MATH HIGH SCHOOL WORKING WITH EXPRESSIONS
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CONTENTS LESSON 1: RATIONAL EXPONENTS... 5 LESSON : SIMPLIFYING RADICALS... 7 LESSON : NUMBER SYSTEM... 9 LESSON 4: POLYNOMIALS... 11 LESSON 5: MULTIPLYING POLYNOMIALS... 1 LESSON 6: FACTORING... 15 LESSON 7: SPECIAL BINOMIALS... 17 LESSON 8: DIVIDING POLYNOMIALS... 1 LESSON 9: OPERATIONS WITH RADICALS... 5 LESSON 10: SOLVING RADICAL EQUATIONS... 7 LESSON 11: PUTTING IT TOGETHER... 9 Copyright 015 Pearson Education, Inc.
CONTENTS LESSON : SIMPLIFYING RADICALS... LESSON : NUMBER SYSTEM... 5 LESSON 4: POLYNOMIALS... 7 LESSON 5: MULTIPLYING POLYNOMIALS... 9 LESSON 6: FACTORING... 41 LESSON 7: SPECIAL BINOMIALS... 4 LESSON 8: DIVIDING POLYNOMIALS... 45 LESSON 9: OPERATIONS WITH RADICALS... 48 LESSON 10: SOLVING RADICAL EQUATIONS... 50 LESSON 11: PUTTING IT TOGETHER... 54 Copyright 015 Pearson Education, Inc. 4
LESSON 1: RATIONAL EXPONENTS 1. Write three things you already know about epressions. Share your work with a classmate. Did your classmate understand what you wrote?. Write your wonderings about working with epressions.. Write a goal stating what you plan to accomplish in this unit. 4. Based on your previous work, write three things you will do differently during this unit to increase your success. For eample, consider ways you will participate in classroom discussions, your study habits, how you will organize your time, what you will do when you have a question, and so on. Copyright 015 Pearson Education, Inc. 5
LESSON : SIMPLIFYING RADICALS 1. Which number is a simplification of this radical epression? 15 A 10 5 B 15 C 5 5 D 6.5. Assuming that all the variables used in those epressions represent positive numbers, determine which equations are true. There may be more than one correct answer. A a+ a = a B = C t t = t t t D p q r = p q r 1 E a b a = a b. 96 Simplify this radical epression to its simplest form by filling in the empty boes. 4. Simplify this radical epression to its simplest form. 144 5. Simplify this radical epression to its simplest form. 6 6. Simplify this radical epression to its simplest form. 18 Copyright 015 Pearson Education, Inc. 7
LESSON : SIMPLIFYING RADICALS 7. Simplify this radical epression to its simplest form. 147 4 A 7 B 7 5. C 49 D 49 8. Match each radical epression with its simplified form. 6 180 448 4 5 16 48 00 6 10 9 6 8 7 6 5 4 Challenge Problem 9. For what value of n are these two radical epressions equivalent? n 64 6 n and? Show your work and eplain your thinking. Copyright 015 Pearson Education, Inc. 8
LESSON : NUMBER SYSTEM 1. You are given the digits and 5. Use both digits to write an integer.. You are given the digits and 5. Use both digits to write a fraction.. You are given the digits and 5. Use both digits to write a sum of two numbers, where the sum is an irrational number. 4. You are given the digits and 5. Use both digits to write an irrational decimal. 5. Which of these epressions are rational numbers? There may be more than one rational epression. A 1 7 + 9 B 1 5 1 C.. D 4 + π E 4 9 6. Decide whether each number is rational, irrational, or whether you can t tell and place it in the appropriate column. Rational Irrational Can't Tell 46 4 6 0.464... 6 4+ 6 0 464. 6 Copyright 015 Pearson Education, Inc. 9
LESSON : NUMBER SYSTEM 7. π 4 is number. A a rational B an irrational Challenge Problem 8. Eplain why non-terminating, repeating decimals are always rational numbers. Use eamples to justify your eplanation. Copyright 015 Pearson Education, Inc. 10
LESSON 4: POLYNOMIALS 1. Which justification eplains why the two epressions are equivalent? 7 ( 4y ) = 14 8y A Associative property B Identity property C Commutative property D Distributive property. Which justification eplains why the two epressions are equivalent? 4 7y + + 9y = 4 + 7y + 9y A Associative property B Identity property C Commutative property D Distributive property. Which justification eplains why the two epressions are equivalent? (4 + 7y ) + 9y = (4 + ) 7y + 9y A Associative property B Identity property C Commutative property D Distributive property 4. Which justification eplains why the two epressions are equivalent? 7 4 + 9y y = (7 4) + y(9 ) A Associative property B Identity property C Commutative property D Distributive property 5. Simplify this polynomial epression. 9 + 4 6. Simplify this polynomial epression. 9 + 7 + Copyright 015 Pearson Education, Inc. 11
