Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Unit 9: Factoring Lesson 1: Introduction to Factoring Factors and Multiples 6 ● 5 = 30 6 ● 8 = 48 _________ _______ ___ and ____ are _________ of _____. ___ and ____ are _________ of _____. _________ is a multiple of ________. _________ is a multiple of ________. _________ is a multiple of ________. _________ is a multiple of ________. Common Factors If two numbers are divisible by the same number, then they are called common factors ___ is a common factor of ______ and ________. Try These on Your Own 1. List the factors of 72. 2. 36 is divisible by _____. 3. _____ is a multiple of 12. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com 4. ______ is a common factor of 32 & 40. Unit 9: Factoring Prime and Composite Numbers Prime Numbers Composite Numbers Greatest Common Factor The Greatest Common Factor is the greatest integer that is a common factor of two or more numbers. Example 1 Example 2 What is the greatest common factor for 18 and 27? What is the greatest common factor for 18 27 3x and 9x2 9x2 3x Example 3 Example 4 What is the greatest common factor of What is the greatest common factor of 6x2 y and 9x3 y2 ? 2x and 3y? 6x2 y 9x3 y2 2x 3y If: 2x + 3y This polynomial is ____________ because Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 1: Introduction to Factoring Practice Part 1: Read each question and answer completely. 1. 35 is divisible by _______ (give 2 solutions) 2. Name 6 factors of 12. 3. ________ is a multiple of 13. (Give 3 solutions) 4. _______ is a common factor of 36 and 48. 5. ______ is a common factor of 144 and 120. 6. ______ is the greatest common factor of 48 and 64. 7. ______ is the greatest common factor of 48 and 60. 8. Is 9 a prime number? Explain why or why not. 9. 15 is a composite number. Explain why. 10. Do you think that the number 1 is prime, composite, or neither? Explain. Part 2: Find the greatest common factor (GCF) for each set of numbers. 1. 12a2 and 14 a 7. 3xy, 2x2y, 2. x2 y3 and 2xy2 8. 3a2b, 27ab2, and 9ab 3. 3ab5 and 4a3b3 9. 2x2y3z, 4. 9rs and 12r2s 10. 15rs3t, 7r2st, and 5r2s2t 5. 4a3b2 and 2a3 11. 18xz, 9x2y4z2, and 3xyz 6. 16x4yz2 and 24x3z 12. 7x2y5, 5x5y4, and 8x3y5 and 4x3y 6xy2z2, and 18xz3 Find the greatest common factor (GCF) for each set of numbers. (2 points each) 1. 13x3yz and 26xy2z2 2. 18r5s8t4, 81r6s5t6, and 27r9s7t5 3. 22abc and 11a2b Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 1: Introduction to Factoring Practice – Answer Key Part 1: Read each question and answer completely. 1. 35 is divisible by 7 and 5 (give 2 solutions) 2. Name 6 factors of 12. (7 & 5 are factors of 35) (1,2,3,4,6,12) (1 x12) (2 x6) (3x4) 3. 26, 39, 52 is a multiple of 13. (Give 3 solutions) (13 x2 = 26) (13x3 = 39) (13 x4 = 52) (You may also have had: 65, 78, 91, 104, 117, 130 …) 4. 6 is a common factor of 36 and 48. (6 x6 = 36) (6 x 8 = 48) (You may also have had 4 because (4 x9 = 36) (4 x 12 = 48). Other answers include: 2 , 3, and 12 5. 12 is a common factor of 144 and 120. (12 x12 = 144) (12 x 10 = 120) (Other answers include: 2, 3, 4, 6, 8, 24) 6. 16 is the greatest common factor of 48 and 64. Factors of 48: Factors of 64: 1, 2, 3, 4,6, 8, 12, 16, 24 ,48 1, 2, 4, 8, 16, 32, 64 The common factors are: 1,2,4, 8, 16 The greatest common factor would be 16. 7. 12 is the greatest common factor of 48 and 60. Factors of 48: Factors of 60: 1, 2, 3, 4,6, 8, 12, 16, 24 ,48 1,2,3,4,5,6,10,12,15,20,30,60 The common factors are: 1,2,3,4,6, 12 The greatest common factor would be 12. 8. Is 9 a prime number? Explain why or why not. 9 is not a prime number. A prime number only has factors of 1 and itself. 9 has the following factors: 1,3,9. Since 3x3=9, 9 is not a prime number. 9. 15 is a composite number. Explain why. 15 is a composite number. A composite number has one or more factors other than itself and 1. (It’s the opposite of a prime number.) The factors of 15 are: 1,3,5,15. Since 3x5=15, we can classify 15 as a composite number. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 10. Do you think that the number 1 is prime, composite, or neither? Explain. Since the only factor of 1 is 1, it is not considered to be prime or composite. The prime numbers less than 50 are: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47. Part 2: Find the greatest common factor (GCF) for each set of numbers. 1. 12a2 and 14 a GCF: 2a The GCF for 12 and 14 is 2. The GCF for a2 and a is a 2. x2 y3 and 2xy2 7. 3xy, 2x2y, and 4x3y GCF: xy The GCF for 3,2, and 4 is 1. (We don’t need to write) The GCF for x, x2 and x3 is x. The GCF for y, y, and y is y 8. 3a2b, 27ab2, and 9ab GCF: xy2 GCF: 3ab The GCF for 1 and 2 is 1. (We don’t need to write.) The GCF for x2 and x is x. The GCF for y3 and y2 is y2 The GCF for 3, 27, and 9 is 3. The GCF for a2 and a is a. The GCF for b, and b2 is b. 3. 3ab5 and 4a3b3 9. 2x2y3z, 6xy2z2, and 18xz3 GCF: ab3 GCF: 2xz The GCF for 3 and 4 is 1. (We don’t need to write.) The GCF for a3 and a is a. The GCF for b5 and b3 is b3. The GCF for 2, 6 and 18 is 2. The GCF for x2 and x is x. Since 18xz3 does not have a y term, we cannot include a y in our GCF. The GCF for z3, z2 and z is z. 4. 9rs and 12r2s 10. 15rs3t, 7r2st, and 5r2s2t GCF: 3rs GCF: rst The GCF for 9 and 12 is 3. The GCF for r2 and r is r. The GCF for s is s. The GCF for 15, 7 and 5 is 1. (We don’t need to write.) The GCF for r2 and r is r. The GCF for s3,s2, and s is s. The GCF for t is t. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 5. 4a3b2 and 2a3 11. 18xz, 9x2y4z2, and 3xyz GCF: 2a3 GCF: 3xz The GCF for 4 and 2 is 2. The GCF for a3 is a3. Since 2a3 does not have a b term, we cannot include a b term in the GCF. The GCF for 18, 9 and 3 is 3. The GCF for x2 and x is x. Since 18xz does not have a y term, we cannot include a y term in the GCF. The GCF for z2 and z is z. 6. 16x4yz2 and 24x3z 12. 7x2y5, 5x5y4, and 8x3y5 GCF: 8x3z GCF: x2y4 The GCF for 16 and 24 is 8. The GCF for x4 and x3 is x3. Since 24x3z does not have a y term, we cannot include a y term in our GCF. The GCF for z2 and z is z. The GCF for 7 and 5, and 8 is 1. (We don’t need to write.) The GCF for x2, x3, and x5 is x2. The GCF for y4 and y5 is y4. 1. 13x3yz and 26xy2z2 2. 18r5s8t4, 81r6s5t6, and 27r9s7t5 GCF: 13xyz GCF: 9r5s5t4 The GCF for 13 and 26 is 13. The GCF for x3 and x is x. The GCF for y2 and y is y. The GCF for z and z2 is z. The GCF for 18, 81 and 27 is 9. The GCF for r5, r6 and r9 is r5. The GCF for s8, s5 and s7 is s5. The GCF for t4, t6, and t5 is t4. 3. 22abc and 11a2b GCF: 11ab The GCF for 22 and 11 is 11. The GCF for a2 and a is a. The GCF for b is b. Since 11a2b does not have a c term, we cannot include a c term in our GCF. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 2: Factoring Polynomials Using the Greatest Common Factor (GCF) The Greatest Common Factor (GCF) is the greatest integer that is a common factor of two or more numbers. Example 1 Factor: 3x2 + 27x3 Example 2 Factor: 2x2y2 - 8xy2 NOTE: Factoring can always be checked using multiplication! Example 3 Factor: 3x2y2 + 2x2 + xy2 Example 4 Factoring can make it easier to simplify some fractions: Simplify: Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 2: Factoring Polynomials by Using the GCF Practice Part 1: The greatest common factor has been factored out ( in bold). Fill in the blanks to complete the factorization of each polynomial. 1. 3x2 + 12xy2 = 3x(______ + ________) 2. 2r2s3 + 5rs6 = rs3(________+_________) 3. 