# CHAPTER 6 Polynomials: Factoring (continued) Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 6.1Multiplying and Simplifying Rational Expressions.

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CHAPTER 6 Polynomials: Factoring (continued) Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 6.1Multiplying and Simplifying Rational Expressions 6.2Division and Reciprocals 6.3Least Common Multiples and Denominators 6.4Adding Rational Expressions 6.5Subtracting Rational Expressions 6.6Solving Rational Equations 6.7Applications Using Rational Equations and Proportions

CHAPTER 6 Polynomials: Factoring Slide 3Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 6.8Complex Rational Expressions 6.9Direct Variation and Inverse Variation

OBJECTIVES 6.3 Least Common Multiples and Denominators Slide 4Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. aFind the LCM of several numbers by factoring. bAdd fractions, first finding the LCD. cFind the LCM of algebraic expressions by factoring

Least Common Multiples To add when denominators are different, we first find a common denominator. To find the LCM of 12 and 30, we factor: 12 = 2 · 2 · 3 30 = 2 · 3 · 5 The LCM is the number that has 2 as a factor twice, 3 as a factor once, and 5 as a factor once: LCM = 2 · 2 · 3 · 5 = 60 6.3 Least Common Multiples and Denominators a Find the LCM of several numbers by factoring. Slide 5Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

To find the LCM, use each factor the greatest number of times that is appears in any one factorization. 6.3 Least Common Multiples and Denominators Finding LCMs Slide 6Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE Solution 48 = 2 · 2 · 2 · 2 · 3 54 = 2 · 3 · 3 · 3 LCM = 2  2  2  2  3  3  3 or 432 6.3 Least Common Multiples and Denominators a Find the LCM of several numbers by factoring. AFind the LCM of 48 and 54. Slide 7Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE Solution 6.3 Least Common Multiples and Denominators b Add fractions, first finding the LCD. BAdd: Slide 8Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE Solution 6x 2 = 2  3  x  x 4x 3 = 2  2  x  x  x LCM = 2  2  3  x  x  x The LCM = 2 2  3  x 3, or 12x 3. 6.3 Least Common Multiples and Denominators c Find the LCM of algebraic expressions by factoring CFind the LCM of 6x 2 and 4x 3 Slide 9Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE a) 16a and 24b b) 24x 4 y 4 and 6x 6 y 2 c) x 2  4 and x 2  2x  8 Solution a) 16a = 2  2  2  2  a 24b = 2  2  2  3  b The LCM = 2  2  2  2  a  3  b The LCM is 2 4  3  a  b, or 48ab 6.3 Least Common Multiples and Denominators c Find the LCM of algebraic expressions by factoring DFor each pair of polynomials, find the least common multiple. (continued) Slide 10Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE b) 24x 4 y 4 = 2 · 2 · 2 · 3 · x · x · x · x · y · y · y · y 6x 6 y 2 = 2 · 3 · x · x · x · x · x · x · y · y LCM = 2 · 2 · 2 · 3 · x · x · x · x · y · y · y · y · x · x Note that we used the highest power of each factor. The LCM is 24x 6 y 4 6.3 Least Common Multiples and Denominators c Find the LCM of algebraic expressions by factoring DFinding the LCM by Factoring (continued) Slide 11Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE c) x 2 – 4 = (x – 2)(x + 2) x 2 – 2x – 8 = (x + 2)(x – 4) LCM = (x – 2)(x + 2)(x – 4) 6.3 Least Common Multiples and Denominators c Find the LCM of algebraic expressions by factoring DFinding the LCM by Factoring Slide 12Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE a) 15x, 30y, 25xyzb) x 2 + 3, x + 2, 7 Solution a) 15x = 3  5  x 30y = 2  3  5  y 25xyz = 5  5  x  y  z LCM = 2  3  5  5  x  y  z The LCM is 2  3  5 2  x  y  z or 150xyz 6.3 Least Common Multiples and Denominators c Find the LCM of algebraic expressions by factoring EFor each group of polynomials, find the least common multiple. (continued) Slide 13Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE a) 15x, 30y, 25xyzb) x 2 + 3, x + 2, 7 Solution b) x 2 + 3 = x + 2 = 7 = b) Since x 2 + 3, x + 2, and 7 are not factorable, the LCM is their product: 7(x 2 + 3)(x + 2). 6.3 Least Common Multiples and Denominators c Find the LCM of algebraic expressions by factoring EFor each group of polynomials, find the least common multiple. Slide 14Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

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