LESSON 4: POLYNOMIALS 7. Look at these two polynomials. A = 9 + 4 7 y B = 7 + y + 4 5y Find the polynomial A B. 8. Simplify this polynomial epression. 9( + 4y ) 7(y ) 9. Simplify this polynomial epression. 0.5(9y + 4) (7y ) 10. Look at these three polynomials. A = (9y y + 4 ) B = 4( + 7y ) (5 + 9y ) C = (9y + 4 7y ) ( + y ) Find the polynomial A B + C. 11. Simplify this polynomial epression. 5( ) 7( ) + ( ) 1. Simplify this polynomial epression. 9( 4y ) 7( 4y + ) + ( y ) 1. Simplify this polynomial epression. (7y 4) + 7(7y 4) 9(7y 4) 14. Simplify this polynomial epression. 4( 7y) 9( 7y) ( + 7y) Challenge Problem 15. What epression do you have to add to 7 for the sum of those two epressions to be 5? Show your work. Copyright 015 Pearson Education, Inc. 1
LESSON 5: MULTIPLYING POLYNOMIALS 1. Identify which of the following epressions are equal to ( + ). There may be more than one equivalent epression. A ( ) B + + 4 C ( + )( + ) D + E 4(1 + ) +. Find the polynomial epression matching each multiplication. ( + 1)(4 4) ( )(1 ) ( 1)( + )( ) ( 5) ( 1)( + 1) 5 + 6 + 1 1 ( ) 4 + 4 10 + 4. Fill in the coefficients for each term in the trinomial epression that is equal to ( )( ). + + 4. Simplify this epression. ( + )( + ) 5. Simplify this epression. ( )( ) 6. Simplify this epression. ( 9)( + 4) 7. Simplify this epression. ( + 9)( 4) 8. Simplify this epression. ( + 5)( + 7) Copyright 015 Pearson Education, Inc. 1
LESSON 5: MULTIPLYING POLYNOMIALS 9. Simplify this epression. (9 + 7 )( ) 10. Simplify this epression. (9 + 4)( 7 + ) 11. Simplify this epression. (7 + )(4 5) Challenge Problem 1. Two multiplied binomials result in a trinomial: (a + b)(c + d) = R + S + T If a = 1 and c =, what are the values of R, b, and d so that S = 1 and T = 10? Show your work and eplain your thinking. Copyright 015 Pearson Education, Inc. 14
LESSON 6: FACTORING 1. Find the factored binomials matching each trinomial epression. b, c, A, and B are all positive numbers. + b + c + b c b c b + c c ( A)( + B) A < B ( A)( + B) A = B ( + A)( + B) ( A)( + B) A > B ( A)( B). Factor this second-degree polynomial. + 8 A ( + )( + 4) B ( )( + 4) C ( + )( 4) D ( )( 4). Find the factored binomials matching each trinomial epression. + 6 + 8 + 11 + 18 + 15 + 56 4 7 + 10 ( + 4)( 6) ( + )( + 4) ( + )( + 9) ( )( 5) ( + 7)( + 8) 4. Find the factored binomials matching each trinomial epression. + 15 + 6 5 14 4 14 + 10 + 17 + 7 4 17 + 15 ( + )( + 1) ( )( 5) ( + 8)( + 9) ( )(4 5) ( + )( 7) 5. Which set of factored binomials is equivalent to this trinomial epression? + 15 + 54 Copyright 015 Pearson Education, Inc. 15
LESSON 6: FACTORING 6. Which set of factored binomials is equivalent to this trinomial epression? + 15 + 44 7. Factor this second-degree polynomial. 8 + 15 8. Factor this second-degree polynomial. + 6 + 5 9. Factor this second-degree polynomial. + 1 + 18 10. Factor this second-degree polynomial. + 6 4 11. Factor this second-degree polynomial. 5 17 14 1. Factor this second-degree polynomial. 5 + 5 Challenge Problem 1. The general form of a second-degree polynomial is a + b + c. This polynomial can be factored into the two binomials (m + p)(n + q). a. Determine whether m and n are positive or negative numbers depending on the sign of a. b. Determine whether p and q are positive or negative numbers depending on the signs of b and c. Copyright 015 Pearson Education, Inc. 16
LESSON 7: SPECIAL BINOMIALS Complete each sentence to make it true. 1. The polynomial a + ab + b is a. A difference of two squares B square of a sum C square of a difference D sum of a square. The polynomial a ab + b is a. A difference of two squares B square of a sum C square of a difference D sum of a square. The polynomial a b is a. A difference of two squares B square of a sum C square of a difference D sum of a square 4. Look at this epression. ( + ) 9 Factor this epression using one of the special products of binomials. Eplain your thinking and show your work. Copyright 015 Pearson Education, Inc. 17
LESSON 7: SPECIAL BINOMIALS 5. Each of these polynomial epressions can be simplified using one of the special products of binomials: Square of a Sum, Square of a Difference, or Difference of Two Squares. Sort each epression to the special product form it belongs to. Square of a Sum Square of a Difference Difference of Two Squares 9 1 + 4 (4 + )(4 ) ( + ) ( 1) 5 + 40 + 16 9 4 6. Write this epression in polynomial form using the formula of a special product of binomials. Describe which special product you are using and show your work. (4 + ) 7. Write this epression in polynomial form using the formula of a special product of binomials. Describe which special product you are using and show your work. (1 ) 8. Factor this polynomial using the formula of a special product of binomials. Describe which special product you are using and show your work. 16 4 9. Factor this polynomial using the formula of a special product of binomials. Describe which special product you are using and show your work. 64 + 144 + 81 10. Miki made an error in her multiplication. Eplain her error and provide the correct answer. ( + )( ) (+)(-) = 9 + 4 Copyright 015 Pearson Education, Inc. 18
LESSON 7: SPECIAL BINOMIALS Challenge Problem 11. Demonstrate that the epression a ab 4a b + b can be simplified to the product (a b ab)(a b + ab) using special products of binomials. Describe all the steps of your work. Copyright 015 Pearson Education, Inc. 19
LESSON 8: DIVIDING POLYNOMIALS 1. Find the factored binomials matching each trinomial epression. 5 5 4 1 5 5. Which is the quotient of this epression? + 5 A + 5 B 1 5 + C + 5 D 5. Which equation demonstrates that polynomials are not closed under division? A + 6 = + B 5 + 7 5 7 = + 1 C 4 = + D 4 + 4 = 1 ( ) Copyright 015 Pearson Education, Inc. 1
LESSON 8: DIVIDING POLYNOMIALS 4. Find the quotient matching each epression. + 4 + 6 + + 5 1 + 1 1 + + 5 1 1 + 5. Find the quotient of this epression. Show your work. + 10+ 16 + 8 6. Find the quotient of this epression. Show your work. 0 + 4 7. Simplify this epression. 5 ( ) 8. Find the quotient of this epression. Show your work. 4 + 18 ( ) 9. Which of these epressions has a quotient equal to ( + )? A B C D 10 + 4 6 10 5 + 8 + 7 1 7 + 10 Copyright 015 Pearson Education, Inc.
LESSON 8: DIVIDING POLYNOMIALS 10. What divisor of 6 16 results in + 4? Challenge Problem 11. m + 5 14 = + q n ( + p) Use your knowledge of polynomial operations to find the values of m, n, p, and q, knowing that p > q. Eplain your thinking. Copyright 015 Pearson Education, Inc.
LESSON 9: OPERATIONS WITH RADICALS 1. Which number is a simplification of this radical epression? 65 80 A 5 1 16 B 1 4 C 5 1 D 1 16. Find the matching epressions. 5 + 6 49 16 4 7 + 7 0 5 45 11 5 6 7 5. 81 Simplify this radical epression to its simplest form by filling in the empty boes. 4. Find the matching epressions. 8 7 5 45 40 90 4 5 6 5 8 4 15 5 14 60 Copyright 015 Pearson Education, Inc. 5
LESSON 9: OPERATIONS WITH RADICALS 5. Simplify this radical epression. Show your work. 54 ( + ) 6. Simplify this radical epression. Show your work. 6 7 16 4 16 ( ) 7. Simplify this radical epression. Show your work. 7 7 + 5 ( ) 8. Simplify this radical epression. Show your work. 6 4 + 9. Simplify this radical epression. Show your work. 8 5 5 4 10. Simplify this radical epression. 4 80 5 4 A 8 5 4 B 5 4 C 4 5 4 D 4 10 Challenge Problem n n 11. a b ma b b What should be the values of m and n for this epression to be equal to ab. Show your work and eplain your thinking. Copyright 015 Pearson Education, Inc. 6
LESSON 10: SOLVING RADICAL EQUATIONS 1. Which answer solves this equation? + = 4 A = B = 16 9 C = 4 D = 16. What is the missing number in this equation so that = 9? 4 + 1 = + A = B = 0 C = 14 D =. Select the epressions that have a valid solution. There may be more than one epression. A = B + 7 = 4 1 C 1 = D 1 = ( + ) E = 5 6 4. Solve this radical epression. Show your work. Use substitution to check your answers. 7 = + 5. Solve each radical epression. Show your work. Use substitution to check your answers. + = 7 Copyright 015 Pearson Education, Inc. 7
