Copyright © 2008 Pearson Education, Inc Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

5.6 Equations Containing Perfect-Square Trinomials and Differences of Squares ■ Perfect-Square Trinomials ■ Differences of Squares ■ More Factoring by Grouping ■ Solving Equations Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Perfect-Square Trinomials To Recognize a Perfect-Square Trinomial Two terms must be squares, such as A2 and B2. There must be no minus sign before A2 or B2. The remaining term must be 2AB or its opposite, 2AB. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Determine whether each of the following is a perfect-square trinomial. a) x2 + 8x + 16 b) t2  9t  36 c) 25x2 + 4  20x Solution a) x2 + 8x + 16 1. Two terms, x2 and 16, are squares. 2. Neither x2 or 16 is being subtracted. 3. The remaining term, 8x, is 2  x  4, where x and 4 are the square roots of x2 and 16. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example continued b) t2  9t  36 1. Two terms, t2 and 36, are squares. But 2. Since 36 is being subtracted t2  9t  36 is not a perfect-square trinomial. c) 25x2 + 4  20x It helps to write it in descending order. 25x2  20x + 4 1. Two terms, 25x2 and 4, are squares. 2. There is no minus sign before 25x2 or 4. 3. Twice the product of the square roots is 2  5x  2, is 20x, the opposite of the remaining term, 20x. Thus 25x2  20x + 4 is a perfect-square trinomial. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Factoring a Perfect-Square Trinomial A2 + 2AB + B2 = (A + B)2; A2 – 2AB + B2 = (A – B)2 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Factor: a) x2 + 8x + 16 b) 25x2  20x + 4 Solution a) x2 + 8x + 16 = (x + 4)2 b) 25x2  20x + 4 = (5x  2)2 We find the square terms and write the square roots with a plus sign between them. Note the sign! Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Factor: 16a2  24ab + 9b2 Solution 16a2  24ab + 9b2 = (4a  3b)2 Check: (4a  3b)(4a  3b) = 16a2  24ab + 9b2 The factorization is (4a  3b)2. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Factor: 12a3 108a2 + 243a Solution Always look for a common factor. This time there is one. We factor out 3a. 12a3 108a2 + 243a = 3a(4a2  36a + 81) = 3a(2a  9)2 The factorization is 3a(2a  9)2. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Differences of Squares An expression that can be written in the form A2  B2 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, x2, 100y2, 36y8 2. The terms in the binomial must have different signs. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Factoring a Difference of Two Squares A2 – B2 = (A + B)(A – B) Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Factor: a) x2  9 b) y2  16w2 c) 25  36a12 d) 98x2  8x8 Solution a) x2  9 = x2  32 = (x + 3)(x  3) A2  B2 = (A + B)(A  B) b) y2  16w2 = y2  (4w)2 = (y + 4w)(y  4w) A2  B2 = (A + B) (A  B) Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example continued c) 25  36a12 = 52  (6a6)2 = (5 + 6a6)(5  6a6) d) 98x2  8x8 Always look for a common factor. This time there is one, 2x2: 98x2  8x8 = 2x2(49  4x6) = 2x2[(72  (2x3)2] = 2x2(7 + 2x3)(7  2x3) Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

More Factoring by Grouping Sometimes when factoring a polynomial with four terms, we may be able to factor further. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Factor: x3 + 6x2 – 25x – 150. Solution x3 + 6x2 – 25x – 150 = x2(x + 6) – 25(x + 6) = (x + 6)(x2 – 25) = (x + 6)(x + 5)(x – 5) Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Factor: x2 + 8x + 16 – y2. Solution x2 + 8x + 16 – y2 = (x2 + 8x + 16) – y2 = (x + 4)2 – y2 = (x + 4 + y)(x + 4 – y) Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Solving Equations We can now solve polynomial equations involving differences of squares and perfect-square trinomials. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Example Solve: x3 + 6x2 = 25x + 150. Solution--Algebraic We have an equation that is a third-degree polynomial, thus it will have 3 or fewer solutions. x3 + 6x2 = 25x + 150 x3 + 6x2 – 25x – 150 = 0 (x + 6)(x2 – 25) = 0 (x + 6)(x + 5)(x – 5) = 0 x + 6 = 0 or x + 5 = 0 or x – 5 = 0 x = –6 or x = –5 or x = 5 The solutions are –6, –5, and 5. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

continued--graphical We let f(x) = x3 + 6x2 we let g(x) = 25x + 150 Look for the intersection of any points on the graphs. The x-coordinates are the solutions. Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley