Discussion Week 5 Angela Mercier 42. m2+23m+24 - I am using the \"ac method\" to solve this problem. -First, check for the GCF, the largest integer that is a factor of two or more integers. There is none. ax2 + bx +c - This is the form for the \"ac method.\" a= 1, b= 23, c= 24 - We want two numbers that multiply to give ac= 24, and add to give you b= 23. None of the possible factors of 24 have a sum of 23, so we can be certain that m 2+23m+24 cannot be factored. It is a prime polynomial. This polynomial is not a perfect square. The rule for factoring a perfect square trinomial is that a 2 + 2ab + b2. My problem did not fit into this form. The prime factors of the number 23 are 23 and 1. 96. 2a2+1+3a - I am using the \"ac method\" to solve this problem. ax2 + bx +c - This is the form for the \"ac method.\" a= 2, b= 3, c= 1 - We want two numbers that multiply to give ac= 2, and add to give you b= 3. 2= (2)(1) 2+1=3 - These are the integers that I will use, (2)(1). 2a2 + 2a + a + 1 - Replace 3a by 2a + a by (2a2 + 2a) + (a + 1) - Grouping like terms. 2a (a +1) + 1 (a+1) - Factor out the common factors. (a+1)(2a+1) - Final Answer! Check by FOIL. 54. 15x2 + 13x -2 - I am using the \"ac method\" to solve this problem. ax2 + bx + c - This is the form for the \"ac method.\" a= 15, b= 13, c= -2 - We want two numbers that multiply to give ac= -30, and add to give you b= 13. -30= (15)(-2) 15-2= 13 15x2 +15x - 2x -2 - These are the integers that I will use, (15)(-2). - Replace 13x by 15x-2x. (15x2 + 15x) + (-2x - 2 ) - Group like terms. 15x ( x+1) -2 ( x+1) - Factor out the common factors. ( x+1 )( 15x-2 ) - Final Answer! Check by FOIL. For the problems on pages 345 and 346, (#18 and 98) factor the polynomials using whatever strategy seems appropriate. State what methods you will use and then demonstrate the methods on your problems, explaining the process as you go. Discuss any particular challenges those particular polynomials posed for the factoring. For the problem on page 353 (#30 below) make sure you use the \"ac method\" regardless of what the book's directions say. Show the steps of this method in your work in a similar manner as how the book shows it in examples. Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.): o Factor o GCF o Prime factors o Perfect square o Grouping Your initial post should be 150250 words in length. Exercise 14 - The ac Method: Leading Coefficient Only Has Two Factors #18 Factoring Completely: GCF Has a Variable and Further Factors #98 30w2 + w + 1 The ac Method: Leading Coefficient Only Has Two Factors #30 INSTRUCTOR GUIDANCE EXAMPLE: Week Five Discussion Factoring Since there are several different types of factoring problems assigned from pages 345346, four types will be demonstrated here to offer a selection, even though individual students will only be working two from these pages. #73. x3 - 2x2 - 9x + 18 x2(x - 2) - 9(x - 2) (x - 2)(x2 - 9) (x - 2)(x - 3)(x -+ 3) Four terms means start with grouping The common factor for each group is (x - 2) Notice the difference of squares in second group Now it is completely factored. #81. 6w2 - 12w - 18 6(w2 - 2w - 3) Every term has a GCF of 6 Common factor is removed, now have a trinomial Need two numbers that add to -2 but multiply to -3 Try with -3 and +1 This works, check by multiplying it back together 6(w - 3)(w + 1) #97. 8vw2 + 32vw + 32v 8v(w2 + 4w + 4) 8v(w + 2)(w + 2) 8v(w + 2)2 Every term has a GCF of 8v The trinomial is in the form of a perfect square Showing the squared binomial Writing the square appropriately #103. -3y3 + 6y2 - 3y -3y(y2 - 2y + 1) -3y(y - 1)(y - 1) -3y(y - 1)2 Every term has a GCF of -3y Another perfect square trinomial Showing the squared binomial Writing the square appropriately Here are two examples of problems similar to those assigned from page 353. 5b2 - 13b + 6 a = 5 and c = 6, so ac = 5(6) = 30. The factor pairs of 30 are 1, 30 2, 15 3, 10 5,6 5b2 - 3b - 10b + 6 b(5b - 3) - 2(5b - 3) (5b - 3)( b - 2) Now factor by grouping. The common binomial factor is (5b - 3). Check by multiplying it back together. -3(-10)=30 while -3+(-10)= -13 so replace -13b by -3b and -10b 3x2 + x - 14 3x2 - 6x + 7x - 14 3x(x - 2) + 7(x - 2) (x - 2)(3x + 7) a = 3 and c = -14, so ac =3(-14)= -42. The factor pairs of - 42 are 1, -42 -1, 42 3, -14 -3, 14 2, -21 -2, 21 6, -7 -6, 7 We see that -6(7) = -42 while -6 + 7 = 1 so replace x with -6x + 7x. Factor by grouping. The common binomial factor is (x - 2). Check by multiplying it back together. b28b + 15 For the problems on pages 345 and 346, (#18 and 98) factor the polynomials using whatever strategy seems appropriate. State what methods you will use and then demonstrate the methods on your problems, explaining the process as you go. Discuss any particular challenges those particular polynomials posed for the factoring. For the problem on page 353 (#30 below) make sure you use the \"ac method\" regardless of what the book's directions say. Show the steps of this method in your work in a similar manner as how the book shows it in examples. Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.): o Factor o GCF o Prime factors o Perfect square o Grouping Your initial post should be 150-250 words in length. Exercise 14 - The ac Method: Leading Coefficient Only Has Two Factors #18 b28b + 15 Revised problem to work ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = 15 and add to give b = 8 3*5 = 15 and 3+5 = 8 I will use integer 3 and 5 to factor, here 3 and 5 are prime factors I will replace 8t by 3t + 5t 3t2 + 3t + 5t + 5 (3t2+3t)+(5t+5) grouping like terms 3t(t+1)+5(t+1) factor out the common factor (3t+5)(t+1) final answer Check by multiplying back together to get the initial equation Factoring Completely: GCF Has a Variable and Further Factors #98 REVISED PROBLEM TO WORK 30w2 + w + 1 ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 first term is -30 which is not a perfect square now, we have to find two numbers that multiplied to give ac = -30 and add to give b = 1 -5*6 = -30 and -5+6 = 1 I will use integer -5 and 6 to factor I will replace w by -5w+6w -30w2 -5w+6w+1 (-30w2-5w)+(6w+1) grouping like terms -5w(6w+1)+1(6w+1) factor out the common factor (-5w+1)(6w+1) final answer Check by multiplying back together to get the initial equation The ac Method: Leading Coefficient Only Has Two Factors #30 ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = -35 and add to give b = -2 -7*5 = -35 and -7+5 = -2 I will use integer -7 and 5 to factor I will replace -2x by -7x + 5x 35x2 -7x+5x - 1 (35x2-7x)+(5x-1) grouping like terms 7x(5x-1)+1(5x-1) factor out the common factor (7x+1)(5x-1) final answer Check by multiplying back together to get the initial equation b28b + 15 For the problems on pages 345 and 346, (#18 and 98) factor the polynomials using whatever strategy seems appropriate. State what methods you will use and then demonstrate the methods on your problems, explaining the process as you go. Discuss any particular challenges those particular polynomials posed for the factoring. For the problem on page 353 (#30 below) make sure you use the \"ac method\" regardless of what the book's directions say. Show the steps of this method in your work in a similar manner as how the book shows it in examples. Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.): o Factor o GCF o Prime factors o Perfect square o Grouping Your initial post should be 150-250 words in length. Exercise 14 - The ac Method: Leading Coefficient Only Has Two Factors #18 b28b + 15 Revised problem to work ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = 15 and add to give b = 8 3*5 = 15 and 3+5 = 8 I will use integer 3 and 5 to factor, here 3 and 5 are prime factors I will replace 8t by 3t + 5t 3t2 + 3t + 5t + 5 (3t2+3t)+(5t+5) grouping like terms 3t(t+1)+5(t+1) factor out the common factor (3t+5)(t+1) final answer Check by multiplying back together to get the initial equation Factoring Completely: GCF Has a Variable and Further Factors #98 REVISED PROBLEM TO WORK 30w2 + w + 1 ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 first term is -30 which is not a perfect square now, we have to find two numbers that multiplied to give ac = -30 and add to give b = 1 -5*6 = -30 and -5+6 = 1 I will use integer -5 and 6 to factor I will replace w by -5w+6w -30w2 -5w+6w+1 (-30w2-5w)+(6w+1) grouping like terms -5w(6w+1)+1(6w+1) factor out the common factor (-5w+1)(6w+1) final answer Check by multiplying back together to get the initial equation The ac Method: Leading Coefficient Only Has Two Factors #30 ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = -35 and add to give b = -2 -7*5 = -35 and -7+5 = -2 I will use integer -7 and 5 to factor I will replace -2x by -7x + 5x 35x2 -7x+5x - 1 (35x2-7x)+(5x-1) grouping like terms 7x(5x-1)+1(5x-1) factor out the common factor (7x+1)(5x-1) final answer Check by multiplying back together to get the initial equation b28b + 15 For the problems on pages 345 and 346, (#18 and 98) factor the polynomials using whatever strategy seems appropriate. State what methods you will use and then demonstrate the methods on your problems, explaining the process as you go. Discuss any particular challenges those particular polynomials posed for the factoring. For the problem on page 353 (#30 below) make sure you use the \"ac method\" regardless of what the book's directions say. Show the steps of this method in your work in a similar manner as how the book shows it in examples. Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.): o Factor o GCF o Prime factors o Perfect square o Grouping Your initial post should be 150-250 words in length. Exercise 14 - The ac Method: Leading Coefficient Only Has Two Factors #18 b28b + 15 Revised problem to work ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = 15 and add to give b = -8 -3*(-5) = 15 and -3-5 = -8 I will use integer -3 and -5 to factor, here -3 and -5 are prime factors I will replace -8b by -3b - 5b b2 -3b -5b + 15 (b2-3b)+(-5b+15) grouping like terms b(b-3) - 5(b-3) factor out the common factor (b-5)(b-3) final answer Check by multiplying back together to get the initial equation Factoring Completely: GCF Has a Variable and Further Factors #98 REVISED PROBLEM TO WORK 3h2t + 6ht +3t ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 3t, we can write the equation by taking GCF common 3t(h2 + 2h +1) first term is 1 which is a perfect square now, we have to find two numbers that multiplied to give ac = 1 and add to give b = 2 1*1 = 1 and 1+1 = 2 I will use integer 1 and 1 to factor I will replace 2h by 1h+1h 3t(h2 + 1h + 1h +1) 3t(h2 + 1h) + (1h +1) grouping like terms (3t)h(h + 1) + 1(1h +1) factor out the common factor 3t(h+1)(h+1) final answer Check by multiplying back together to get the initial equation The ac Method: Leading Coefficient Only Has Two Factors #30 ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = -35 and add to give b = -2 -7*5 = -35 and -7+5 = -2 I will use integer -7 and 5 to factor I will replace -2x by -7x + 5x 35x2 -7x+5x - 1 (35x2-7x)+(5x-1) grouping like terms 7x(5x-1)+1(5x-1) factor out the common factor (7x+1)(5x-1) final answer Check by multiplying back together to get the initial equation b28b + 15 For the problems on pages 345 and 346, (#18 and 98) factor the polynomials using whatever strategy seems appropriate. State what methods you will use and then demonstrate the methods on your problems, explaining the process as you go. Discuss any particular challenges those particular polynomials posed for the factoring. For the problem on page 353 (#30 below) make sure you use the \"ac method\" regardless of what the book's directions say. Show the steps of this method in your work in a similar manner as how the book shows it in examples. Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.): o Factor o GCF o Prime factors o Perfect square o Grouping Your initial post should be 150-250 words in length. Exercise 14 - The ac Method: Leading Coefficient Only Has Two Factors #18 b28b + 15 Revised problem to work ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = 15 and add to give b = 8 3*5 = 15 and 3+5 = 8 I will use integer 3 and 5 to factor, here 3 and 5 are prime factors I will replace 8t by 3t + 5t 3t2 + 3t + 5t + 5 (3t2+3t)+(5t+5) grouping like terms 3t(t+1)+5(t+1) factor out the common factor (3t+5)(t+1) final answer Check by multiplying back together to get the initial equation Factoring Completely: GCF Has a Variable and Further Factors #98 REVISED PROBLEM TO WORK 30w2 + w + 1 ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 first term is -30 which is not a perfect square now, we have to find two numbers that multiplied to give ac = -30 and add to give b = 1 -5*6 = -30 and -5+6 = 1 I will use integer -5 and 6 to factor I will replace w by -5w+6w -30w2 -5w+6w+1 (-30w2-5w)+(6w+1) grouping like terms -5w(6w+1)+1(6w+1) factor out the common factor (-5w+1)(6w+1) final answer Check by multiplying back together to get the initial equation The ac Method: Leading Coefficient Only Has Two Factors #30 ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = -35 and add to give b = -2 -7*5 = -35 and -7+5 = -2 I will use integer -7 and 5 to factor I will replace -2x by -7x + 5x 35x2 -7x+5x - 1 (35x2-7x)+(5x-1) grouping like terms 7x(5x-1)+1(5x-1) factor out the common factor (7x+1)(5x-1) final answer Check by multiplying back together to get the initial equation b28b + 15 For the problems on pages 345 and 346, (#18 and 98) factor the polynomials using whatever strategy seems appropriate. State what methods you will use and then demonstrate the methods on your problems, explaining the process as you go. Discuss any particular challenges those particular polynomials posed for the factoring. For the problem on page 353 (#30 below) make sure you use the \"ac method\" regardless of what the book's directions say. Show the steps of this method in your work in a similar manner as how the book shows it in examples. Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.): o Factor o GCF o Prime factors o Perfect square o Grouping Your initial post should be 150-250 words in length. Exercise 14 - The ac Method: Leading Coefficient Only Has Two Factors #18 b28b + 15 Revised problem to work ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = 15 and add to give b = 8 3*5 = 15 and 3+5 = 8 I will use integer 3 and 5 to factor, here 3 and 5 are prime factors I will replace 8t by 3t + 5t 3t2 + 3t + 5t + 5 (3t2+3t)+(5t+5) grouping like terms 3t(t+1)+5(t+1) factor out the common factor (3t+5)(t+1) final answer Check by multiplying back together to get the initial equation Factoring Completely: GCF Has a Variable and Further Factors #98 REVISED PROBLEM TO WORK 30w2 + w + 1 ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 first term is -30 which is not a perfect square now, we have to find two numbers that multiplied to give ac = -30 and add to give b = 1 -5*6 = -30 and -5+6 = 1 I will use integer -5 and 6 to factor I will replace w by -5w+6w -30w2 -5w+6w+1 (-30w2-5w)+(6w+1) grouping like terms -5w(6w+1)+1(6w+1) factor out the common factor (-5w+1)(6w+1) final answer Check by multiplying back together to get the initial equation The ac Method: Leading Coefficient Only Has Two Factors #30 ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = -35 and add to give b = -2 -7*5 = -35 and -7+5 = -2 I will use integer -7 and 5 to factor I will replace -2x by -7x + 5x 35x2 -7x+5x - 1 (35x2-7x)+(5x-1) grouping like terms 7x(5x-1)+1(5x-1) factor out the common factor (7x+1)(5x-1) final answer Check by multiplying back together to get the initial equation b28b + 15 For the problems on pages 345 and 346, (#18 and 98) factor the polynomials using whatever strategy seems appropriate. State what methods you will use and then demonstrate the methods on your problems, explaining the process as you go. Discuss any particular challenges those particular polynomials posed for the factoring. For the problem on page 353 (#30 below) make sure you use the \"ac method\" regardless of what the book's directions say. Show the steps of this method in your work in a similar manner as how the book shows it in examples. Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.): o Factor o GCF o Prime factors o Perfect square o Grouping Your initial post should be 150-250 words in length. Exercise 14 - The ac Method: Leading Coefficient Only Has Two Factors #18 b28b + 15 Revised problem to work ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = 15 and add to give b = -8 -3*(-5) = 15 and -3-5 = -8 I will use integer -3 and -5 to factor, here -3 and -5 are prime factors I will replace -8b by -3b - 5b b2 -3b -5b + 15 (b2-3b)+(-5b+15) grouping like terms b(b-3) - 5(b-3) factor out the common factor (b-5)(b-3) final answer Check by multiplying back together to get the initial equation Factoring Completely: GCF Has a Variable and Further Factors #98 REVISED PROBLEM TO WORK 3h2t + 6ht +3t ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 3t, we can write the equation by taking GCF common 3t(h2 + 2h +1) first term is 1 which is a perfect square now, we have to find two numbers that multiplied to give ac = 1 and add to give b = 2 1*1 = 1 and 1+1 = 2 I will use integer 1 and 1 to factor I will replace 2h by 1h+1h 3t(h2 + 1h + 1h +1) 3t(h2 + 1h) + (1h +1) grouping like terms (3t)h(h + 1) + 1(1h +1) factor out the common factor 3t(h+1)(h+1) final answer Check by multiplying back together to get the initial equation The ac Method: Leading Coefficient Only Has Two Factors #30 ax2 + bx + c This is the form of \"ac method\" Every term has a GCF of 1 now, we have to find two numbers that multiplied to give ac = -35 and add to give b = -2 -7*5 = -35 and -7+5 = -2 I will use integer -7 and 5 to factor I will replace -2x by -7x + 5x 35x2 -7x+5x - 1 (35x2-7x)+(5x-1) grouping like terms 7x(5x-1)+1(5x-1) factor out the common factor (7x+1)(5x-1) final answer Check by multiplying back together to get the initial equation