LESSON 10: SOLVING RADICAL EQUATIONS 6. Solve each radical epression. Show your work. Use substitution to check your answers. 7 = 7. Solve each radical epression. Show your work. Use substitution to check your answers. 1= 7 8. Solve each radical epression. Show your work. Use substitution to check your answers. = 7 9. Solve each radical epression. Show your work. Use substitution to check your answers. = Challenge Problem 10. Remember how the distance formula uses the Pythagorean theorem to find the distance from one point ( 1, y 1 ) to another (, y ). 1 1 d = ( ) + ( y y ) Eplain why, irrespectively of the value of the coordinates of the two points (even negative values), it will always be possible to find the distance between the two points (i.e., there cannot be an impossible radical value). Copyright 015 Pearson Education, Inc. 8
LESSON 11: PUTTING IT TOGETHER 1. Read through your Self Check and think about your work in this unit. Write three things you have learned about working with epressions. Share your work with a classmate. Does your classmate understand what you wrote?. Use your notes from class and your thoughts about the unit to review the concepts and properties encountered in the unit. For each concept and property, include a description and one or more eamples. Concept or Property positive integer eponent Description For any nonzero number a and a positive integer n: a n = a a a a (a occurs n times) Eamples = = 8 Review these concepts and properties: negative integer eponent 0 as an eponent fraction as an eponent properties of eponents rational numbers irrational numbers real numbers polynomial epression factoring second-degree polynomials special products of binomials. Review the notes you took during the lessons about epressions. Add any additional ideas you have about this topic to your notes. If you are still confused about a topic, make sure to research your questions and add more information to your notes. Ask for help from a classmate, review the related lessons, look at the resources in the Concept Corner, talk with your teacher, and so on, to help you clear up any confusion. 4. Complete any eercises from this unit you have not finished. Copyright 015 Pearson Education, Inc. 9
MATH HIGH SCHOOL WORKING WITH EXPRESSIONS FOR
LESSON : SIMPLIFYING RADICALS N.RN. 1. C N.RN.. B 5 5 = C t t = t t t 1 E a b a = a b N.RN.. 4 6 N.RN. 4. 1 N.RN. 5. N.RN. 6. 7 8 N.RN. 7. A 7 N.RN. 8. 6 180 448 4 5 16 48 00 6 6 6 5 8 7 9 4 10 Copyright 015 Pearson Education, Inc.
LESSON : SIMPLIFYING RADICALS Challenge Problem N.RN. 9. If the two radical epressions are equivalent, you have: n 6 n 64 = This means that: n 6 n 64 = n ( ) = n n ( ) n+ 1 n+ 1 n+ 1 = You can compare the terms inside the radicals to figure out the value of n. The value of n should solve both equations: n+ 1 6 = n+ 1 = 64 The equations are equivalent when n = 5. Therefore the two radical epressions are 5 64 6 5 and. Copyright 015 Pearson Education, Inc. 4
LESSON : NUMBER SYSTEM N.RN. 1. 5 or 5 or 5 or 5 N.RN.. N.RN.. N.RN. 4. 5 or 5 + 5 or 5+ or + 5 5. or 5. or 5. or 5. or. 5 or. 5 or. 5 or. 5 or. 5 or. 5 N.RN. 5. B 1 5 1 C.. E 4 9 N.RN. 6. Rational Irrational Can't Tell 46 4 6 464. 6 4+ 6 0 6 0.464... N.RN. 7. B an irrational Copyright 015 Pearson Education, Inc. 5
LESSON : NUMBER SYSTEM Challenge Problem N.RN. 8. A rational number is a number that can be represented as a fraction of two integers. If number is a repeating number, you can always convert this number into a fraction of two integers. To find those integers, use eponents of 10 to create a subtraction operation using two numbers derived from that both only have the repeating digits in their decimal portion. For eample, take the repeating decimal number =. 5871. The repeating portion of this number is 871. Create two numbers that only have the repeating portion as their decimal portion by multiplying by eponents of 10 (e.g. 105 and 10). Subtracting these two numbers will get rid of the decimal portion. The outcome of the subtraction will be an integer. 5 10 =,58,71. 871 10 = 5. 871 5 10 10 =,58,406 (10 5 10) =,58,406,58,406,58,406 1,69,0 = = = 10 5 10 99,990 49,9 95 1,69,0 So,. 5871= 49,995 As with this eample, all repeating decimals can be converted to fractions. Copyright 015 Pearson Education, Inc. 6
