 6.75 -(-3x - 9x - 1.5x) = x 4 + 6x - 2.5 ――――――――― 3 AVOID COMMON ERRORS Check. 2 8.5x 2 + 7.5x - 2.5 -(8.5x 2 + 25.5x + 4.25) ――――――――― 8. -18x - 6.75 1 2 _ (x + 250x + 100x) ÷ 2 x + 25x + 9 2x + 400 1 2 x + 25x + 9 ⟌ x 3 + 250x 2 + 100x + 0 2 -(x 3 + 50x 2 + 18x) 3 ( 2 ) Check. 1 2 x + 25x + 9 (2x + 400) - 9918x - 3600 2 3 = x + 200x 2 + 50x 2 + 10, 000x + 18x + 3600 ––––––––––––––––––––– _ (_ ―――――――― 200x 2 + 82x + 0 -(200x 2 + 10, 000x + 3600) ――――――――――― ) - 9918x - 3600 = x 3 + 250x 2 + 100x -9918x - 3600 Given a polynomial p(x), use synthetic division to divide by x - a and obtain the quotient and the (nonzero) remainder. Write the result in the form p(x) = (x - a)(quotient) + p(a). You may wish to carry out a check. 9. (7x 3 - 4x 2 - 400x - 100) ÷ (x - 8) (x - 8)(7x 2 + 52x + 16) + 28 -4 -400 -100 56 416 128 ――――――――― 7 52 16 28 8 7 = 7x 3 - 56x 2 + 52x 2 - 416x + 16x - 128 + 28 = 7x 3 - 4x 2 - 400x - 100 -0.25 8 0 -28.5 -9 10 -2 0.5 7 -0.5 ――――――――――――― 8 -2 -28 -2 9.5 11. (2.5x 3 + 6x 2 - 5.5x - 10) ÷ (x + 1) -1 IN3_MNLESE389885_U3M06L5 374 = 8x 4 + 2x 3 - 2x 3 - 0.5x 2 - 28x 2 - 7x - 2x - 0.5 + 9.5 = 8x 4 - 28.5x 2 - 9x + 10 (x + 1)(2.5x 2 + 3.5x - 9) - 1 6 -5.5 -10 -2.5 -3.5 9 ――――――――― 2.5 3.5 -9 -1 2.5 Module 6 (x + 0.25) (8x 3 - 2x 2 - 28x - 2) + 9.5 © Houghton Mifflin Harcourt Publishing Company 10. (8x 4 - 28.5x 2 - 9x + 10) ÷ (x + 0.25) = 2.5x 3 + 2.5x 2 + 3.5x 2 + 3.5x - 9x - 9 - 1 = 2.5x 3 + 6x 2 - 5.5x - 10 374 Lesson 5 8/7/14 7:25 AM Dividing Polynomials 374 HARDBOUND SE Determine whether the given binomial is a factor of the polynomial p(x) . If so, find the remaining factors of p(x). PAGE 288 BEGINS HERE 12. p(x) = x 3 + 2x 2 - x - 2; (x + 2) -2 13. p(x) = 2x 4 + 6x 3 - 5x - 10; (x + 2) -2 2 -1 -2 -2 0 2 ―――――― 1 0 -1 0 1 x + 2 is a factor. 6 0 -5 -10 -4 -4 8 -6 ―――――――――― 2 2 -4 3 -16 2 x + 2 is not a factor. x - 1 = (x + 1)(x - 1) 2 So, p(x) = x 3 + 2x 2 - x - 2 = (x + 1)(x - 1)(x + 2). 14. p(x) = x - 22x 2 + 157x - 360; (x - 8) 3 8 1 -22 157 -360 8 -112 360 ―――――――――― 1 -14 45 0 x - 8 is a factor. x 2 - 14x + 45 = (x - 5)(x - 9) So, p(x) = x 3 - 22x 2 + 157x - 360 = (x - 5)(x - 9)(x - 8). 15. p(x) = 4x 3 - 12x 2 + 2x - 5; (x - 3) © Houghton Mifflin Harcourt Publishing Company 3 4 -12 2 -5 12 0 6 ―――――――― 4 0 2 1 x - 3 is not a factor. 16. The volume of a rectangular prism whose dimensions are binomials with integer coefficients is modeled by the function V(x) = x 3 - 8x 2 + 19x - 12 . Given that x - 1 and x - 3 are two of the dimensions, find the missing dimension of the prism. 1 1 -8 19 -12 1 -7 12 ―――――――― 1 -7 12 0 This gives the expression x 2 -7x + 12. 3 1 -7 12 3 -12 ――――― 1 -4 0 This gives the expression x - 4, which is the missing dimension. Module 6 IN3_MNLESE389885_U3M06L5.indd 375 375 Lesson 6.5 375 Lesson 5 3/11/16 4:48 AM 17. Given that the height of a rectangular prism is x + 2 and the volume is x 3 - x 2 - 6x, write an expression that represents the area of the base of the prism. -2 1 -1 -6 0 -2 6 0 INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Emphasize the conditions that must be met ―――――――― 1 -3 0 0 to use synthetic division: The divisor must be a linear binomial with a leading coefficient of 1. The dividend must be written in standard form with 0 representing any missing terms. So, the area can be represented by x 2 - 3x. 18. Physics A