# FACTORING ax 2 bx c WITH a 1

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1 296 (6 20) Chapter 6 Factoring 6.4 FACTORING a 2 b c WITH a 1 In this section The ac Method Trial and Error Factoring Completely In Section 6.3 we factored trinomials with a leading coefficient of 1. In this section we will use a slightly different technique to factor trinomials with leading coefficients not equal to 1. The ac Method If the leading coefficient of a trinomial is not 1, we can again use grouping to factor the trinomial. However, the procedure is slightly different. Consider the trinomial First find the product of the leading coefficient and the constant term. In this case it is 2 6, or 12. Now find two numbers with a product of 12 and a sum of 7. The pairs of numbers with a product of 12 are 1 and 12, 2 and 6, and 3 and 4. Only 3 and 4 have a product of 12 and a sum of 7. Now replace 7 by 3 4 and factor by grouping: Replace 7 by 3 4. (2 3) (2 3)2 Factor out the common factors. (2 3)( 2) Factor out 2 3. The strategy for factoring a trinomial is summarized in the following bo. The steps listed here actually work whether or not the leading coefficient is 1. This is actually the method that you learned in Section 6.3 with a 1. This method is called the ac method. Strategy for Factoring a 2 b c by the ac Method To factor the trinomial a 2 b c: 1. Find two numbers that have a product equal to ac and a sum equal to b. 2. Replace b by two terms using the two new numbers as coefficients. 3. Factor the resulting four-term polynomial by grouping. E X A M P L E 1 The ac method Factor each trinomial. a) b) a) Because 2 ( 6) 12, we need two integers with a product of 12 and a sum of 1. We can list the possible pairs of integers with a product of 12: 1 and 12 2 and 6 3 and 4 1 and 12 2 and 6 3 and4 Only 3 and 4 have a sum of 1. Replace by 3 4 and factor by grouping: Replace by 3 4. (2 3) (2 3)2 Factor out the common factors. (2 3)( 2) Factor out 2 3. Check by FOIL.

2 6.4 Factoring a 2 b c with a 1 (6 21) 297 b) Because 10 ( 3) 30, we need two integers with a product of 30 and a sum of 13. The product is negative, so the integers must have opposite signs. We can list all pairs of factors of 30 as follows: 1 and 30 2 and 15 3 and 10 5 and 6 1 and 30 2 and 15 3 and 10 5 and 6 The only pair that has a sum of 13 is 2 and 15: Replace 13 by (5 1)2 (5 1)3 Factor out the common factors. (5 1)(2 3) Factor out 5 1. Check by FOIL. helpful hint The ac method is more systematic than trial and error. However, trial and error can be faster and easier, especially if your first or second trial is correct. Trial and Error After you have gained some eperience at factoring by the ac method, you can often find the factors without going through the steps of grouping. For eample, consider the polynomial The factors of 3 2 can only be 3 and. The factors of 6 could be 2 and 3 or 1 and 6. We can list all of the possibilities that give the correct first and last terms, without regard to the signs: (3 3)( 2) (3 2)( 3) (3 6)( 1) (3 1)( 6) Because the factors of 6 have unlike signs, one binomial factor is a sum and the other binomial is a difference. Now we try some products to see if we get a middle term of 7: (3 3)( 2) Incorrect. (3 3)( 2) Incorrect. Actually, there is no need to try (3 3)( 2) or (3 6)( 1) because each contains a binomial with a common factor. As you can see from the above products, a common factor in the binomial causes a common factor in the product. But has no common factor. So the factors must come from either (3 2)( 3) or (3 1)( 6). So we try again: (3 2)( 3) Incorrect. (3 2)( 3) Correct. Even though there may be many possibilities in some factoring problems, it is often possible to find the correct factors without writing down every possibility. We can use a bit of guesswork in factoring trinomials. Try whichever possibility you think might work. Check it by multiplying. If it is not right, then try again. That is why this method is called trial and error. E X A M P L E 2 Trial and error Factor each trinomial using trial and error. a) b)