LESSON 4: POLYNOMIALS A.APR.1 1. A.APR.1. A.APR.1. A.APR.1 4. D Distributive property C Commutative property A Associative property D Distributive property A.APR.1 5. 9 + 4 = (9 4) + = 5 + = + 5 A.APR.1 6. 9 + 7 + = (7 + ) + (9 ) = 10 + 7 = 7 + 10 A.APR.1 7. A B = (9 + 4 7 y) (7 + y + 4 5y ) = 4 + ( 7 7) y + (9 4 ) + ( y + 5y ) = 4 14 y + 5 + y = y + 5 y 14 + 4 A.APR.1 8. 9( + 4y ) 7(y ) = 9 + 6y 7y + 14 = + 9y A.APR.1 9. 0.5(9y + 4) (7y ) = 4.5y + 14y + 6 = 4.5y 14y + 8 A.APR.1 10. To make the operation simpler, you can first simplify each polynomial before adding and subtracting. A = (9y y + 4 ) = (7y + 4 ) = 14y + 8 B = 4( + 7y ) (5 + 9y ) = 8 + 8y 5 9y = (8 5 ) + ( 9y + 8y ) = + 19y C = (9y + 4 7y ) ( + y ) = 18y + 8 14y y = (8 ) + (18y 14y y ) = 5 + y You can then find the polynomial A B + C using the simplified epression. A B + C = (14y + 8 ) ( + 19y ) + (5 + y ) = (8 + 5) + (14 19 + 1)y = 10 4y Copyright 015 Pearson Education, Inc. 7
LESSON 4: POLYNOMIALS A.APR.1 11. 5( ) 7( ) + ( ) = (5 7 + )( ) = 1( ) = A.APR.1 1. 9( 4y ) 7 ( 4y + ) + ( y ) = 9( 4y ) 7( 4y ) + 4y = (9 7 + 1)( 4y ) = ( 4y ) = 6 1y A.APR.1 1. (7y 4) + 7(7y 4) 9 (7y 4) = ( + 7 9)(7y 4) = 0 = 0(7y 4) = 0 A.APR.1 14. 4( 7y) 9( 7y) ( + 7y) = 4( 7y) 9( 7y) + ( 7y) = (4 9 + )( 7y) = ( 7y) = 9 + 1y = 1y 9 Challenge Problem A.APR.1 15. Let s call A the missing epressions. The sum of the two epressions is 5. Therefore you can write: ( 7) + A = 5 A = 5 ( 7) A = + ( + 5) + 7 A = + 7 + 7 The missing epressions is + 7 + 7 Copyright 015 Pearson Education, Inc. 8
LESSON 5: MULTIPLYING POLYNOMIALS A.APR.1 1. A ( ) C ( + )( + ) E 4 (1 + ) + A.APR.1. ( + 1)(4 4) ( )(1 ) ( 1)( + )( ) ( 5) ( 1)( + 1) 1( ) 4 + 4 5 + 6 10 + 4 + 1 A.APR.1. 4 + 14 1 A.APR.1 4. + 5 + 6 A.APR.1 5. 5 + 6 A.APR.1 6. 5 6 A.APR.1 7. + 5 6 A.APR.1 8. + 17 + 5 A.APR.1 9. 18 6 41 5 1 4 A.APR.1 10. 6 4 + 18 8 + 8 A.APR.1 11. 8 4 7 10 Copyright 015 Pearson Education, Inc. 9
LESSON 5: MULTIPLYING POLYNOMIALS Challenge Problem A.APR.1 A.REI. 1. If you multiply the two binomials, you will find: (a + b)(c + d) = ac + (ad + bc) + bd Therefore: R = ac S = ad + bc T = bd Since a = 1, c =, S = 1, and T = 10, you have: R = (1)() = 1 = (1)d + b() = b + d 10 = bd From this you can deduce that R =. The easiest way to approimate the values of b and d is to graph the two equations remaining and find the point of intersection of the corresponding graphs. Since graphing only provides an approimation of the intersection, you will need to verify that the coordinates obtained solve both equations. If you change the variables in the equations for b and d with and y, respectively, you obtain: y = 1 y = 10 There are two possible points of intersections: ( 10, 1) and ( 0.5, 0). Both coordinate pairs solve the equations and are therefore correct values. As a result, there are two possible sets of binomials that solve the problem: ( 10)( 1) = 1 + 10 ( 0.5)( 0) = 1 + 10 Copyright 015 Pearson Education, Inc. 40
LESSON 6: FACTORING A.SSE. 1. + b + c + b c b c b + c c ( + A)( + B) ( A)( + B) A < B ( A)( + B) A > B ( A)( B) ( A)( + B) A = B A.SSE..a. B ( )( + 4) A.SSE..a. + 6 + 8 + 11 + 18 + 15 + 56 4 7 + 10 ( + )( + 4) ( + )( + 9) ( + 7)( + 8) ( + 4)( 6) ( )( 5) A.SSE..a 4. + 15 + 6 5 14 4 14 + 10 + 17 + 7 4 17 + 15 ( + )( + 1) ( + )( 7) ( )( 5) ( + 8)( + 9) ( )(4 5) A.SSE..a 5. ( + 6) and ( + 9) A.SSE..a 6. ( + 4) and ( + 11) A.SSE..a 7. ( )( 5) A.SSE..a 8. ( + 1)( + 5) A.SSE..a 9. ( + )( + 9) A.SSE..a 10. ( + 4)( 6) A.SSE..a 11. ( + )( 5 7) A.SSE..a 1. 5( + 1)( 1) Copyright 015 Pearson Education, Inc. 41