Van de Graaff generator is a machine that produces very high voltages by using small, safe levels of electric current. One machine has a current that can be modeled by l(t) = t + 2 , where t > 0 represents time in seconds. The power of the system can be modeled by P(t) = 0.5t 3 + 6t 2 + 10t. Write an expression that represents the voltage of the system. Recall that V = __Pl . -2 Discuss the characteristics of synthetic division that make it synthetic: there are no variables, only coefficients; and addition is used instead of subtraction. 0.5 6 10 0 -1 -10 0 ――――――― 0.5 5 0 0 INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Point out that students should review how to The voltage can be represented by 0.5t 2 + 5t. 19. Geometry The volume of a hexagonal pyramid is modeled by the function 2 V(x) = __13 x 3 + __43 x + __23 x - __13 . Given the height x + 1, use polynomial division to find an expression for the area of the base. © Houghton Mifflin Harcourt Publishing Company • ©Ted Kinsman/Science Photo Library (Hint: For a pyramid, V = __13 Bh.) 1 3 (x + 4x 2 + 2x - 1). V(x) = _ 3 -1 1 4 2 -1 -1 -3 1 ―――――― 1 3 -1 0 So, the area of the base can be represented by x 2 + 3x - 1 . 20. Explain the Error Two students used synthetic division to divide 3x 3 - 2x - 8 by x - 2. Determine which solution is correct. Find the error in the other solution. A. 2 3 0 -2 -8 6 12 20 ――――――― 3 6 10 12 B. -2 0 -2 -8 -6 12 -20 ――――――― 3 -6 10 -28 3 do synthetic division and long division when the dividend is missing terms for some powers of the variable. Emphasize that they must include the missing terms written as zero times the power of the variable to complete the division. Student A is correct. Student B used the incorrect sign of a. Module 6 IN3_MNLESE389885_U3M06L5 376 376 Lesson 5 09/06/15 1:43 PM Dividing Polynomials 376 PEERTOPEER DISCUSSION Instruct one student in each pair to solve a polynomial division problem using long division, while the other student solves it by using synthetic division. Then have students switch roles and repeat the exercise for a new division problem with a polynomial that does not have a linear factor. Then have them discuss their results and any preferences they have for one method or the other. HARDBOUND SE H.O.T. Focus on Higher Order Thinking PAGE 289 BEGINS HERE 21. Multi-Step Use synthetic division to divide p(x) = 3x 3 - 11x 2 - 56x - 50 by (3x + 4). Then check the solution. ( Rewrite 3x + 4 as 3 x + 4 -_ 3 3 -11 _4 ). 3 -56 -50 -4 20 48 ――――――――― 3 -15 -36 -2 The quotient needs to be divided by 3. 3x - 15x - 36 __ =x 2 3 Check. JOURNAL ( 2 - 5x - 12 ) 3x 2 - 15x - 36 -2 (3x + 4) __ Have students describe a mnemonic device that can help them remember the steps in synthetic division. 3 = (3x + 4)(x 2 - 5x - 12) - 2 = 3x 3 + 4x 2 - 15x 2 - 20x - 36x - 48 - 2 = 3x 3 - 11x 2 - 56x - 50 22. Critical Thinking The polynomial ax 3 + bx 2 + cx + d is factored as 3(x - 2)(x + 3)(x - 4). What are the values of a and d? Explain. © Houghton Mifflin Harcourt Publishing Company a = 3; d = 72; The value of a is the product of the GCF and the coefficients of the x-terms of each factor. The value of d is the product of the GCF and the constant terms of each factor. 