3 298 (6 22) Chapter 6 Factoring study tip Have you ever used ecuses to avoid studying? ( Before I can study, I have to do my laundry and go to the bank. ) Since the average attention span for one task is approimately 20 minutes, it is better to take breaks from studying to run errands and do laundry than to get everything done before you start studying. a) Because 2 2 factors only as 2 and 3 factors only as 1 3, there are only two possible ways to get the correct first and last terms, without regard to the signs: (2 1)( 3) and (2 3)( 1) Because the last term of the trinomial is negative, one of the missing signs must be, and the other must be. The trinomial is factored correctly as (2 1)( 3). Check by using FOIL. b) There are four possible ways to factor : (3 1)( 6) (3 2)( 3) (3 6)( 1) (3 3)( 2) Because the last term in is positive and the middle term is negative, both signs in the factors must be negative. Because has no common factor, we can rule out (3 6)( 1) and (3 3)( 2). So the only possibilities left are (3 1)( 6) and (3 2)( 3). The trinomial is factored correctly as (3 2)( 3). Check by using FOIL. Factoring by trial and error is not just guessing. In fact, if the trinomial has a positive leading coefficient, we can determine in advance whether its factors are sums or differences. Using Signs in Trial and Error 1. If the signs of the terms of a trinomial are, then both factors are sums: ( 2)( 3). 2. If the signs are, then both factors are differences: ( 2)( 3). 3. If the signs are or, then one factor is a sum and the other is a difference: 2 6 ( 3)( 2) and 2 6 ( 3)( 2). In the net eample we factor a trinomial that has two variables. E X A M P L E 3 Factoring a trinomial with two variables by trial and error Factor 6 2 7y 2y 2. We list the possible ways to factor the trinomial: (3 2y)(2 y) (3 y)(2 2y) (6 2y)( y) (6 y)( 2y) Because the last term of the trinomial is positive and the middle term is negative, both factors must contain subtraction symbols. To get the middle term of 7y, we use the first possibility listed: 6 2 7y 2y 2 (3 2y)(2 y) Factoring Completely You can use the latest factoring technique along with the techniques that you learned earlier to factor polynomials completely. Remember always to first factor out the greatest common factor (if it is not 1).

4 6.4 Factoring a 2 b c with a 1 (6 23) 299 E X A M P L E 4 Factoring completely Factor each polynomial completely. a) b) 12 2 y 6y 6y a) ( ) Factor out the GCF, 2. 2(2 1)( 3) Factor Check by multiplying. b) 12 2 y 6y 6y 6y(2 2 1) Factor out the GCF, 6y. To factor by the ac method, we need two numbers with a product of 2 and a sum of 1. Because there are no such numbers, is prime and the factorization is complete. Usually, our first step in factoring is to factor out the greatest common factor (if it is not 1). If the first term of a polynomial has a negative coefficient, then it is better to factor out the opposite of the GCF so that the resulting polynomial will have a positive leading coefficient. E X A M P L E 5 Factoring out the opposite of the GCF Factor each polynomial completely. a) b) 3a 2 2a 21 a) The GCF is 3. Because the first term has a negative coefficient, we factor out 3: ( ) Factor out 3. 3(3 1)(2 5) Factor b) The GCF for 3a 2 2a 21 is 1. Because the first term has a negative coefficient, factor out 1: 3a 2 2a 21 1(3a 2 2a 21) Factor out 1. 1(3a 7)(a 3) Factor 3a 2 2a 21. WARM-UPS True or false? Answer true if the correct factorization is given and false if the factorization is incorrect. Eplain your answer (2 1)( 1) (2 1)( 3) (3 1)( 3) (3 7)(5 2) (2 9)( 1) (2 3)( 3) (2 9)(2 1) (4 1)(2 5) (5 1)(4 1) (3 1)(4 3)