8a2bc4 - 6ab4c4 = 2abc4(___________-___________) 4. 2x2y5 + 4xy3 + 6x2y7 = 2xy3(___________+ ____________+ ____________) 5. 3r3s3t5 – 2r2s7t4 + s5t3 = s3t3( ___________ - ___________+______________) Part 2: Factor Completely. 1. 7a2b5 + 21ab4 6. 9x2y – 27xy3 + 3xy 2. 81x2yz – 27xy2z3 7. 12a4b5c + 2a2c + 8a2bc 3. 3a2b3c + 12a3b2c2 8. 22g2h – 11gh2 + g2h2 4. 5s5t2 – 25s4 9. 2w5xy – 3w2x2y + 2w2x 5. 14x4yz2 + 2x4y3z – 7x3yz 10. 8s2t5u2 – 12st2u2 + 2st4u Part 3: Simplify each fraction by simplifying the numerator. 1. 2. Factor completely. ( 2 points each) 1. 4x3y5z – 6x3y5 3. Simplify the fraction. 2. 8r3st + 24r4s2t2 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com (2 points) Unit 9: Factoring Lesson 2: Factoring Polynomials by Using the GCF Practice- Answers Part 1: The greatest common factor has been factored out ( in bold). Fill in the blanks to complete the factorization of each polynomial. 1. 3x2 + 12xy2 = 3x( x 4y2) + 2. 2r2s3 + 5rs6 = rs3( 2r + 5s3) 3. 8a2bc4 - 6ab4c4 = 2abc4( 4. 2x2y5 + 4xy3 + 6x2y7 = 4a - 3b3) 2xy3( 5. 3r3s3t5 – 2r2s7t4 + s5t3 = s3t3( xy2+ 2+ 3xy4 ) 3r3t2 - 2r2s4t + s2) Part 2: Factor Completely. 1. 7a2b5 + 21ab4 6. 9x2y – 27xy3 + 3xy Identify the GCF: 7ab4 Identify the GCF: 3xy 7ab4(ab + 3) 3xy(3x – 9y2 + 1) 2. 81x2yz – 27xy2z3 7. 12a4b5c + 2a2c + 8a2bc Identify the GCF: 27xyz Identify the GCF: 2a2c 27xyz(3x – yz2) 2a2c(6a2b5 + 1 + 4b) 3. 3a2b3c + 12a3b2c2 8. 22g2h – 11gh2 + g2h2 Identify the GCF: 3a2b2c Identify the GCF: gh 3a2b2c(b + 4ac) gh(22g – 11h + gh) 4. 5s5t2 – 25s4 9. 2w5xy – 3w2x2y + 2w2x Identify the GCF: 5s4 Identify the GCF: w2x 5s4(st2 – 5) w2x(2w3y – 3xy + 2) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 5. 14x4yz2 + 2x4y3z – 7x3yz 10. 8s2t5u2 – 12st2u2 + 2st4u Identify the GCF: x3yz Identify the GCF: 2st2u x3yz(14xz + 2xy2 – 7) 2st2u(4st3u – 6u + t2) Part 3: Simplify each fraction by simplifying the numerator. 1. 2. Identify the GCF of the numerator: 2xy 2(2 + ) = 2 + 2 Identify the GCF of the numerator: b3 ( + 1 − − ) = − − + 1 1. 4x3y5z – 6x3y5 2. 8r3st + 24r4s2t2 Identify the GCF: 2x3y5 Identify the GCF: 8r3st 2x3y5(2z – 3) 8r3st(1 + 3rst) 3.Simplify the fraction. (2 points) Identify the GCF of the numerator: ab2 (6 + 5 ") = 6 + 5 " Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 3: Factoring Trinomials What is a Trinomial? A trinomial is a polynomial that has _____ terms. x2 –x - 12 Before we start, let’s review multiplication of polynomials. Multiply: (x+2)(x+3) = Multiply: (x – 5) (x+6) When factoring trinomials in the form of: x2 + bx + c, you must find two numbers whose product is c and whose sum is b. NOTE: (Always take the sign in front of b and c when determining your numbers. Example 1 Example 2 Factor: x2 + 6x + 8 Factor: x2 -9x + 20 2#’s: Product = ____ ( ) ( Sum = ____ 2#’s: Product = ____ ) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Sum = ____ Unit 9: Factoring Example 3 Example 4 Factor: x2 – 5x – 24 2#’s: Product = _____ Let’s first multiply: Sum = _______ (x – 3) (x + 3) Factor: x2 - 81 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 3: Factoring Trinomials Practice Part 1: Factor by filling in the missing parts. 1. x2 + 8x + 12 Product: _____ Sum: ______ Product: _____ Sum: ______ Product: _____ Sum: ______ Product: _____ Sum: ______ (x + ____) (x + _____) 2. x2 -7x + 12 (x + ____) (x + _____) 3. x2 -x - 12 (x + ____) (x + _____) 2 4. y + y - 6 (y + ____) (y + _____) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Part 2: Factor completely. 1. x2 + 15x + 36 5. x2 - 49 2. x2 + 2x – 24 6. x2 - 36 3. x2 – 14x + 48 7. x2 + 14x + 49 4. x2 – 7x – 30 8. x2 – 17x + 72 Part 3: Thinking Questions 1. What polynomial, when factored, gives (x + 5)(x -6)? 2. What polynomial when factored gives (x – 8) (x+8)? 3. One factor of the trinomial: x2 + 85x + 750 is (x+10), what is the other factor? 4. One factor of the trinomial: x2 – 10 x – 375 is (x – 25), what is the other factor? 5. Explain why the trinomial: x2 – 7x + 14 cannot be factored. Factor Completely. (2 points each) 1. x2 + 16x + 55 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com 2. x2 + 8x + 11 Unit 9: Factoring Lesson 3: Factoring Trinomials Practice Part 1: Factor by filling in the missing parts. 1. x2 + 8x + 12 (x + 2 ) (x + Product: 12 6 ) 1 ● 12 1 + 12 = 13 2●6 2+6=8 4●3 4+3=7 Product: 12 2. x2 -7x + 12 (x + -4) (x + -3) Most commonly written as: (x-4) (x-3) (x + -4) (x + 3) Most commonly written as: X X Sum: -7 You know that both factors must be negative since the sum is negative and the product is positive. -4 ● -3 -4 + (-3) = -7 -6 ● -2 -6 + -2 = -8 -1 ● -12 -1 + -12 = -13 X Product: -12 3. x2 -x - 12 Sum: 8 X Sum: -1 Since the product is negative, you know that one factor must be positive and one must be negative. Since the sum is negative also, the larger integer will be negative. -12 ● 1 -12 + 1 = -11 X -6 ● 2 -6+ 2 = -4 X -4 ● 3 -4 + 3 = -1 (x-4) (x+3) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Product: -6 Sum: 1 Since the product is negative, we know that one integer must be positive and one must be negative. Also, since the sum is positive, the larger number must be positive. 4. y2+ y - 6 (y + 3 ) (y + -2) 6 ● (-1) 6 + (-1) = 5 3 ● (-2) 3 + (-2) = 1 X Most commonly written as: (y+3)(y-2) Part 2: Factor completely. 1. x2 + 15x + 36 5. x2 – 49 Need a Sum of 15 and Product of 36. 12 ●3 = 36 and 12+ 3 = 15 Therefore, we can use: 12 and 3 (x+12)(x+3) Need a Sum of 0 and Product of -49. (There is no x term, therefore, the sum is 0) The product is -49, so one integer must be positive and one must be negative. Since the sum is 0, the integer should be the same. 7 ● -7 = 49 7 + (-7) = 0 Therefore, we can use: 7 and -7 (x+7)(x-7) This is a Difference of Two Squares 2. x2 + 2x – 24 6. x2 – 36 Need a Sum of 2 and Product of -24. The product is negative, so one integer must be positive and one must be negative. Since the sum is positive, the largest integer should be positive. 6 ● (-4) = -24 and 6 + (-4) = 2 Need a Sum of 0 and Product of -36. (There is no x term, therefore, the sum is 0) The product is negative, so one integer must be positive and one must be negative. Since the sum is 0, the integer should be the same. 6 ● (-6) = -36 and 6 +(-6) = 0 Therefore, we can use: 6 and -4 (x+6)(x-4) Therefore, we can use: 6 and -6 (x+6)(x-6) This is a Difference of Two Squares Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 3. x2 – 14x + 48 7. x2 + 14x + 49 Need a Sum of -14 and Product of 48. Need a Sum of 14 and Product of 49. The product is positive and the sum is negative. Therefore, both integers must be negative. Since both the sum and product are positive, both integers will be positive. -6 ● (-8) = 48 and (-6) + (-8) = -14 Therefore, we can use: -6 and -8 7 ●7 = 49 and 7 + 7 = 14 Therefore, we can use: 7 and 7 (x+7)(x+ 7) or (x+7)2 (x-6)(x-8) This is a perfect square binomial. 4. x2 – 7x – 30 8. x2 – 17x + 72 Need a Sum of -7 and Product of -30. Need a Sum of -17 and Product of 72. The product is negative, so one integer must be positive and one must be negative. Since the sum is negative, the largest integer should be negative. Since the product is positive and the sum is negative, both integers must be negative. -10 ● 3 = -30 and -10 + 3 = -7 -9 ● (-8) = 72 and -9 + (-8) = -17 Therefore, we can use: -9 and -8 Therefore, we can use: -10 and 3 (x-9)(x-8) (x+3)(x-10) Part 3: Thinking Questions 1. What polynomial, when factored, gives (x + 5)(x -6)? In order to find the original polynomial, we can multiply the two binomials. Use FOIL. (x+5) (x-6) x(x) + x(-6) + 5(x) + 5(-6) x2 – 6x + 5x - 30 x2 – x – 30 is the original trinomial that factors into (x+5) (x-6) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 2. What polynomial when factored gives (x – 8) (x+8)? This set of binomials is a Difference of Perfect Squares. Notice that they are opposites. Therefore, the original trinomial is: x2 – 64. 