23. Analyze Relationships Investigate whether the set of whole numbers, the set of integers, and the set of rational numbers are closed under each of the four basic operations. Then consider whether the set of polynomials in one variable is closed under the four basic operations, and determine whether polynomials are like whole numbers, integers, or rational numbers with respect to closure. Use the table to organize. Whole Numbers Integers Rational Numbers Polynomials Addition Yes Yes Yes Yes Subtraction No Yes Yes Yes Multiplication Yes Yes Yes Yes Division (by nonzero) No No Yes No Polynomials are similar to integers with respect to closure. They are closed under each operation except division. Module 6 IN3_MNLESE389885_U3M06L5.indd 377 377 Lesson 6.5 377 Lesson 5 3/11/16 4:48 AM Lesson Performance Task HARDBOUND SE PAGE 290 BEGINS HERE The table gives the attendance data for all divisions of NCAA Women’s Basketball. Students may try to include the “3 divisions” in their calculation, perhaps by dividing the number of teams or attendance by 3. Explain to students that a division is a group of schools that compete against each other and there are 3 such groups, or divisions, for college basketball. Explain that this number is irrelevant for this calculation. What is important is that the numbers in the table represent the total number of teams and the total attendance for all of women’s collegiate basketball. NCAA Women’s Basketball Attendance 2006–2007 Years since 2006–2007 0 2007–2008 1 Season Attendance Number of teams (in thousands) for in all 3 divisions all 3 divisions 1003 10,878.3 1013 11,120.8 2008–2009 2 1032 11,160.3 2009–2010 3 1037 11,134.7 2010–2011 4 1048 11,160.0 2011–2012 5 1055 11,201.8 Enter the data from the second, third, and fourth columns of the table and perform linear regression on the data pairs (t, T) and cubic regression on the data pairs (t, A) where t = years since the 2006–2007 season, T = number of teams, and A = attendance (in thousands). INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Have students highlight the remainder in the A(t) . Carry out the division to Then create a model for the average attendance per team: A avg(t) = ___ T(t) remainder write A avg(t) in the form quadratic quotient + _______ . () Tt Use an online computer algebra system to carry out the division of A(t) by T(t). Models: final function Aavg(t). Ask them if the remainder is zero or nonzero and what a nonzero remainder means. Have students discuss the significance of a remainder for a function that describes attendance per team. If the attendance is 315.8 fans per team, ask students if the 0.8 represents an actual person or if it results from a limitation of the model. T(t) = 10.57t + 1005 A(t) = 13.80t 3 - 121.6t 2 + 329.8t + 10,880 Online computer algebra system result: 12,981,600 A avg(t) = 1.30558t 2 - 135.64t + 12,927.9 - __ 10.57t + 1005 © Houghton Mifflin Harcourt Publishing Company Module 6 378 AVOID COMMON ERRORS Lesson 5 EXTENSION ACTIVITY IN3_MNLESE389885_U3M06L5.indd 378 Have students research the attendance for a specific NCAA women’s basketball team for a single season. Have them compare that number to the value calculated from the model Aavg(t) for the same year. Ask the students how accurate the model is in predicting the attendance for that team and what some of the sources of error might be 2/29/16 10:37 PM Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem. Dividing Polynomials 378