5 300 (6 24) Chapter 6 Factoring 6.4 EXERCISES Reading and Writing After reading this section, write out the answers to these questions. Use complete sentences. 1. What types of polynomials did we factor in this section? 2. What is the ac method of factoring? t 2 9t t 2 5t How can you determine if a 2 b c is prime? 4. What is the trial-and-error method of factoring? Find the following. See Eample Two integers that have a product of 20 and a sum of Two integers that have a product of 36 and a sum of Two integers that have a product of 12 and a sum of 4 8. Two integers that have a product of 8 and a sum of 7 Each of the following trinomials is in the form a 2 b c. For each trinomial, find two integers that have a product of ac and a sum of b. Do not factor the trinomials. See Eample y 2 11y z 2 19z w 2 w t 2 17t 4 Factor each trinomial using the ac method. See Eample h 2 7h t 2 7t t 2 8t a 2 14a b 2 13b y 2 16y y 2 11y Complete the factoring ( 2)( ) ( 3)( ) (5 1)( ) (4 1)( ) 45. 6a 2 17a 5 (3a 1)( ) 46. 4b 2 16b 15 (2b 5)( ) Factor each trinomial using trial and error. See Eamples 2 and a 2 11a y 2 10y w 2 8w z 2 13z y y y y a 2 13ab b a 2 27ab 5b 2 Factor each polynomial completely. See Eamples 4 and w 3 w w 3 w w 2 2w y 12y w 2 11w z 3z 18z 74. a 2 b 2ab 15b

6 6.4 Factoring a 2 b c with a 1 (6 25) y 2 y 2 9y y 2 y 2 3y a 2 2ab 15b a 2 b 2 2a 2 b 15a t 3 t 2 2t t 2 6t t 4 2t 3 4t t 3 14t 2 4t y 8y 2 3y y 9y w 2 7w w 2 w a 3 22a 2 b 6ab a 2 b 21ab 2 3b 3 Solve each problem. 89. Height of a ball. If a ball is thrown upward at 40 feet per second from a rooftop 24 feet above the ground, then its height above the ground t seconds after it is thrown is given by h 16t 2 40t 24. Rewrite this formula with the polynomial on the right-hand side factored completely. Use the factored version of the formula to find h when t ft/sec h 16t 2 40t 24 FIGURE FOR EXERCISE Worker efficiency. In a study of worker efficiency at Wong Laboratories it was found that the number of components assembled per hour by the average worker t hours after starting work could be modeled by the function N(t) 3t 3 23t 2 8t. a) Rewrite the formula by factoring the right-hand side completely. b) Use the factored version of the formula to find N(3). c) Use the accompanying graph to estimate the time at which the workers are most efficient. Number of components Time (hours) FIGURE FOR EXERCISE 90 d) Use the accompanying graph to estimate the maimum number of components assembled per hour during an 8-hour shift. GETTING MORE INVOLVED 91. Eploration. Find all positive and negative integers b for which each polynomial can be factored. a) 2 b 3 b) 3 2 b 5 c) 2 2 b Eploration. Find two integers c (positive or negative) for which each polynomial can be factored. Many answers are possible. a) 2 c b) 2 2 c c) c 93. Cooperative learning. Working in groups, cut two large squares, three rectangles, and one small square out of paper that are eactly the same size as shown in the accompanying figure. Then try to place the si figures net to one another so that they form a large rectangle. Do not overlap the pieces or leave any gaps. Eplain how factoring can help you solve this puzzle. FIGURE FOR EXERCISE Cooperative learning. Working in groups, cut four squares and eight rectangles out of paper as in the previous eercise to illustrate the trinomial Select one group to demonstrate how to arrange the 12 pieces to form a large rectangle. Have another group eplain how factoring the trinomial can help you solve this puzzle. 1

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### 4.4 Factoring ax 2 + bx + c 4.4 Factoring ax 2 + bx + c From the last section, we now know a trinomial should factor as two binomials. With this in mind, we need to look at how to factor a trinomial when the leading coefficient is

### Big Bend Community College. Beginning Algebra MPC 095. Lab Notebook Big Bend Community College Beginning Algebra MPC 095 Lab Notebook Beginning Algebra Lab Notebook by Tyler Wallace is licensed under a Creative Commons Attribution 3.0 Unported License. Permissions beyond 250) 960-6367 Factoring Polynomials Sometimes when we try to solve or simplify an equation or expression involving polynomials the way that it looks can hinder our progress in finding a solution. Factorization