LESSON 6: FACTORING Challenge Problem A.SSE. 1. You can multiply the binomial epressions into a trinomial form. (m + p)(n + q) = mn + (p + q) + pq Since a + b + c = (m + p)(n + q), you write the following relationships: a = mn b = p + q c = pq From these relationships you can determine the sign of each coefficient m, n, p, and q depending on the signs of a, b, and c. a. a is a positive number Since a = mn, when a > 0, then m and n are both either positive or negative. a is a negative number Since a = mn, when a > 0, then m and n have opposite signs (m is negative if n is positive and vice versa). b. b and c are positive numbers b = mp + nq > 0 c = pq > 0 For both equations to be true, both p and q must be positive numbers. Both m and n must also be positive numbers. b is a positive number and c is a negative number b = mp + nq > 0 c = pq < 0 For both equations to be true, p and q must have opposite signs and the absolute value of the positive number must be greater than the absolute value of the negative number. Both m and n must also be positive numbers. p < 0 and q > 0 and q > p or p > 0 and q < 0 and p > q b is a negative number and c is a positive number b = mp + nq > 0 c = pq > 0 For both equations to be true, both p and q must be negative numbers. Both m and n must also be positive numbers. b and c are negative numbers. b = mq + np < 0 c = pq < 0 For both equations to be true, p and q must have opposite signs and the absolute value of the negative number must be greater than the absolute value of the positive number. Both m and n must also be positive numbers. p < 0 and q > 0 and q < p or p > 0 and q < 0 and p < q Copyright 015 Pearson Education, Inc. 4
LESSON 7: SPECIAL BINOMIALS A.SSE. 1. B square of a sum A.SSE.. C square of a difference A.SSE.. A difference of two squares A.SSE. 4. This epression is a difference of two squares, a b, where: a = + and b = This special product of binomials can also be written in factor form as: (a + b)(a b). Therefore the epression can be factored: ( + + )( + ) ( + 6) A.SSE..a 5. Square of a Sum Square of a Difference Difference of Two Squares ( + ) 5 + 40 + 16 9 1 + 4 (4 + )(4 ) ( 1) 9 4 A.SSE..a 6. This epression is the square of a sum (a + b), where: a = 4 and b = Therefore the polynomial form of this epression is: a + ab + b : 16 + 4 + 9 A.SSE. 7. This epression is the square of a difference (a b), where: a = 1 and b = Therefore the polynomial form of this epression is: a ab + b 1 6 + 9 A.SSE..a 8. This polynomial is the difference of two squares a b, where: a = 4 and b = Therefore the polynomial can be factored to: (a + b)(a b) (4 + )(4 ) Copyright 015 Pearson Education, Inc. 4
LESSON 7: SPECIAL BINOMIALS A.SSE..a 9. This polynomial is in the form a + ab + b, which is the polynomial form associated with the square of a sum (a + b). In this particular epression you can have two possible sets of values for a and b. a = 64 and b = 81 As a result: a =±8 and b =±9 Since ab >0, it means that a and b are either both positive or both negative numbers. Therefore the values of a and b can either be: a = 8 and b = 9 or a = 8 and b = 9 The following two sets of factors are therefore valid: (8 + 9) or ( 8 9) A.SSE. 10. Miki confused the sum of squares with the difference of two squares. The product of and is negative, so the 4 must be subtracted from the square of when multiplying. ( + )( ) (+)(-) = 9 + 4 Challenge Problem A.SSE. 11. The epression can be rearranged to bring forward the polynomial equivalent to the square of a difference a ab + b : (a ab + b ) 4a b The first portion of the equation can then be factored as the square of the difference of a b: (a b) 4a b Since 4a b = (ab), the epression can then be rewritten: (a b) (ab) This epression is the sum of two squares, which can be factored: (a b ab)(a b + ab) Copyright 015 Pearson Education, Inc. 44
LESSON 8: DIVIDING POLYNOMIALS A.APR.1 A.APR.6 1. 5 5 4 5 1 5 A.APR.1. B 1 5 + A.APR.1 A.APR.6. B 5 + 7 5 7 = + 1 A.APR.1 A.APR.6 4. + 4 + 6 + + 5 1 1 + 1 1 1 + + + 5 A.APR.1 5. The polynomial +10 + 16 can be factored to ( + )( + 8). Therefore, the epression can be simplified to: + 10+ 16 ( + )( + 8) = = + + 8 + 8 The quotient of the epression is ( + ). A.APR.1 A.APR.6 6. The polynomial 0 can be factored to ( + 4)( 5). Therefore, the epression can be simplified to: 0 ( + 4)( 5) = = 5 + 4 + 4 The quotient of the epression is ( 5). A.APR.1 7. 5 5 ( ) = = = = 6 ( ) 4 4 4 8 Copyright 015 Pearson Education, Inc. 45