3. One factor of the trinomial: x2 + 85x + 750 is (x+10), what is the other factor? We are given the trinomial and one of the factors. We know that we must find two numbers whose product is 750 and sum is 85. One of the numbers is 10. So, 10 times what is 750? 10 ● 75 = 750 AND 10 + 75 = 85. Therefore, the other factor is: (x+75) 4. One factor of the trinomial: x2 – 10 x – 375 is (x – 25), what is the other factor? We are given the trinomial and one of the factors. We know that we must find two numbers whose product is -375 and sum is -10. One of the numbers is -25. So, -25 times what is -375? Or, -375/-25 = 15 So, -25 ● 15 = -375 and -25 + 15 = 10 Therefore, the other factor is: (x+15) 5. Explain why the trinomial: x2 – 7x + 14 cannot be factored. We are looking for two integers that have a product of 14 and a sum of -7. The two integers must both be negative since the product is positive and the sum is negative. Let’s try: -1 ● -14 = 14 but -1 + -14 = -15; -7 ● -2 = 14 but -7 + -2 = -9 There are no other factors of 14 to try. Therefore, this trinomial cannot be factored. Factor Completely. (2 points each) 1. x2 + 16x + 55 2. x2 + 8x + 11 Need a Sum of 16 and Product of 55 Need a Sum of 8 and Product of 11. Since both the sum and product are positive, then both integers must be positive. Since both the sum and product are positive, then both integers must be positive. 11 ● 5 = 55 and 11 + 5 = 16 Therefore, we can use: 11 and 5 11 is a prime integer. Only 1 ● 11 = 11. Since 1 + 11 ≠ 8, we can conclude that this trinomial is prime and cannot be factored. (x+ 11)(x+ 5) There is no way to factor this trinomial. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Factoring – Quiz # 1 Part 1: Complete each problem. Express your answer in simplest terms. (Use the GCF) 1. Factor the following polynomial using the GCF. 3x2 – 18x + 33 2. Factor the following polynomial using the GCF. 6a5 – 12a4 + 15a3 3. Factor the following polynomial using the GCF. 8x2y3z + 4xy4 z3 – 12x4y5z2 4. Simplify the following fraction: 5. Simplify the following fraction: 6x2y7z5 + 3x2y3z6 + 2xy3 z8 xy3z5 Part 2: Factor each polynomial. 1. x2 + 9x + 18 2. x2 – x – 20 3. y2 – 9y + 14 4. r2 + 2r – 48 5. z2 – 6z + 5 Part 3: Short Answer. 1. Explain the steps used to factor the following polynomial: 3x4 + 6x2 – 3x 2. Is the following polynomial factorable? Explain why or why not. x2 + 5x + 5 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Factoring Quiz #1 – Answer Key Part 1: Complete each problem. Express your answer in simplest terms. (1 point each) 3x2 – 18x + 33 1. Factor the following polynomial using the GCF. Step 1: Identify the GCF for 3x2, 18x and 33. The GCF is 3. Factor a 3 out of each term: 3(x2 – 6x + 11) Final Answer: 3(x2 – 6x + 11) 6a5 – 12a4 + 15a3 2. Factor the following polynomial using the GCF. Step 1: Identify the GCF for 6a5, 12a4 and 16a3. The GCF is 3a3. Factor 3a3 out of each term. 3a3(2a2 – 4a + 5) Final Answer: 3a3(2a2 – 4a + 5) 8x2y3z + 4xy4 z3 – 12x4y5z2 3. Factor the following polynomial using the GCF. 2 3 4 3 4 5 2 Step 1: Identify the GCF for 8x y z, 4xy z , and 12x y z . The GCF is 4xy3z. Factor 4xy3z out of each term. 4xy3z(2x+yz2 – 3x3y2z) Final Answer: 4xy3z(2x+yz2 – 3x3y2z) 4. Simplify the following fraction: Step 1: Identify the GCF for the numerator: The GCF is: 2a2b3. Factor this out of each term: 2a2b3 (4+ 3ab) 2a2b3 Step 2: Simplify: 2a2b3 (4+ 3ab) 2a2b3 = 4+3ab Final Answer: 4+3ab Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 5. Simplify the following fraction: 6x2y7z5 + 3x2y3z6 + 2xy3 z8 xy3z5 Step 1: Identify the GCF for the numerator: The GCF is: xy3z5 Factor this out of each term: xy3z5 (6xy4+3xz+2z3) xy3z5 Step 2: Simplify: xy3z5 (6xy4+3xz+2z3) xy3z5 Final Answer: 6xy4 + 3xz + 2z3 Part 2: Factor each polynomial. (1 point each) 1. x2 + 9x + 18 We need two factors whose product is 18 and sum is 9. 6 & 3 6●3 = 18; 6+3 = 9 Final Answer: (x+6)(x+3) 2. x2 – x – 20 We need two factors whose product is -20 and sum is -1. (Since the product is negative, we need a positive number and a negative number. The larger number should be negative since the sum is negative. 4 & -5 4●(-5) = -20; 4+(-5) = -1 Final Answer: (x+4)(x-5) 3. y2 – 9y + 14 We need two factors whose product is 14 and sum is -9. (Since the product is positive and the sum is negative, we must have two numbers that are negative. -7& -2 -7●(-2) = 14; -7+(-2) = - 9 Final Answer: (y-7)(y-2) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 4. r2 + 2r – 48 We need two factors whose product is -48 and sum is 2. (Since the product is negative, we need a positive and a negative number. The larger number will be positive since the sum is positive. -6 & 8 -6●8 = -48; -6+8 = 2 Final Answer: (r-6)(r+8) 5. z2 – 6z + 5 We need two factors whose product is 5 and sum is -6. (Since the product is positive and the sum is negative, we must have two number that are negative.) -5 & -1 -5●(-1) = 5; -5+(-1) = -6 Final Answer: (z-5)(z-1) Part 3: Short Answer. (2 points each) 1. Explain the steps used to factor the following polynomial: 3x4 + 6x2 – 3x First you must look to see if there is a greatest common factor. The GCF is 3x Then factor out the 3x. 3x(x3 + 2x – 1) This is the final answer. ***Look for key words: greatest common factor*** 2. Is the following polynomial factorable? Explain why or why not. x2 + 5x + 5 This polynomial is NOT factorable. We would need to find two positive numbers whose product is 5 and whose sum is 5. The only factors of 5 are 5 and 1. Since 5+1 = 6 and not 5, there is no way to factor this polynomial using integers. This quiz is worth 14 points. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 4: Factoring Trinomials: A Special Situation Using two different factoring methods. In the previous lesson, we factored trinomials that had a lead coefficient of 1. Of course, many trinomials have a lead coefficient that is greater than 1. We will explore one method of factoring these types of trinomials. Example 1 Factor: 3x2 – 3x – 36 Notice how the lead coefficient is not 1. However, each term is divisible by 3. Step 1: Note: If a trinomial looks difficult to factor, first check to see if a Greatest Common Factor (GCF) can be factored out. Example 2 Factor: 4x3 + 4x2 – 24x Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 4: Factoring Trinomials (A Special Situation) Practice Part 1: Factor Completely 1. 2x2 + 10x – 48 7. 2y5 – 16y4 + 32y3 2. 5y2 – 40y + 60 8. 3x2 - 243 3. 6x3 + 54x2 + 120x 9. 2b3 – 2b2 – 24b 4. 3s3 – 108s 10. 4x3 – 20x2 + 24x 5. 5x4 – 50x3 + 125x2 11. 6w4 – 18w3 – 24w2 6. 2a4 + 12a3 + 18a2 12. 3x2 – 9x - 54 Part 2: Thinking Questions 1. Give at least one example of a trinomial in the form of: x2 +bx +c that can be factored. 2. Give at least one example of a trinomial without a lead coefficient of one that can be factored. 3. Give an example of a prime trinomial. Factor each trinomial. (3 points each) 1. 3x3 +18x2 – 81x Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com 2. 