### 2.4. Factoring Quadratic Expressions. Goal. Explore 2.4. Launch 2.4 2.4 Factoring Quadratic Epressions Goal Use the area model and Distributive Property to rewrite an epression that is in epanded form into an equivalent epression in factored form The area of a rectangle

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### Factoring Special Polynomials 6.6 Factoring Special Polynomials 6.6 OBJECTIVES 1. Factor the difference of two squares 2. Factor the sum or difference of two cubes In this section, we will look at several special polynomials. These

### called and explain why it cannot be factored with algebra tiles? and explain why it cannot be factored with algebra tiles? Factoring Reporting Category Topic Expressions and Operations Factoring polynomials Primary SOL A.2c The student will perform operations on polynomials, including factoring completely first- and second-degree

### 8-8 Differences of Squares. Factor each polynomial. 1. x 9 SOLUTION: 2. 4a 25 SOLUTION: 3. 9m 144 SOLUTION: 4. 2p 162p SOLUTION: 5. Factor each polynomial. 1.x 9 SOLUTION:.a 5 SOLUTION:.9m 1 SOLUTION:.p 16p SOLUTION: 5.u 81 SOLUTION: Page 1 5.u 81 SOLUTION: 6.d f SOLUTION: 7.0r 5n SOLUTION: 8.56n c SOLUTION: Page 8.56n c SOLUTION:

### SPECIAL PRODUCTS AND FACTORS CHAPTER 442 11 CHAPTER TABLE OF CONTENTS 11-1 Factors and Factoring 11-2 Common Monomial Factors 11-3 The Square of a Monomial 11-4 Multiplying the Sum and the Difference of Two Terms 11-5 Factoring the

### Factoring. 472 Chapter 9 Factoring Factoring Lesson 9- Find the prime factorizations of integers and monomials. Lesson 9- Find the greatest common factors (GCF) for sets of integers and monomials. Lessons 9-2 through 9-6 Factor polynomials.

### Polynomial and Synthetic Division. Long Division of Polynomials. Example 1. 6x 2 7x 2 x 2) 19x 2 16x 4 6x3 12x 2 7x 2 16x 7x 2 14x. 2x 4. _.qd /7/5 9: AM Page 5 Section.. Polynomial and Synthetic Division 5 Polynomial and Synthetic Division What you should learn Use long division to divide polynomials by other polynomials. Use synthetic

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### SOLVING QUADRATIC EQUATIONS BY THE NEW TRANSFORMING METHOD (By Nghi H Nguyen Updated Oct 28, 2014)) SOLVING QUADRATIC EQUATIONS BY THE NEW TRANSFORMING METHOD (By Nghi H Nguyen Updated Oct 28, 2014)) There are so far 8 most common methods to solve quadratic equations in standard form ax² + bx + c = 0. Factoring A Quadratic Polynomial If we multiply two binomials together, the result is a quadratic polynomial: This multiplication is pretty straightforward, using the distributive property of multiplication

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### Tool 1. Greatest Common Factor (GCF) Chapter 4: Factoring Review Tool 1 Greatest Common Factor (GCF) This is a very important tool. You must try to factor out the GCF first in every problem. Some problems do not have a GCF but many do. When

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### AIP Factoring Practice/Help The following pages include many problems to practice factoring skills. There are also several activities with examples to help you with factoring if you feel like you are not proficient with it. There

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### Factoring Trinomials of the Form Section 4 6B: Factoring Trinomials of the Form A x 2 + Bx + C where A > 1 by The AC and Factor By Grouping Method Easy Trinomials: 1 x 2 + Bx + C The last section covered the topic of factoring second SECTION 11.3 Solving Quadratic Equations b Graphing 11.3 OBJECTIVES 1. Find an ais of smmetr 2. Find a verte 3. Graph a parabola 4. Solve quadratic equations b graphing 5. Solve an application involving Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}