LESSON 8: DIVIDING POLYNOMIALS A.APR.1 A.APR.6 8. You can first simplify the epression by dividing the numerator by : + 18 The resulting polynomial + 18 can be factored to ( )( + 6). Therefore, the epression can be simplified to: + 18 ( )( + 6) = = + 6 The quotient of the epression is ( + 6). A.APR.1 A.APR.6 9. B 10 5 A.APR.1 A.APR.6 10. You must find the divisor A that solves this equation: 6 16 = + 4 A This equation can be rewritten: 6 16 A = + 4 ( 8 16) = + 4 ( + 4)( 4) = + 4 = ( 4) = 8 Therefore, the divisor of this equation is ( 8) Copyright 015 Pearson Education, Inc. 46
LESSON 8: DIVIDING POLYNOMIALS Challenge Problem A.APR.1 A.APR.6 11. This equation can be rewritten as: m + 5 14 = n ( + p)( + q) The right factored epression can be epressed as a polynomial: m + 5 14 = n ( + (p + q) + pq) = + n + (p + q) 1 + n + pq n Since the greatest eponent of the left epression is, the greatest eponent of the right epression must also be, therefore ( + n) = and n = 1. m + 5 14 = + (p + q) + pq From this equation, these equalities can be deduced: m = 1 p + q = 5 pq = 14 If p + q = 5, then p = 5 q and (5 q)q = 14 (5 q)q = q + 5q = 14 The possible values of q solving this equation are either 7 or. p = 5 q If q = 7 then p =. If q = then p = 7. Since p > q, the only possible combination is p = 7 and q =. The polynomial operation with all the coefficients replaced by their actual values is therefore: + 5 14 = ( + 7) Copyright 015 Pearson Education, Inc. 47
LESSON 9: OPERATIONS WITH RADICALS N.RN. 1. B N.RN.. 1 4 5 + 6 49 16 4 7 + 7 0 5 45 11 6 7 5 5 N.RN.. N.RN. 4. 8 7 5 45 40 90 4 5 6 5 8 14 15 5 60 4 N.RN. 5. N.RN. 6. 54 ( + ) = 5( 6 ) = 0 ( )= ( ) 6 7 16 4 16 6 4 = 6( 4) = 1 6 N.RN. 7. 7( 7 + 5)= 7+ 5 7 = 1+ 5 N.RN. 8. 6 4 + = 4 + 4 = ( ) + = 1 + = 1 Copyright 015 Pearson Education, Inc. 48
LESSON 9: OPERATIONS WITH RADICALS N.RN. 9. 8 5 5 4 = ( 5) = 4 5 N.RN. 10. C 4 4 5 Challenge Problem N.RN. A.SSE. 11. Using the properties of eponents, you can rewrite the epression in a form that match ab : 1 n n n a b ma b = b ( 1 mab ) b = ( 1 mab ) 1 1 n For this epression to be equal to ab, you must fulfill these two conditions: (1 m) = and 1 1 n = Therefore: m = n = 1 n Since b = b 1 n 1 1., if n = 1, then n b = b = b Therefore when m and n are replaced in the epression by their respective values, the epression becomes: ab ( ) ab b This epression is indeed equal to ab when it is simplified. Copyright 015 Pearson Education, Inc. 49
LESSON 10: SOLVING RADICAL EQUATIONS A.REI. 1. B = 16 9 A.REI.. A = A.REI.. A = B C + 7 = 4 1 1 = A.REI. 4. 7 = + 7 = 4 16 = = You need to substitute 16 7 = 16 + 7 = 16 + 7 = 4+ 7 = 7 = ( ) 16 for in the equation to check your answer. This answer correctly solves the equation. = 16 Copyright 015 Pearson Education, Inc. 50
LESSON 10: SOLVING RADICAL EQUATIONS A.REI. 5. + = 7 ( + ) = ( 7 ) + = 7 = 10 = 5 You need to substitute 5 for in the equation to check your answer. You also need to remember that square roots can only be used on positive numbers. The square root of a negative real number is impossible to calculate. ( 5) + = 1 ( 5) 7 = 1 This solution is etraneous and should not be included in your answer. Since = 5 was the only possible solution, this equation cannot be solved. There are no solutions to this problem. A.REI. 6. 7 = ( 7 ) = 7 = 9 = 16 You need to substitute 16 for in the equation to check your answer. 16 7 = 9 = = This answer correctly solves the equation. = 16 Copyright 015 Pearson Education, Inc. 51
LESSON 10: SOLVING RADICAL EQUATIONS A.REI. 7. 1= 7 = 7 ( + 1) ( ) = 4 = 576 You need to substitute 576 for in the equation to check your answer. 576 1= 7 4 1= 7 8 1= 7 7 = 7 This answer correctly solves the equation. = 576 A.REI. 8. = 7 ( ) = 7 = 7 = 7 7 = You need to substitute 7 7 7 7 = 7 7 1 = = 7 ( ) 14 for in the equation to check your answer. This answer correctly solves the equation. = 7 Copyright 015 Pearson Education, Inc. 5