2x3 – 8x2 – 8x Unit 9: Factoring Lesson 4: Factoring Trinomials (A Special Situation) Answers Part 1: Factor Completely 1. 2x2 + 10x – 48 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2 is the GCF for all terms. 2(x2 + 5x – 24) Step 2: Factor the expression inside of parenthesis. We need a positive and negative number that have a product of -24 and a sum of 5 (8 & 3, 8 must be positive since we need a +5 for a sum). Final Answer: 2(x+ 8) (x-3) 7. 2y5 – 16y4 + 32y3 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 3 2y is the GCF for all terms. 2y3(y2 – 8y + 16) Step 2: Factor the expression inside of parenthesis. We need 2 negative numbers that have a product of 16 and a sum of -8. (-4 & -4). 2y3(y-4) (y-4) Final Answer: 2y3(y-4)2 2. 5y2 – 40y + 60 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 5 is the GCF for all terms. 5(y2 - 8x + 12) Step 2: Factor the expression inside of parenthesis. We need 2 negative numbers that have a product of 12 and a sum of -8 (-6 & -2). 8. 3x2 – 243 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 3 is the GCF for all terms. 3(x2 - 81) Step 2: Factor the expression inside of parenthesis. This is a difference of two squares. So, what number squared is 81? (9) Final Answer: 3(x+ 9) (x-9) Final Answer: 5(y-6) (y-2) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 3. 6x3 + 54x2 + 120x 9. 2b3 – 2b2 – 24b Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 6x is the GCF for all terms. 6x(x2 + 9x + 20) Step 2: Factor the expression inside of parenthesis. We need 2 positive numbers that have a product of 20 and a sum of 9 (4 & 5) Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2b is the GCF for all terms. 2b(b2 - b – 12) Step 2: Factor the expression inside of parenthesis. We need a positive and negative number that have a product of -12 and a sum of -1 (4 & 3, 4 must be negative since we need a -1 for a sum). Final Answer: 6x(x+ 4) (x+ 5) Final Answer: 2b(b-4) (b+3) 4. 3s3 – 108s 10. 4x3 – 20x2 + 24x Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 3s is the GCF for all terms. 3s(s2 - 36) Step 2: Factor the expression inside of parenthesis. This is a difference of two squares. What number when squared equals 36? (6) Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 4x is the GCF for all terms. 4x(x2 – 5x+ 6) Step 2: Factor the expression inside of parenthesis. We need two negative numbers that have a product of 6 and a sum of -5. (-3 & -2) Final Answer: 3s(s+ 6) (s-6) Final Answer: 4x(x-3) (x-2) 5. 5x4 – 50x3 + 125x2 11. 6w4 – 18w3 – 24w2 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2 5x is the GCF for all terms. 5x2(x2 -10x + 25) Step 2: Factor the expression inside of parenthesis. We need two negative numbers whose product is 25 and sum is -10 (-5 & -5) This is a perfect square. Final Answer: 5x2(x - 5) (x-5) or Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2 6w is the GCF for all terms. 6w2(w2 -3w -4) Step 2: Factor the expression inside of parenthesis. We need a positive and negative number whose product is -4 and whose sum is -3. (-4 & 1) 4 must be negative since the sum is negative. 5x2(x-5)2 Final Answer: 6w2(w -4) (w +1) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 6. 2a4 + 12a3 + 18a2 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2 2a is the GCF for all terms. 2a2(a2 + 6a + 9) Step 2: Factor the expression inside of parenthesis. This is a perfect square. What number can you square and get 9, and add to get 6? (3) 12. 3x2 – 9x – 54 Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 3 is the GCF for all terms. 3(x2 – 3x - 18) Step 2: Factor the expression inside of parenthesis. What positive and negative numbers have a product of -18 and a sum of -3? (-6 & 3) 6 must be negative since the sum is negative. Final Answer: 2a2(a+3)2 Final Answer: 3(x -6) (x +3) Part 2: Thinking Questions 1. Give at least one example of a trinomial in the form of: x2 +bx +c that can be factored. These answers may vary. The easiest way to find a trinomial that can be factored is to work backwards and start with two binomials. You can choose any binomials. For example, I’ll choose (x+4) (x+2) now multiply using foil. (x)(x) +(2)x+(4)x+4(2) x2 +6x + 8 is a trinomial that can be factored. We know this because we started with the two factors and multiplied. 2. Give at least one example of a trinomial without a lead coefficient of one that can be factored. These answers may vary. The easiest way to find a trinomial that can be factored is to work backwards and start with two binomials. You can choose any binomials. Since you want a lead coefficient that is greater than 1, make sure that one of your binomials has a coefficient of x that is greater than 1. For example, I’ll choose (2x+4) (x+2) now multiply using foil. (2x)(x) +(2x)2+(4)x+4(2) 2x2 +8x + 8 is a trinomial that can be factored. We know this because we started with the two factors and multiplied. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 3. Give an example of a prime trinomial. A prime trinomial cannot be factored. For example: x2 + 2x + 2 - We would need two positive numbers whose product is 2 and whose sum is 2. The only way to make a product of 2 is (2 ●1) Since 2+1 = 3, we know that this cannot be factored. The best way to make up a prime trinomial is to use a prime number as your constant. Then make sure the coefficient of x is not the sum of the factors of your prime number. 1. 3x3 +18x2 – 81x Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 3x is the GCF for all terms. 3x(x2 + 6x - 27) Step 2: Factor the expression inside of parenthesis. We need a positive and negative number whose product is -27 and whose sum is 6. (9 & -3) (9 must be positive since the sum is positive) Final Answer: 3x(x+9) (x - 3) 2. 2x3 – 8x2 – 8x Step 1: Since the lead coefficient is greater than 1, see if there is a GCF for all terms. 2x is the GCF for all terms. 2x(x2- 4x - 4) Step 2: Factor the expression inside of parenthesis. We need a positive and negative number whose product is -4 and whose sum is -4. The inside of the parenthesis cannot be factored any further because there’s no positive and negative number whose product is -4 and whose sum is -4. Final Answer: 2x(x2 – 4x – 4) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 5: Factoring Trinomials in the Form of: ax2 + bx + c If you are factoring a trinomial with a lead coefficient that is greater than 1 and there is NO GCF to factor out, then we will need to use a guess and check method in order to factor the trinomial. Example 1 Factor: 2x2 +9x + 4 Example 2 Factor: 3x2 + 7x - 6 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Example 3 Factor: 4x2 -4x - 3 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 5: Factoring Trinomials in the Form of: ax2 + bx + c Part 1: Factor completely. 1. 3x2 + 5x – 2 6. 4x2 + 4x +1 2. 2x2 – 3x – 2 7. 6x2 -5x - 6 3. 5x2 – 14x – 3 8. 4x2 – x - 3 4. 3x2 + 13x + 4 9. 36x2 + 12x + 4 5. 2x2 – 9x + 4 10. 8x2 – 14x + 6 Part 2: Mixed Factoring! Factor each trinomial completely using ANY method (or combination of methods.) 1. x2 + 9x + 18 9. 24x2 + 80x + 24 2. 3x3 – 33x2 + 84x 10. 6x3 -24x2 -126x 3. 2x2 + 9x – 5 11. 5x3 -35x2 -40x 4. 10x2 + 5x – 15 12. 18x2 -21x -9 5. x2 –x – 56 13. 10x2 -32x +6 6. 6x2 – 10x – 4 14. 12x3 +36x2 + 27x 7. 3x2 – 33x + 54 15. 4x2 - 1 8. 2x3 + 20x2 + 42x 16. 27x2 -12 Factor Completely. (2 points each) 1. 6x2 + 3x – 3 3. 24x2 +52x - 20 2. 8x2 +14x +6 4. 