# Copyright © 2015, 2011, 2007 Pearson Education, Inc. 1 1 Chapter 6 Factoring.

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. 1 1 Chapter 6 Factoring

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 2 Factoring 6.1Greatest Common Factor and Factoring by Grouping 6.2Factoring Trinomials 6.3Factoring Special Products and Factoring Strategies 6.4Solving Equations by Factoring CHAPTER 6

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 3 Factoring Special Products and Factoring Strategies 1.Factor perfect square trinomials. 2.Factor a difference of squares. 3.Factor a difference of cubes. 4.Factor a sum of cubes. 5.Use various strategies to factor polynomials. 6.3

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 4 Factoring Perfect Square Trinomials a 2 + 2ab + b 2 = (a + b) 2 a 2 – 2ab + b 2 = (a – b) 2

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 5 Example Factor. 9a 2 + 6a + 1 Solution This trinomial is a perfect square because it is in the form a 2 + 2ab + b 2, where a = 3a and b = 1. 9a 2 + 6a + 1 a 2 = (3a) 2 = 9a 2.b 2 = 1 2 = 1 2ab = (2)(3a)(1) = 6a 9a 2 + 6a + 1 = (3a + 1) 2

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 6 Example Factor. 16x 2 – 56x + 49 Solution This trinomial is a perfect square. 16x 2 – 56x + 49 = (4x – 7) 2 Use a 2 – 2ab + b 2 = (a – b) 2, where a = 4x and b = 7.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 7 Example Factor. 9x 2 – 6xy + y 2 Solution = (3x – y) 2 9x 2 – 6xy + y 2 Use a 2 – 2ab + b 2 = (a – b) 2, where a = 3x and b = y.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 8 Example Factor. Solution Use a 2 – 2ab + b 2 = (a – b) 2, where a = 3a and b = 7. Factor out the monomial GCF, ab.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 9 Factoring a Difference of Squares a 2 – b 2 = (a + b)(a – b) Warning: A sum of squares, a 2 + b 2, is prime and cannot be factored.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 10 Example Factor. 9x 2 – 16y 2 Solution This binomial is a difference of squares because 9x 2 – 16y 2 = (3x) 2 – (4y) 2. To factor it, we use the rule a 2 – b 2 = (a + b)(a – b). a 2 – b 2 = (a + b)(a – b) 9x 2 – 16y 2 = (3x) 2 – (4y) 2 = (3x + 4y)(3x – 4y)

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 11 Example Factor. Solution The terms in this binomial have a monomial GCF, 2a 7. Factor out the GCF. Factor 36a 2 – 49, using a 2 – b 2 = (a + b)(a – b) with a = 6a and b = 7.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 12 Example Factor. n 4 – 625 Solution This binomial is a difference of squares, where a = n 2 and b = 25. (n 2 + 25)(n 2 – 25)n 4 – 625 = Use a 2 – b 2 = (a + b)(a – b). Factor n 2 – 25, using a 2 – b 2 = (a + b)(a – b) with a = n and b = 5. = (n 2 + 25)(n + 5)(n – 5) Be sure you have factored completely. Warning: A sum of two squares is prime and cannot be factored.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 13 Factoring a Difference of Cubes a 3 – b 3 = (a – b)(a 2 + ab + b 2 ) Warning: The trinomial a 2 + ab + b 2 is not a perfect square and cannot be factored. Remember a perfect square trinomial has the form a 2 + 2ab + b 2.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 14 Example Factor. 8x 3 – 729 Solution This binomial is a difference of cubes. a 3 – b 3 = (a – b) (a 2 + a b + b 2 ) 8x 3 – 729 = (2x) 3 – (9) 3 = (2x – 9)((2x) 2 + (2x)(9) + (9) 2 ) = (2x – 9)(4x 2 + 18x + 81) Note: The trinomial may seem like a perfect square. However, to be a perfect square, the middle term should be 2ab. In this trinomial, we only have ab, so it cannot be factored.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 15 Example Factor. 216x 3 – 64 Solution This binomial is a difference of cubes. This binomial has a GCF, 8. a 3 – b 3 = (a – b) (a 2 + a b + b 2 ) 8(27x 3 – 8) = 8(3x) 3 – (2) 3 = 8(3x – 2)[(3x) 2 + (3x)(2) + (2) 2 ] = 8(3x – 2)(9x 2 + 6x + 4) = 8(27x 3 – 8)216x 3 – 64

