# Sect Greatest Common Factor and Factoring by Grouping

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1 Sect Greatest Common Factor and Factoring by Grouping Our goal in this chapter is to solve non-linear equations by breaking them down into a series of linear equations that we can solve. To do this, we factor the non-linear polynomial. We might be given a problem like this: Solve: 18x x 2 32x 80 = 0 We factor 18x x 2 32x 80 by the techniques developed in this chapter to get: (3x 4)(3x + 4)(2x + 5) = 0 If we know that a product is zero, then at least one of the factors has to be zero. So, we can break the problem into: (3x 4) = 0 (3x + 4) = 0 and (2x + 5) = 0 and solve 3x = 4 3x = 4 2x = 5 x = 4 x = 4 and x = Thus, this problem has three solutions: { 5 2, 4 3, 4 3 } Factoring is the reverse of what we did in chapter five; it helps us write a sum as a product which in turn, will allow us to solve equations like the one above. Concepts #1 - #3: Factoring out the G.C.F. The Greatest Common Factor (G.C.F.) is the largest factor that can divide into the terms of an expression evenly with no remainder. When we factor out the G.C.F., we are using the distributive property backwards to undistribute the G.C.F. For instance, 12x 12y has a common factor of 12, so we can rewrite 12x 12y as 12(x y). Factor out the G.C.F. Ex. 1a) 9x 9y Ex. 1b) 7a 5ax Ex. 1c) 27x + 27 Ex. 1d) 8x 16 Ex. 1e) 36x 24y a) The G.C.F. = 9, so 9x 9y = 9(x y). b) The G.C.F. = a, so 7a 5ax = a(7 5x). c) Since the first term is negative, we will factor out a negative with the G.C.F. The G.C.F. = 27, so 27x + 27 = 27(x 1). d) The G.C.F. = 8, so 8x 16 = 8(x 2). e) The G.C.F. = 12, so 36x 24y = 12(3x + 2y). 87

2 To see how to find the G.C.F. of variables with powers, let s take a closer look at example #1 d and e. Let s write 8 as 2 3, 16 as 2 4, 36 as , 24 as 2 3 3, and 12 as So, the problems then would look like: 8x 16 = 8(x 2) 36x 24y = 12(3x + 2y) 2 3 x 2 4 = 2 3 (x 2) x 2 3 3y = 2 2 3(3x + 2y) Notice that the two with Notice that the factors of 2 and 3 the lower power was the with the lowest powers were used G.C.F. for the G.C.F. So, to find the G.C.F. of variables, list the variables that appear in all the terms and then choose the lowest power of each respective variables as the powers for the G.C.F. Factor out the G.C.F. Ex. 2a x 2 xy Ex. 2b x 3 y 6 x 4 y 5 Since x is the only variable Both terms have x and y in and the lowest power is one, common. The lowest power then the G.C.F. = x. Thus, of x is 3 and of y is 5, so the x 2 xy = x(x y) G.C.F. = x 3 y 5. Thus, x 3 y 6 x 4 y 5 = x 3 y 5 (y x). Ex. 3 8x 2 y 3 12xy 4 The G.C.F. of 8 and 12 is four. Both terms have x and y in common. The lowest power of x is 1 and of y is 3. So, the G.C.F. = 4xy 3. Thus, 8x 2 y 3 12xy 4 = 4xy 3 (2x 3y) Ex. 4 9xy x 2 y The G.C.F. = 9xy. So, 9xy x 2 y = 9xy(y 3x) Ex. 5 3axy 2 2ax 2 yz + ax 2 y 2 z 3 The G.C.F. = axy. Notice we did not include z since there is no z in the first term. Thus, 3axy 2 2ax 2 yz + ax 2 y 2 z 3 = axy(3y 2xz + xyz 3 ) Ex. 6 7x 3 y 4 14x 4 y x 4 y 3 The G.C.F = 7x 3 y 3. So, 7x 3 y 4 14x 4 y x 4 y 3 = 7x 3 y 3 (y 2xy + 3x) 88

3 89 Concept #4 Factoring out a Binomial Factor Ex. 7 3x(2x 5) + 4(2x 5) The binomial (2x 5) is the G.C.F., so factor it out the same way. 3x(2x 5) + 4(2x 5) = (2x 5)(3x + 4) Ex. 8 12x(4x 3) 16y(4x 3) The G.C.F. = 4(4x 3), so 12x(4x 3) 16y(4x 3) = 4(4x 3)(3x 4y). Ex. 9 6x(2x + 7) (2x + 7) 2 The G.C.F. = (2x + 7). Thus, 6x(2x + 7) (2x + 7) 2 = (2x + 7)(6x (2x + 7)) Now, simplify inside the parenthesis. = (2x + 7)(6x 2x 7) = (2x + 7)(4x 7) Ex (x 3) 3 (y + 2) (x 3) 2 (y + 2) 3 G.C.F. = 7(x 3) 2 (y + 2) 2, so 21(x 3) 3 (y + 2) (x 3) 2 (y + 2) 3 = 7(x 3) 2 (y + 2) 2 [3(x 3) 4(y + 2)] Now, simplify inside the brackets. = 7(x 3) 2 (y + 2) 2 [3x 9 4y 8] = 7(x 3) 2 (y + 2) 2 [3x 4y 17] Concept #5 Factoring by Grouping Whenever a factoring problem has four or more terms, we put the terms into groups and factor out the G.C.F. of each group. We then see if the groupings have any pieces in common. If so, we then factor out the common pieces. Let s look at some examples. Factor the following by grouping: Ex. 11 my + 6y 5m 30 Let s put the first two terms in one grouping and the last two terms in the other grouping: my + 6y 5m 30 = my + 6y + 5m 30 = (my + 6y )+ ( 5m 30) The G.C.F. of the first grouping is y and G.C.F. of the second