18x2 -12x + 2 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Lesson 5: Factoring Trinomials in the Form of: ax2 + bx + c - Answers Part 1: Factor completely. For all of the problems in Part 1, the lead coefficient is greater than 1. However, there is no GCF, so we must use the guess and check method. 1. 3x2 + 5x – 2 6. 4x2 + 4x +1 Using the model (ax + b)(cx+d), we need: Using the model (ax + b)(cx+d), we need: ac=3 (3 &1 )and bd=-2 (-2 &1 or -1 &2) ac=4 (4 &1 or 2 &2 )and bd=1 (1 &1) (3x-2)(x+1) =3x2 +3x -2x -2 or (4x+1)(x+1) = 4x2 +4x+x+1 or 4x2 +5x +1 3x2 +x - 2 (3x+1)(x-2) = 3x2 – 6x +x -2 or 3x2 -5x - 2 (2x+1)(2x+1) = 4x2 +2x +2x +1 or 4x2 +4x +1 (3x -1)(x+2) = 3x2 +6x – x -2 or 3x2 +5x – 2 (3x+2)(x-1) = 3x2 – 3x + 2x – 2 or 3x2 -x – 2 Final Answer: (2x+1)(2x+1) or (2x+1)2 Final Answer: (3x-1)(x+2) 2. 2x2 – 3x – 2 7. 6x2 -5x - 6 Using the model (ax + b)(cx+d), we need: ac=2 (2 &1 )and bd=-2 (-2 &1 or -1 &2) (2x-2)(x+1) = 2x2 +2x -2x -2 or 2x2 -2 (2x+1)(x-2) = 2x2 -4x +x -2 or 2x2 -3x -2 (2x -1)(x+2) = 2x2 +4x – x -2 or 2x2 +3x - 2 (2x+2)(x-1) = 2x2 -2x +2x - 2 or 2x2 -2 Final Answer: (2x+1)(x-2) Using the model (ax + b)(cx+d), we need: ac=6 (6 &1 or 3 & 2 )and bd=-6 (-6 &1 or -1 &6 or -3 &2 or -2 &3) (6x-6)(x+1) = 6x2 +6x -6x -6 or 6x2 -6 (6x-1)(x+6) = 6x2 +36x –x -6 or 6x2 +35x - 6 (6x -3)(x+2) = 6x2 +12x -3x -6 or 6x2 +9x -6 (6x-2)(x+3) = 6x2 +18x -2x -6 or 6x2 +16x -6 (3x-6)(2x+1) = 6x2 +3x-12x -6 or 6x2 -9x -6 (3x-1)(2x+6) = 6x2 +18x -2x -6 or 6x2 +16x - 6 (3x -3)(2x+2) = 6x2 +6x -6x -6 or 6x2 -6 (3x-2)(2x+3) = 6x2 +9x -4x -6 or 6x2 +5x – 6 (3x+2)(2x-3) = 6x2 -9x +4x -6 or 6x2 -5x -5 Final Answer: (3x+2)(2x-3) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 3. 5x2 – 14x – 3 8. 4x2 – x – 3 Using the model (ax + b)(cx+d), we need: Using the model (ax + b)(cx+d), we need: ac=5 (5 &1 )and bd=-3 (-3 &1 or -1 &3) ac=4 (4 &1 or 2 & 2)and bd=-3 (-3 &1 or -1 &3) (5x-3)(x+1) = 5x2 +5x – 3x -3 or 5x2 +2x -3 (4x-3)(x+1) = 4x2 +4x – 3x -3 or 4x2 +x - 3 (5x+1)(x-3) = 5x2 -15x +x -3 or 5x2 -14x -3 After multiplying this problem, I notice that I need to make x a negative x and then I would have the correct factors. Therefore, if I try (5x -1)(x+3) = 5x2 +15x –x -3or 5x2 +14x - 3 (5x+3)(x-1) = 5x2 -5x +3x -3 or 5x2 -2x -3 (4x +3) (x-1) I may be able to arrive at my answer and not have to try all of the other factors. Final Answer: (5x+1)(x-3) (4x+3)(x-1) = 4x2 -4x +3x -3 or 4x2 –x -3 Final Answer: (4x+3)(x-1) 4. 3x2 + 13x + 4 9. 36x2 + 12x + 4 Using the model (ax + b)(cx+d), we need: ac=3 (3 &1 )and bd= 4 (2 &2 or 4 &1 or 1 & 4) 2 (3x+2)(x+2) =3x +6x +2x+4 or 2 3x +8x +4 The first thing I notice is that there is a GCF that I can factor out. 4 is the GCF. We then have: 4(9x2 + 3x +1) Now I can factor inside of the parenthesis: (3x+4)(x+1) = 3x2 +3x +4x +4 or 3x2 7x +4 4(9x2 + 3x +1) (3x +1)(x+4) = 3x2 +12x +x+4 or 3x2 +13x +4 Using the model (ax + b)(cx+d), we need: ac=9 (9 &1 or 3 & 3 ) and bd= 1 (1 &1) Final Answer: (3x+1)(x+4) Since my sum is 3x, I am not going to start with 9 &1, I will start with 3 & 3. (3x +1)(3x+1) = 9x2 + 3x +3x +1 or 9x2 + 6x +1 This cannot be factored, so our final answer is: Final Answer: 4(9x2 + 3x +1) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 5. 2x2 – 9x + 4 Using the model (ax + b)(cx+d), we need: ac=2 (2 &1 )and bd= 4 – Since we have -9x, the factors of 4 must both be negative. (-2 &-2)(-4&-1) (-1&-4) (2x – 2)(x-2) = 2x2 -4x -2x +4 or 2x2 -6x +4 (2x -1)(x-4) = 2x2 -8x –x + 4 or 2x2 -9x +4 I can stop since I have found the factors. Final Answer: (2x-1)(x-4) 10. 8x2 – 14x + 6 Notice that 2 is the GCF for all 3 terms. We’ll factor a 2 out first. 2(4x2 – 7x + 3) Using the model (ax + b)(cx+d), we need: ac=4 (4 &1 )(2 & 2)and bd= 3. Since we have -7x, the factors of 3 must both be negative (-3 &-1) (2x -3)(2x-1) = 4x2 -2x – 6x + 3 or 4x2 -8x +3 (4x-3)(x-1) = 4x2 – 4x – 3x + 3 or 4x2 – 7x + 3. Final Answer: 2(4x-3)(x-1) Part 2: Mixed Factoring! Factor each trinomial completely using ANY method (or combination of methods.) 1. x2 + 9x + 18 9. 24x2 + 80x + 24 Since the lead coefficient is 1, I need to think of two positive numbers that have a product of 18 and a sum of 9. (6&3) Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 8 is the GCF, so first we’ll factor out an 8. Final Answer: (x+6)(x+3) 8(3x2 + 10x + 3) The lead coefficient inside of the parenthesis is still greater than 1, so: Using the model (ax + b)(cx+d), we need: ac=3 (3 &1 ) and bd= 3.(3&1)(1&3) Factor: 3x2 + 10x + 3 (3x +3) (x+1) = 3x2 + 3x+3x + 3 or 3x2 + 6x + 3 (3x+1)(x+3) = 3x2 + 9x + x + 3 or 3x2 + 10x + 3 Final Answer: 8(3x+1)(x+3) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 2. 3x3 – 33x2 + 84x 10. 6x3 -24x2 -126x Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 3x is the GCF, so first we’ll factor out a 3x. Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 6x is the GCF, so first we’ll factor out a 6x. 3x(x2 - 11x + 28) 6x(x2 - 4x - 21) The lead coefficient inside of the parenthesis is 1, so: We need two negative numbers whose product is 28 and sum is -11. (-7 & -4) Factor: x2 - 11x + 28 The lead coefficient inside of the parenthesis is 1, so: We need a positive and negative number whose product is -21 and whose sum is -4. (-7&3 - Since 4 is negative, the largest number (7) must be negative. (x-7)(x-4) Factor: x2 - 4x -21 Final Answer: 3x(x-7)(x-4) (x -7)(x+3) Final Answer: 6x(x-7)(x+3) 3. 2x2 + 9x – 5 11. 5x3 -35x2 -40x Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. There is no GCF. Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. The GCF is 5x, so I will first factor out 5x. Using the model (ax + b)(cx+d), we need: 5x(x2 – 7x – 8) ac=2 (2 &1 ) Now the lead coefficient inside of the parenthesis is 1, so I will factor: x2 – 7x – 8. I need a positive and negative number whose product is -8 and whose sum is -7. (-8 & 1) (8 must be negative since 7 is negative. and bd= -5.(-5&1)(-1&5) (2x – 5)(x+1) = 2x2 +2x – 5x -5 or 2x2 -3x – 5 (2x – 1)(x +5) = 2x2 + 10x – x -5 or 2x2 +9x -5 (x – 8)(x+1) Final Answer: (2x-1)(x+5) Final Answer: 5x (x-8)(x+1) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 4. 10x2 + 5x – 15 12. 18x2 -21x -9 Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. The GCF is 5. Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 3 is the GCF. We’ll first factor out a 3. 5(2x2 + x – 3) 3(6x2 – 7x – 3) The lead coefficient inside of the parenthesis is greater than 1, so: Since the lead coefficient inside of the parenthesis is greater than 1, we need to: Using the model (ax + b)(cx+d), we need: Use the model (ax + b)(cx+d), we need: ac=2 (2 &1 ) ac=6 (2 &3)(6&1) and bd= -3.(-3&1)(-1&3) (2x –3)(x+1) = 2x2 +2x – 3x -3 or 2x2 -x – 3 We need a sum of positive x, so let’s switch the signs: 2 (2x+3)(x-1) = 2x – 2x + 3x – 3 or Factor: 6x2 – 7x – 3 (3x -3)(2x +1) = 6x2 +3x – 6x – 3 or 6x2 – 3x – 3 (3x+1)(2x-3) = 6x2 – 9x + 2x -3 or 6x2 -7x -3 2 2x +x - 3 Final Answer: 5(2x+3)(x-1) Final Answer: 3(3x+1)(2x-3) 5. x2 –x – 56 Since the lead coefficient is 1, we need to find a positive and negative number whose product is 56 and whose sum is -1. (-8&7) (Since x is negative, our largest number (8) must be negative because -8 + 7 = -1. (x-8)(x+7) and bd= -3.(-3&1)(-1&3) 13. 10x2 -32x +6 Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 2 is the GCF. 2(5x2 – 16x + 3) The lead coefficient inside of the parenthesis is still greater than one, so: Using the model (ax + b)(cx+d), we need: Final Answer: (x-8)(x+7) ac=5 (5 &1 ) and bd= 3.(-3&-1)(-1&-3)(Both factors of 3 must be negative since 16 is negative) (5x – 1)(x-3) = 5x2 – 15x – x +3 or 5x2 -16x +3 Final Answer: 2(5x-1)(x-3) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 6. 6x2 – 10x – 4 14. 12x3 +36x2 + 27x Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 2 is the GCF. Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 3x is the GCF. 2(3x2 – 5x – 2) 3x(4x2 + 12x + 9) The lead coefficient inside of the parenthesis is greater than 1, so: The lead coefficient inside of the parenthesis is greater than 1, so: Using the model (ax + b)(cx+d), we need: Using the model (ax + b)(cx+d), we need: ac=3 (3 &1 ) ac=4 (2 &2)(4&1) and bd= -2.(-2&1)(-1&2) (3x + 1)(x – 2) = 3x2 – 6x + x – 2 or 3x2 – 5x – 2 and bd= 9.(9&1)(3&3) I know that 2 x 3 = 6 and 6+6 = 12, so let’s try 2 &2 for ac, and 3 & 3 for bd. (2x+3)(2x+3) = 4x2 + 6x +6x +9 or 4x2 + 12x + 9 Final Answer: 2(3x+1)(x-2) Final Answer: 3x(2x+3)(2x+3) or 3x(2x+3)2 7. 3x2 – 33x + 54 Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 3 is the GCF. 3(x2 – 11x + 18) Now we will factor inside of the parenthesis. I need two negative numbers whose product is 18 and sum is -11. (-9&-2) 15. 4x2 – 1 Since the middle term is missing, I know that this is a difference of two squares. What number can you square to get 4 (2) and what number can you square to get 1 (1)? (2x+1)(2x-1) Multiply to check: 4x2 – 2x + 2x – 1 or 4x2 -1 (x -9)(x-2) Final Answer: (2x +1)(2x-1) Final Answer: 3 (x-9)(x-2) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 8. 2x3 + 20x2 + 42x 16. 27x2 -12 Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 2x is the GCF. Since the lead coefficient is greater than 1, I will first see if there is a GCF for both terms. 3 is the GCF. 2x(x2 + 10x + 21) 3(9x2 – 4) Now factor inside of the parenthesis. The lead coefficient is 1, so we need two positive numbers whose product is 21 and whose sum is 10. (7&3) Inside of the parenthesis we have no middle term, so we have a difference of two squares. What can we square to get 9 (3) and what can we square to get 4 (2)? (x+7)(x+3) (3x +2)(3x-2) Final Answer: 2x (x+7)(x+3) Final Answer: 3(3x+2)(3x-2) Factor Completely. (2 points each) 1. 6x2 + 3x – 3 3. 24x2 +52x – 20 Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 3 is the GCF. Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 4 is the GCF. 3(2x2 + x - 1) 4(6x2 + 13x – 5) Now factor inside of the parenthesis. The lead coefficient is greater than 1, so: (Now factor inside of the parenthesis. The lead coefficient is greater than 1, so: Using the model (ax + b)(cx+d), we need: Using the model (ax + b)(cx+d), we need: ac=2 (2 &1) ac=6 (6 &1)(3&2) and bd= -1.(1&-1)(-1&1) and bd= -5.(5&-1)(-1&5) Factor: 2x2 + x – 1 Factor: 6x2 + 13x - 5 (2x -1)(x+1)= 2x2 +2x – x -1 or 2x2 + x -1 (3x -1)(2x + 5) = 6x2 + 15x -2x – 5 or 6x2 + 13x -5 Final Answer: 3 (2x-1)(x+1) Final Answer: 4(3x-1)(2x+5) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 2. 8x2 +14x +6 4. 18x2 -12x + 2 Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 2 is the GCF. Since the lead coefficient is greater than 1, I will first see if there is a GCF for all three terms. 2 is the GCF. 2(4x2 +7 x +3) 2(9x2 -6x +1) Now factor inside of the parenthesis. The lead coefficient is greater than 1, so: Now factor inside of the parenthesis. The lead coefficient is greater than 1, so: Using the model (ax + b)(cx+d), we need: Using the model (ax + b)(cx+d), we need: ac=4 (4 &1)(2&2) ac=9 (9 &1)(3&3) and bd= 3.(1&3)(3&1) and bd= 1.(-1&-1) Factor: 4x2 +7 x+ 3 Factor: 9x2 – 6x +1 (4x +3)(x+1) = 4x2 +4x +3x+3 or 4x2 +7x +3 (3x -1)(3x-1) = 9x2 -3x -3x +1 or 9x2 – 6x +1 Final Answer: 2(4x+3)(x+1) Final Answer: 2 (3x-1)(3x-1) or 2(3x-1)2 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Unit 9: Factoring Chapter Test Part 1: Factor Completely. Remember to factor out the greatest common factor first if it is other than 1. 1. x2 + 12x + 32 7. 3y2 + 21y + 36 2. y2 – 2y – 35 8. 2x3 – 18x 3. n2-8n + 16 9. 12x4y2 + 18x3y3 – 24x2y4 4. x2 – 13x + 36 10. 6x2 + 5x - 4 5. 8x4 + 2x2 – 4x 11. 4p2 – 7p - 2 6. 2x3 + 6x2 – 20x 12. 4x2 + 10x – 6 Part 2: Answer each problem completely. 1. What polynomial when factored gives: (3x+ 2y)( 2x – 5y) 2. One factor of the polynomial x2 + 27x + 180 is (x+15). What is the other factor? 3. Find two values for b. Explain how you determined your answer. x2 + bx – 12 4. Find a value for c that will make this a perfect square trinomial. x2 + 18x + c Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Unit 9: Factoring Chapter Test – Answer Key Part 1: Factor Completely. Remember to factor out the greatest common factor first if it is other than 1. (1 point each) 1. x2 + 12x + 32 7. 3y2 + 21y + 36 We need to find two positive integers whose product is 32 and whose sum is 12. 8&4 8●4 = 32 8+4 = 12 Since the lead coefficient is greater than1, we need to check to see if there is a GCF that can be factored out. The GCF is 3. 3(y2 + 7y + 12) Final Answer: (x+8)(x+4) Now we need to factor the polynomial inside of the parenthesis. We need two positive numbers whose product is 12 and whose sum is 7. 4&3 Final Answer: 3(x+4)(x+3) 2. y2 – 2y – 35 8. 2x3 – 18x We need to find two integers whose product is -35 and whose sum is -2. Since the product is negative, one integer is positive and one negative. The larger integer is negative since the sum is negative. Since the lead coefficient is greater than1, we need to check to see if there is a GCF that can be factored out. The GCF is 2x. -7 & 5 Now we need to factor the polynomial inside of the parenthesis. Since there is no middle term and 9 is a perfect square, we know this factors as a difference of two squares: -7●5 = -35 -7+5 = -2 Final Answer: (y-7)(y+5) 2x(x2 - 9) Final Answer: 2x(x+3)(x-3) 3. n2-8n + 16 9. 12x4y2 + 18x3y3 – 24x2y4 We need to find two integers whose product is 16 and whose sum is -8. Since the product is positive and the sum is negative, both integers must be negative. First we need to see if there is a GCF. The GCF is 6x2y2. Factor out 6x2y2. -4 & -4 There is no way to factor inside of the parenthesis. Therefore: -4●(-4) = 16 -4+(-4) = -8 Final Answer: (n-4)(n-4) or (n-4)2 6x2y2 (2x2+ 3xy – 4y2) Final Answer: 6x2y2 (2x2+ 3xy – 4y2) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 4. x2 – 13x + 36 10. 6x2 + 5x – 4 We need to find two integers whose product is 36 and whose sum is -13. Since the product is positive and the sum is negative, both integers must be negative. Since the lead coefficient is greater than 1, look to see if there is a GCF. There is no GCF for this polynomial, so we must use the guess and check method for factoring. -9 & -4 Factors of 6: 6,1 & 3,2 -9●(-4) = 36 -9+(-4) = -13 Final Answer: (x-9)(x-4) Factors of -4: (2,-2) & (-4,1) & (-1,4) (2x -1) (3x+4) = 6x2 + 8x – 3x – 4 or 6x2 +5x -4 Final Answer: (2x -1) (3x+4) 5. 8x4 + 2x2 – 4x First we need to check to see if there is a GCF that can be factored out. 2x is the GCF. 2x(4x3 + x – 2) The inside of the parenthesis cannot be factored any further, therefore: Final Answer: 2x(4x3 + x – 2) 11. 4p2 – 7p – 2 Since the lead coefficient is greater than 1, look to see if there is a GCF. There is no GCF for this polynomial, so we must use the guess and check method for factoring. Factors of 4: 2,2 & 4,1 Factors of -2: (1,-2) & (-2,1) (4p +1) (p-2) = 4p2 – 8p + p – 2 or 4p2 -7x -2 Final Answer: (4p +1) (p-2) 6. 2x3 + 6x2 – 20x 12. 