LESSON 10: SOLVING RADICAL EQUATIONS A.REI. 9. = ( ) = ( ) 9 = 9 = 0 8 = 0 = 0 You need to substitute 0 for in the equation to check your answer. 0 0 = 0 ( ) Since you cannot divide by 0, this solution is etraneous and should not be included in your answer. Since this was the only possible solution, this equation cannot be solved. There are no solutions to this problem. Challenge Problem A.REI. 10. Square roots of real numbers can only be calculated if the number is positive. Square roots of negative numbers are impossible. The distance between two points is epressed as the square root of the sum of the squares of the differences between the coordinates and the y coordinates of the two points. Since the differences are squared, the square of the differences will always be positive numbers even when the differences themselves are negative. Since both squares of the differences are positive, the sum of those squares will also be positive. Therefore the number inside the square root will always be positive and a distance can always be calculated irrespectively of the coordinates of the points. Copyright 015 Pearson Education, Inc. 5
LESSON 11: PUTTING IT TOGETHER 8.EE.5 8.EE.6. Definitions and eamples will vary. Here are some eamples. Concept or Property negative integer eponent Description Eamples For any nonzero number a and a positive integer n: 1 1 1 = = = n a = 1 8 n a 0 as an eponent For any nonzero number a: a 0 = 1 0 0 = 1 0 = 0 = 0 = 1,900 0 = 1 fraction as an eponent properties of eponents For a > 0: a 1 n = n a m, n, and p are positive integers: a m + n = a m a n a m n a = a m n (a m ) n = a m n a b m n p a = b mp np 1 = 4 + = (4 4) (4 4 4) = 4 4 4 4 = = 4 4 (4 ) = (4 4 4) = (4 4 4) (4 4 4) = 4 6 = 4 4 5 4 4 4 4 = 4 5 = 5 1 8 rational numbers Numbers that can be epression as a ratio between two integers such that the second integer is not zero. All integers are included in rational numbers since they are all divisible by 1. All decimals that terminate are rational since they are divisible by a factor of 10. All decimals that repeat after some point are rational. All roots of perfect numbers are rational numbers (because they are integers). The system of rational numbers is closed under addition, subtraction, multiplication, and division. 1, 0.56, and 57. are rational numbers because they can all be represented by a fraction: 1 1 = 1 56 056. = 100 89 57. = 165 4 = = = 1 7 = = = 1 5 5 5 7,776 = 6 = 6 = 6 1 Copyright 015 Pearson Education, Inc. 54
LESSON 11: PUTTING IT TOGETHER 8.EE.5 8.EE.6. Concept or Property irrational numbers Description Numbers that cannot be written as a fraction (a ratio of two integers). In decimal form, irrational numbers never end or repeat. All square roots of numbers that are not perfect squares are irrational. Special numbers such as π are also irrational. real numbers A set of numbers containing all rational numbers and all irrational numbers. polynomial epression factoring seconddegree polynomials All the numbers on the number line are real numbers. An epression of one or more algebraic terms with whole-number (positive integers) eponents. Since the eponent has to be a positive integer, epression containing inverse of eponents or roots are not polynomial epressions. Second-degree polynomials can sometimes be factored into two binomials with coefficients. The epression a + b + c can be factored into (m + p)(n + q), where mn = a, pq = c, and (pn + mq) = b. Eamples = 141415. π =.141596559 polynomial epressions: + + 1 not polynomial epressions: + 4 1 + 5 1 + 10 = ( + )( 5) Copyright 015 Pearson Education, Inc. 55
LESSON 11: PUTTING IT TOGETHER 8.EE.5 8.EE.6. Concept or Property special products of binomials Description Eamples There are three special products of binomials: The square of a sum: (a + b) = (a + b)(a + b) = a + ab + b The square of a difference: (a b) = (a b)(a b) = a ab + b The difference of two squares: (a + b) (a b) = a b The square of a sum: ( + ) = + ()() + (5) = 4 + 1 + 9 5 = 5 The square of a difference: ( ) = ()() + ( 1) = 4 1 + 9 1 = 1 The difference of two squares: ( + )( ) = (5)( 1) = 4 9 5 = 5 Copyright 015 Pearson Education, Inc. 56