4 grouping is 5, so (my + 6y )+ ( 5m 30) = y(m + 6) 5(m + 6) Now, the two pieces have (m + 6) in common, so we will factor that out: y(m + 6) 5(m + 6) = (m + 6)(y 5) Always be sure to factor out the G.C.F. of all the terms before factoring by grouping. Ex uv 3 5v 3 30uv v 2 The G.C.F. = 5v 2, so 10uv 3 5v 3 30uv v 2 = 5v 2 [2uv v 6u + 3] Now, let s put the first two terms in one grouping and the last two terms in the other grouping: 5v 2 [2uv v 6u + 3] = 5v 2 [(2uv v) + ( 6u + 3)] The G.C.F. of the first grouping is v and G.C.F. of the second grouping is 3, so 5v 2 [(2uv v) + ( 6u + 3)] = 5v 2 [v(2u 1) 3(2u 1)] Now, the two pieces have (2u 1) in common, so we will factor that out: 5v 2 [v(2u 1) 3(2u 1)] = 5v 2 (2u 1)(v 3) Ex. 13 w z + 5w + 4wz w z + 5w + 4wz = (w z) + (5w + 4wz) = 1(w z) + w(5 + 4z) Here, we are stuck. When this happens, you want to try a different arrangement. Perhaps group 5w with w 2 instead: w z + 5w + 4wz = w 2 + 5w + 20z + 4wz = (w 2 + 5w) + (20z + 4wz) = w(w + 5) + 4z(5 + w) (switch w & 5) = w(w + 5) + 4z(w + 5) = (w + 5)(w + 4z) Ex. 14 5p 3 10p p p 3 10p p + 33 = (5p 3 10p 2 ) + (15p + 33) = 5p 2 (p 2) + 3(5p + 11) Here, we are stuck. So, let s try grouping 5p 3 and 15p: 90

5 5p 3 10p p + 33 = 5p p 10p = (5p p) + ( 10p ) = 5p(p 2 + 3) (10p 2 33) No, this does not work either. Our last possibility is to pair 5p 3 with 33: 5p 3 10p p + 33 = 5p p p = (5p ) + ( 10p p) = 1(5p ) 5p(2p 3) This does not work either. This means that 5p 3 10p p + 33 cannot be factored. If a problem cannot be factored, then we say that it is prime. So, 5p 3 10p p + 33 is prime. Ex. 15 8ax 5 y 2 24bx 5 y 2 a + 3b 8ax 5 y 2 24bx 5 y 2 a + 3b = (8ax 5 y 2 24bx 5 y 2 ) + ( a + 3b) = 8x 5 y 2 (a 3b) (a 3b) = (a 3b)(8x 5 y 2 1) 91

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### In this section, you will develop a method to change a quadratic equation written as a sum into its product form (also called its factored form). CHAPTER 8 In Chapter 4, you used a web to organize the connections you found between each of the different representations of lines. These connections enabled you to use any representation (such as a graph,

### What are the place values to the left of the decimal point and their associated powers of ten? The verbal answers to all of the following questions should be memorized before completion of algebra. Answers that are not memorized will hinder your ability to succeed in geometry and algebra. (Everything

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### 1.3 Algebraic Expressions 1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts,

### 6.4 Special Factoring Rules 6.4 Special Factoring Rules OBJECTIVES 1 Factor a difference of squares. 2 Factor a perfect square trinomial. 3 Factor a difference of cubes. 4 Factor a sum of cubes. By reversing the rules for multiplication

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### Factoring, Solving. Equations, and Problem Solving REVISED PAGES 05-W4801-AM1.qxd 8/19/08 8:45 PM Page 241 Factoring, Solving Equations, and Problem Solving 5 5.1 Factoring by Using the Distributive Property 5.2 Factoring the Difference of Two Squares 5.3 Factoring

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### Algebraic expressions are a combination of numbers and variables. Here are examples of some basic algebraic expressions. Page 1 of 13 Review of Linear Expressions and Equations Skills involving linear equations can be divided into the following groups: Simplifying algebraic expressions. Linear expressions. Solving linear Factoring... 1 Polynomials...1 Addition of Polynomials... 1 Subtraction of Polynomials...1 Multiplication of Polynomials... Multiplying a monomial by a monomial... Multiplying a monomial by a polynomial... The Mathematics 11 Competency Test Monomial Factors The first stage of factoring an algebraic expression involves the identification of any factors which are monomials. We will describe the process by