4x2 + 10x – 6 First we need to check to see if there is a GCF that can be factored out. 2x is the GCF. Since the lead coefficient is greater than 1, look to see if there is a GCF. 2 is the GCF. 2x(x2 + 3x - 10) 2(2x2 + 5x – 3) Now we must factor the inside of the parenthesis. We need a positive and negative integer whose product is -10 and whose sum is 3. The larger integer will be positive since the sum is positive. 5 & -2 Now factor the inside of the parenthesis using the guess and check method. Final Answer: 2x(x+5)(x-2) 2(2x-1) (x+3) =2(2x2 +6x - x – 3) or 2(2x2 +5x -3) Factors of 2: 2,1 Factors of -3: (1,-3) & (-3,1) Final Answer: 2(2x-1)(x+3) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Part 2: Answer each problem completely. (2 points each) 1. What polynomial when factored gives: (3x+ 2y)( 2x – 5y) We can multiply to figure out the polynomial: Use the foil method. (3x+2y)(2x-5y) 3x(2x) + 3x(-5y) + 2y(2x) + 2y(-5y) 6x2 – 15xy + 4xy – 10y2 Use FOIL Multiply the terms 6x2 – 11xy – 10y2 Simplify the middle terms. Final Answer: 6x2 – 11xy – 10y2 2. One factor of the polynomial x2 + 27x + 180 is (x+15). What is the other factor? We know that our two chosen integers must have a product of 180 and a sum of 27. Since we know that one factor is (x+15), we know that one of our integers is 15. The other integer must be 12 since 180/15 = 12. Also, 15+ 12 = 27, so this confirms our choice. Therefore the other factor is: (x+ 12) Final Answer: (x+12) 3. Find two values for b. Explain how you determined your answer. x2 + bx – 12 We know that we need two integers, one positive and one negative, whose product is -12. The sum of these two integers is the variable b in the polynomial. Integers whose product is -12: (-4)(3); (-3)(4); (-6)(2); (-2)(6); (-12)(1); (12)(-1) b could be equal to: -1, 1, -4, 4,-11, 11 (If you add the integers together, you will end up with their sum, which is the variable b.) 4. Find a value for c that will make this a perfect square trinomial. x2 + 18x + c Letter C must be a perfect square. The square root of this number multiplied by 2 is 18. The only number that can represent letter c is 81. The factors would be (x+9)2 This test is worth 20 points Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Cumulative Test on Polynomials and Factoring Part 1: Complete each problem by circling the correct answer. For problems 1-8, simplify the expression. State the answer in standard form. 1. (3x2+8x – 11) + (2x2 – 6x +4) A. 5x2 + 14x + 15 C. 5x2 + 2x – 7 B. 5x2 + 2x + 7 D. 5x2 + 2x – 15 2. (4x4 + 2x2 – 3x + 1) + (3x3 – 2x2 + 8x – 6) A. 4x4 + 3x3 + 5x – 5 C. 4x4 + 3x2 + 4x2 – 3x +8x – 5 B. 7x7 + 5x – 5 D. 4x4 + 3x3 + 2x2 + 5x – 5 3. (6x2 – 3x + 9) – (4x2 + 8x – 6) A. 2x2 + 5x + 3 C. 2x2 + 11x + 15 B. 2x2 – 11x + 15 D. 10x2 + 5x + 3 4. 3(2x2 – 4x +2) – 2(x2 + 4) A. 4x2 – 12x + 14 C. 8x2 – 12x – 2 B. 8x2 – 12x + 14 D. 4x2 – 12x – 2 5. (5r-3)2 A. 25r2 – 9 C. 5r2 + 9 B. 25r2 – 30r + 9 D. 25r2 + 9 6. (x+6)(x-6) A. x2 + 12x – 36 C. x2 + 36 B. x2 – 36 D. x2 -6x – 36 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 7. (3x-2)(2x+1) A. 6x2 + 7x – 2 C. 6x2 – x + 2 B. 6x2 + x – 2 D. 6x2 – x – 2 8. (x+4)2(3x-2) A. 3x3 + 22x2 +32x – 32 C. 3x2 + 10x – 8 B. 3x3 – 2x2 + 48x – 32 D. 3x3 + 26x2 + 48x – 32 For problems 9-10, factor the polynomial. 9. x2 -3x -70 A. (x+10)(x-7) C. (x+7)(x-10) B. (x+5)(x-14) D. (x+14)(x-5) 10. 6x2 – 2x – 8 A. (3x-4)(2x+2) B. (3x+2)(2x-4) C. (2x-2)(3x+4) D. (2x+4)(3x-2) Part 2: Short Answer 1. A square table measures 3x +2y units. Write an expression in simplified form that represents the area of the table. 2. A rectangular sheet of paper has a width of x units. The area of the paper is x2+8x. Write an expression that represents the length of the rectangular sheet of paper. 3. Factor the polynomial completely. 2x2 + 11x + 12 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 4. Factor the polynomial completely. 9x3 +6x2 -48x 5. Factor the polynomial completely. y2 – 121 6. A rectangle has the following dimensions: width: (2x + y) length: (5x – 2y) Express the area of the rectangle as a polynomial in standard form. 7. A triangle has the following dimensions: Side 1: x2 +2x -1 Side 2: x2 + 2x -1 Side 3: 2x2 +x – 3 Write an expression that represents the perimeter of the triangle. 8. A rectangle has a length of 8x -4 and a width of 4x +3. The perimeter of the rectangle is 58 units. Find the width of the rectangle. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Part 3: Extended Response. 1. You have a rectangular piece of wood in which you are making a carnival game. You plan to cut out 2 circles that children can use to try to throw a bean bag through at the carnival. Complete each bullet below given the following dimensions: Length of rectangular piece of wood: (5x+2) Width of rectangular piece of wood: (3x+1) Radius of each circle: (x+1) • Write an algebraic expression for the area of the rectangular piece of wood before the circles are cut. • • Write an algebraic expression for the area of 1 circle. Write an algebraic expression for the area of the wood after both circles are cut out. Explain how you determined your answer. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Cumulative Test on Polynomials and Factoring - Answer Key Part 1: Complete each problem by circling the correct answer. (1 point each) For problems 1-8, simplify the expression. State the answer in standard form. 1. (3x2+8x – 11) + (2x2 – 6x +4) Write the expression vertically. 3x2 + 8x – 11 + 2x2 – 6x + 4 5x2 +2x - 7 A. 5x2 + 14x + 15 C. 5x2 + 2x – 7 B. 5x2 + 2x + 7 D. 5x2 + 2x – 15 2. (4x4 + 2x2 – 3x + 1) + (3x3 – 2x2 + 8x – 6) Write the expression vertically. 4x4 + 0x3 +2x2 – 3x + 1 + 3x3 -2x2 + 8x – 6 4x4 + 3x3 +0x2 +5x - 5 A. 4x4 + 3x3 + 5x – 5 C. 4x4 + 3x2 + 4x2 – 3x +8x – 5 B. 7x7 + 5x – 5 D. 4x4 + 3x3 + 2x2 + 5x – 5 3. (6x2 – 3x + 9) – (4x2 + 8x – 6) Step 1: Rewrite as an addition problem. (6x2 – 3x + 9) + (-4x2 – 8x + 6) Write the new expression vertically. 6x2 – 3x + 9 + -4x2 – 8x + 6 2x2 – 11x + 15 A. 2x2 + 5x + 3 C. 2x2 + 11x + 15 B. 2x2 – 11x + 15 D. 10x2 + 5x + 3 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 4. 3(2x2 – 4x +2) – 2(x2 + 4) Step 1: Distribute the 3 throughout the first set of parenthesis and the -2 throughout the second set of parenthesis: (6x2 – 12x + 6) + (-2x2 – 8) Step 2: Write the new expression vertically. 6x2 – 12x + 6 + -2x2 +0x -8 4x2 – 12x -2 A. 4x2 – 12x + 14 C. 8x2 – 12x – 2 B. 8x2 – 12x + 14 D. 4x2 – 12x – 2 5. (5r-3)2 Step 1: Use the Difference of a Square rule. (a-b)2 = a2 – 2ab + b2 (5r-3)2 = (5r)2 – 2(5r)(3) + 32 (5r-3)2 = 25r2 – 30r +9 A. 25r2 – 9 C. 5r2 + 9 B. 25r2 – 30r + 9 D. 25r2 + 9 6. (x+6)(x-6) Step 1: Use the Difference of Two Square Rule: (a+b)(a-b) = a2 – b2 (x+6)(x-6) = x2 – 62 (x+6)(x-6) = x2 - 36 A. x2 + 12x – 36 C. x2 + 36 B. x2 – 36 D. x2 -6x – 36 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 7. (3x-2)(2x+1) Step 1: Using FOIL multiply the two binomials. (3x-2)(2x+1) = 3x(2x) + 3x(1) + (-2)(2x)+ (-2)(1) 6x2 + 3x – 4x – 2 Step 2: Combine like terms. 6x2 – x – 2 Final Answer: 6x2 – x - 2 A. 6x2 + 7x – 2 C. 6x2 – x + 2 B. 6x2 + x – 2 D. 6x2 – x – 2 8. (x+4)2(3x-2) Step 1: Using the Square of A sum rule, multiply (x+4)2 (x+4)2 = x2 + 8x + 16 (a2 + 2ab + b2) Step 2: Multiply (3x-2) (x2 + 8x + 16) Use the extended distributive property. 3x(x2) + 3x(8x) + 3x(16) + (-2)(x2) + (-2)(8x) + (-2)(16) 3x3 + 24x2 + 48x – 2x2 - 16x – 32 Step 3: Rewrite with like terms together. 