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 16 Factoring a Sum of Cubes a 3 + b 3 = (a + b)(a 2 – ab + b 2 ) Warning: In the rule for the sum of cubes, the trinomial factor a 2 – ab + b 2 cannot be factored.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 17 Example Factor. 125x 3 + 64 Solution This binomial is a sum of cubes. a 3 + b 3 = (a + b) (a 2  a b + b 2 ) 125x 3 + 64 = (5x) 3 + (4) 3 = (5x + 4)((5x) 2  (5x)(4) + (4) 2 ) = (5x + 4)(25x 2  20x + 16)

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 18 Example Factor. 6x +162xy 3 Solution The terms in this binomial have a monomial GCF, 6x. = 6x(1 + 27y 3 )6x +162xy 3 = 6x(1 + 3y)[(1) 2 – (1)(3y) + (3y) 2 ] = 6x(1 + 3y)(1 – 3y + 9y 2 ) Factor out the monomial GCF, 6x. Simplify.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 19 Factoring a Polynomial To factor a polynomial, first factor out any monomial GCF and then consider the number of terms in the polynomial. I. Four terms: Try to factor by grouping. II. Three terms: Determine if the trinomial is a perfect square. A. If the trinomial is a perfect square, consider its form. 1. If it is in the form a 2 + 2ab + b 2, the factored form is (a + b) 2. 2. If it is in the form a 2  2ab + b 2, the factored form is (a  b) 2. B.If the trinomial is not a perfect square, consider its form. 1. If it is in the form x 2 + bx + c, find two factors of c whose sum is b, and write the factored form as (x + first number)(x + second number).

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 20 Factoring a Polynomial continued 2. If it is in the form ax 2 + bx + c, where a  1, use trial and error. Or find two factors of ac whose sum is b; write these factors as coefficients of two like terms that, when combined, equal bx; and factor by grouping. III.Two terms: Determine if the binomial is a difference of squares, sum of cubes, or difference of cubes. A. If it is a difference of squares, a 2 – b 2, the factors are conjugates and the factored form is (a + b)(a – b). Note that a sum of squares cannot be factored. B. If it is a difference of cubes, a 3 – b 3, the factored form is (a – b)(a 2 + ab + b 2 ). C. If it is a sum of cubes, a 3 + b 3, the factored form is (a + b)(a 2 – ab + b 2 ). Note: Always check to see if any of the factors can be factored further.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 21 Example Factor completely. Solution Remove the GCF of 7x 3 The binomial x 2 + 9 is a sum of squares, which cannot be factored using real numbers.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 22 Example Factor completely. Solution There is no monomial GCF. Because there are four terms, we factor by grouping.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 23 Example Factor completely. Solution There is no monomial GCF. Because there are three terms, we determine whether it is a perfect square trinomial. It is a perfect square trinomial with a = 3x and b = 5y.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 24 Example Factor completely. 12x 2 – 8x – 15 Solution There is no GCF. It is in the form ax 2 + bx + c, where a 1, so we use trial and error. After trying various combinations, we find the following correct combination. = (6x + 5)(2x – 3 )

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 25 Example Factor completely. Solution There is no monomial GCF. Determine whether it is a difference of squares or the sum or difference of cubes. It is the difference of cubes.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 26 Example Factor completely. Solution There is no monomial GCF. Determine whether it is a difference of squares or the sum or difference of cubes. It is the difference of squares.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 27 Example Factor. 5x 3 – 10x 2 – 120x Solution 5x(x 2 – 2x – 24) Factor out the monomial GCF, 5x. = 5x(x + 4)(x – 6) Factor x 2 – 2x – 24.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 28 Example Factor. 150x 3 y – 120x 2 y 2 + 24xy 3 Solution 6xy(25x 2 – 20xy + 4y 2 ) 6xy(5x – 2y) 2 Factor out the monomial GCF, 6xy. Factor the perfect square trinomial.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 29 Example Factor. 8a 4 – 72n 2 Solution 8a 4 – 72n 2 = 8(a 4 – 9n 2 ) Factor out the monomial GCF, 8. a 4 – 9n 2 is a difference of squares = 8(a 2 – 3n)(a 2 + 3n)

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 30 Example Factor. x 5 – 2x 3 – 27x 2 + 54 Solution No common monomial, factor by grouping. (x 5 – 2x 3 ) + (– 27x 2 + 54) x 3 (x 2 – 2) – 27(x 2 – 2) (x 2 – 2)(x 3 – 27) Difference of cubes (x 2 – 2)(x – 3)(x 2 + 3x + 9)

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 31 Example Factor. Solution The expression is a difference of squares, where a = h + 2 and b = 6k.

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