Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Chapter 5 Polynomials and Factoring

5-2 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Factoring Perfect-Square Trinomials and Differences of Squares Recognizing Perfect-Square Trinomials Factoring Perfect-Square Trinomials Recognizing Differences of Squares Factoring Differences of Squares Factoring Completely 5.4

5-3 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Recognizing Perfect-Square Trinomials A trinomial that is the square of a binomial is called a perfect-square trinomial. A 2 + 2AB + B 2 = (A + B) 2 ; A 2 – 2AB + B 2 = (A – B) 2 Reading the right sides first, we see that these equations can be used to factor perfect-square trinomials. Note that in order for the trinomial to be the square of a binomial, it must have the following: 1. Two terms, A 2 and B 2, must be squares, such as 9, x 2, 100y 2, 25w 2 2. Neither A 2 or B 2 is being subtracted. 3. The remaining terms is either 2  A  B or  2  A  B where A and B are the square roots of A 2 and B 2.

5-4 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Determine whether each of the following is a perfect-square trinomial. a) x 2 + 8x + 16b) t 2  9t  36c) 25x  20x Solution a) x 2 + 8x Two terms, x 2 and 16, are squares. 2. Neither x 2 or 16 is being subtracted. 3. The remaining term, 8x, is 2  x  4, where x and 4 are the square roots of x 2 and 16.

5-5 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example b) t 2  9t  Two terms, t 2 and 36, are squares. But 2. Since 36 is being subtracted t 2  9t  36 is not a perfect-square trinomial. c) 25x  20x It helps to write it in descending order. 25x 2  20x Two terms, 25x 2 and 4, are squares. 2. There is no minus sign before 25x 2 or Twice the product of the square roots is 2  5x  2, is 20x, the opposite of the remaining term,  20x. Thus 25x 2  20x + 4 is a perfect-square trinomial.

Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Factoring a Perfect-Square Trinomial A 2 + 2AB + B 2 = (A + B) 2 ; A 2 – 2AB + B 2 = (A – B) 2

5-7 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor: a) x 2 + 8x + 16b) 25x 2  20x + 4 Solution a) x 2 + 8x + 16 = x  x  = (x + 4) 2 A A B + B 2 = (A + B) 2 b) 25x 2  20x + 4 = (5x) 2  2  5x  = (5x  2) 2 A 2 – 2 A B + B 2 = (A – B) 2

5-8 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor: 16a 2  24ab + 9b 2 Solution 16a 2  24ab + 9b 2 = (4a) 2  2(4a)(3b) + (3b) 2 = (4a  3b) 2 Check: (4a  3b)(4a  3b) = 16a 2  24ab + 9b 2 The factorization is (4a  3b) 2.

5-9 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor: 12a 3  108a a Solution Always look for a common factor. This time there is one. We factor out 3a. 12a 3  108a a = 3a(4a 2  36a + 81) = 3a[(2a) 2  2(2a)(9) ] = 3a(2a  9) 2 The factorization is 3a(2a  9) 2.

5-10 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Recognizing Differences of Squares An expression, like 25x 2  36, that can be written in the form A 2  B 2 is called a difference of squares. Note that for a binomial to be a difference of squares, it must have the following. 1. There must be two expressions, both squares, such as 9, x 2, 100y 2, 36y 8 2. The terms in the binomial must have different signs.

5-11 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Determine whether each of the following is a difference of squares. a) 16x 2  25b) 36  y 5 c)  x Solution a) 16x 2  The first expression is a square: 16x 2 = (4x) 2 The second expression is a square: 25 = The terms have different signs. Thus, 16x 2  25 is a difference of squares, (4x) 2  5 2.

5-12 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example b) 36  y 5 1. The expression y 5 is not a square. Thus, 36  y 5 is not a square. c)  x The expressions x 12 and 49 are squares: x 12 = (x 6 ) 2 and 49 = The terms have different signs. Thus,  x is a difference of squares, 7 2  (x 6 ) 2. It is often useful to rewrite  x in the equivalent form 49  x 12.

Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Factoring a Difference of Squares A 2 – B 2 = (A + B)(A – B).

5-14 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor: a) x 2  9b) y 2  16w 2 c) 25  36a 12 d) 98x 2  8x 8 Solution a) x 2  9 = x 2  3 2 = (x + 3)(x  3) A 2  B 2 = (A + B)(A  B) b) y 2  16w 2 = y 2  (4w) 2 = (y + 4w)(y  4w) A 2  B 2 = (A + B) (A  B)

5-15 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example c) 25  36a 12 = 5 2  (6a 6 ) 2 = (5 + 6a 6 )(5  6a 6 ) d) 98x 2  8x 8 Always look for a common factor. This time there is one, 2x 2 : 98x 2  8x 8 = 2x 2 (49  4x 6 ) = 2x 2 [(7 2  (2x 3 ) 2 ] = 2x 2 (7 + 2x 3 )(7  2x 3 )

5-16 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Factoring Completely Sometimes, a complete factorization requires two or more steps. Factoring is complete when no factor can be factored further.

5-17 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example Factor 5x 4  125. Solution We have 5x 4  125 = 5(x 4  25) = 5[(x 2 ) 2  5 2 ] = 5(x 2  5)(x 2 + 5) The factorization is 5(x 2  5)(x 2 + 5).

Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Tips for Factoring 1.Always look first for a common factor! If there is one, factor it out. 2.Be alert for perfect-square trinomials and for binomials that are differences of squares. Once recognized, they can be factored without trial and error. 3.Always factor completely. 4.Check by multiplying.