3x3 + 24x2 – 2x2 + 48x – 16x – 32 Step 4: Combine like terms: 3x3 + 22x2 + 32x - 32 A. 3x3 + 22x2 +32x – 32 C. 3x2 + 10x – 8 B. 3x3 – 2x2 + 48x – 32 D. 3x3 + 26x2 + 48x – 32 Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring For problems 9-10, factor the polynomial. 9. x2 -3x -70 Since the lead coefficient is 1, we need to think of two numbers that have a product of -70 and a sum of -3. One integer must be positive and one must be negative. The larger integer will be negative since the sum is -3. **Since this is multiple choice – we should take a look at the answers given. The answers are using factors of 10 and 7 and 5 and 14. Of those two choices, let’s see which one works. ***I know that the larger integer must be negative, since the sum is -3; therefore, I can eliminate A because the larger integer, 10 is positive. For the same reason, I can eliminate D. Let’s try letter C. We’ll multiply: (x+7)(x-10) = x2 – 10x + 7x – 70 x2 – 3x – 70 This is the answer we need; therefore, letter C is the correct choice. A. (x+10)(x-7) C. (x+7)(x-10) B. (x+5)(x-14) D. (x+14)(x-5) 10. 6x2 – 2x – 8 Since this is multiply choice, we should use the answers to our advantage. It may be best to guess one of the answer choices below and check by multiplying. We would have to use the guess and check strategy anyway to factor this polynomial since the lead coefficient is greater than 1. A: 2(3x-4)(x+1) (6x – 8)(x+1) - Distribute the 2 throughout the first parenthesis 6x2 +6x – 8x -8 = 6x2 – 2x – 8 Our first try works!!! A. 2(3x-4)(x+1) B. 2(x-2) (3x+2) C. 2(x-1)(3x+4) D. 2(x+2)(3x-2) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Part 2: Short Answer (2 points each) 1. A square table measures 3x +2y units. Write an expression in simplified form that represents the area of the table top. A square table has four sides with the same measurement (3x+2y). Area of a square is: A = l● w. or A = s2 A = (3x+2y)2 We can use the Square of a Sum rule in order to multiply and simplify this expression. (a+b)2 = a2 + 2ab +b2 (3x+2y)2 = (3x)2 + 2(3x)(2y) + (2y)2 A = 9x2 + 12xy + 4y2 2. A rectangular sheet of paper has a width of x units. The area of the paper is x2+8x. Write an expression that represents the length of the rectangular sheet of paper. width= x length = ? Area = x2 + 8x The formula for area of a rectangle is: A = lw Substitute our given information into this formula: x2 + 8x = Lx Now we need to solve for L by dividing by x. x2 + 8x /x = l/x ****You may also, have solved this problem mentally by thinking: 2 x + 8x = L x An x can be factored out of the numerator: x(x+ 8) = L X Now we can simplify the fraction: x(x+ 8) = L X x + 8 = L We end up with an expression of x + 8 = L x(?) = x2 + 8x What can I multiply by x in order to get x2+ 8x? x(x +8) = x2 + 8x Length = x+8 Let’s check: A = lw A = (x+8)(x) A = x2 + 8x This expression matches the original problem; therefore, our answer is correct Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 3. Factor the polynomial completely. 2x2 + 11x + 12 Factors of 2: (2●1) Factors of 12: (12 ●1) (6●2) (3●4) Guess and check: We need the inside terms to add up to 11. (2x + 6)(x+2) = 2x2 + 4x + 6x + 12 = 2x2+ 10x + 12 (2x + 3)(x+4) = 2x2 + 8x + 3x + 12 = 2x2 + 11x + 12 Final Answer: (2x+3)(x+4) 4. Factor the polynomial completely. 9x3 +6x2 -48x First factor out the GCF which is 3x. 3x(3x2 + 2x – 16) Now factor: 3x2 + 2x – 16 Factors of 3: 3,1 Factors of -16: -4(4), -8(2) or -2(8) We must use the guess and check method: 3x (3x+8) (x-2) Let’s check: (3x+8) (x-2) = 3x2 – 6x + 8x – 16 = 3x2 + 2x – 16 (Yeah ! This worked) Final answer: 3x(3x+8)(x-2) 5. Factor the polynomial completely. y2 – 121 Since y2 and 121 are both perfect squares and there is no middle term, I recognize this product as a Difference of Two Squares. I know that 112 = 121. Therefore, my factors are: (y+11)(y-11) Final Answer: (y+11)(y-11) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring 6. A rectangle has the following dimensions: width: (2x + y) length: (5x – 2y) Express the area of the rectangle as a polynomial in standard form. The formula for area of a rectangle is: A = lw A = (2x+y)(5x-2y) Use FOIL to multiply. A = 2x(5x) +2x(-2y) + y(5x) + y(-2y) A = 10x2 – 4xy + 5xy – 2y2 A = 10x2 + xy – 2y2 Final Answer: 10x2 + xy – 2y2 7. A triangle has the following dimensions: Side 1: x2 +2x -1 Side 2: x2 + 2x -1 Side 3: 2x2 +x – 3 Write an expression that represents the perimeter of the triangle. To find the perimeter of the triangle, we must add all sides together. I will set it up vertically. x2 + 2x – 1 x2 + 2x – 1 + 2x2 + x – 3 4x2 + 5x – 5 The perimeter of the triangle is: 4x2 + 5x - 5 8. A rectangle has a length of 8x -4 and a width of 4x +3. The perimeter of the rectangle is 58 units. Find the width of the rectangle. The formula for perimeter of a rectangle is: P = 2L + 2w P = 2(8x-4) + 2(4x+3) P = 16x – 8 + 8x + 6 Distribute. P = 16x + 8x – 8 + 6 Write like terms together. P = 24x – 2 Expression that represents the perimeter. 58 = 24x – 2 Now substitute 58 for P since we know the perimeter is 58 units. 58 +2 = 24x -2 + 2 Add 2 to both sides 60 = 24x 60/24 = 24x/24 Divide by 24 on both sides. 2.5 = x This is the value for x 4(2.5) + 3 = 13 Substitute 2.5 for x into the expression for width (4x+3) The width is 13 units. Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring Part 3: Extended Response. (4 points) 1. You have a rectangular piece of wood in which you are making a carnival game. You plan to cut out 2 circles that children can use to try to throw a bean bag through at the carnival. Complete each bullet below given the following dimensions: Length of rectangular piece of wood: (5x+2) Width of rectangular piece of wood: (3x+1) Radius of each circle: (x+1) • Write an algebraic expression for the area of the rectangular piece of wood before the circles are cut. Area of a rectangle: A = lw A = (5x+2)(3x+1) A = 15x2 + 5x + 6x + 2 or A = 15x2 + 11x + 2 Use FOIL to multiply The Area of the Wood before circles are cut is: 15x2 + 11x + 2 • Write an algebraic expression for the area of 1 circle. The formula for area of a circle is: A = πr2 A = π (x+1)2 A = π (x2 + 2x+ 1) Use square of a sum rule to multiply the expression. The area of one circle is: π (x2 + 2x+ 1) or 3.14(x2 + 2x + 1) Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com Unit 9: Factoring • Write an algebraic expression for the area of the wood after both circles are cut out. Explain how you determined your answer. Step 1: We need to find the expression that represents the total area of both circles that will be cut out. Since the area of 1 circle is: 3.14(x2 + 2x + 1) , we need to multiply this by 2. 3.14(2)(x2 + 2x +1) or 6.28(x2 + 2x + 1) Step 2: Let’s distribute the 6.28 6.28x2 + 12.56x + 6.28 This represents the total area of both circles that will be cut out. Step 3: Subtract the area of the circles from the total area of the wood. 15x2 + 11x + 2 – (6.28x2 + 12.56x + 6.28) Step 4: Rewrite as an addition problem. 15x2 + 11x + 2 + (-6.28x2 - 12.56x - 6.28) Step 4: Add. Set up vertically. 15x2 + 11x + 2 + -6.28x2 – 12.56x – 6.28 8.72x2 – 1.56x - 4.28 The expression that represents the area of the board after two circles are cut is: 8.72x2 – 1.56x - 4.28. I first multiplied the expression that represents the area of the circle by 2 since there are two circles. Then I subtracted the expression that represents the area of two circles, from the area of the rectangle. This test is worth 30 points Copyright © 2009-2010 Karin Hutchinson – Algebra-class.com