3. Example Determine whether each of the following is a perfect-square trinomial. a) x 2 + 8x + 16b) t 2  9t  36c) 25x 2 + 4  20x Solution a) x 2 + 8x + 16 1. 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. b) t 2  9t  36 1. 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 2 + 4  20x It helps to write it in descending order 25x 2  20x + 4 1. Two terms, 25x 2 and 4, are squares. 2. There is no minus sign before 25x 2 or 4. 3. Twice the product of 5x and 2, is 20x, the opposite of the remaining term,  20x. Thus 25x 2  20x + 4 is a perfect-square trinomial. = (x + 4) 2 = (5x – 2) 2

4. Example Factor: 16a 2  24ab + 9b 2 Solution 16a 2  24ab + 9b 2 = (4a  3b) 2 Example Factor: 12a 3  108a 2 + 243a Solution Always look for the greatest common factor. This time there is one. We factor out 3a. 12a 3  108a 2 + 243a = 3a(4a 2  36a + 81) = 3a(2a  9) 2 Note that in order for a term to be a perfect square, its coefficient must be a perfect square and the power(s) of the variable(s) must be even.

5. 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.

6. 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) c) 25  36a 12 = (5 + 6a 6 )(5  6a 6 ) d) 98x 2  8x 8 = 2x 2 (49  4x 6 ) Greatest Common Factor = 2x 2 (7 + 2x 3 )(7  2x 3 )

7. A 2 + B 2 is Prime!!

8. More Factoring by Grouping Sometimes when factoring a polynomial with four terms, we may be able to factor further. Example Factor: x 3 + 6x 2 – 25x – 150. Solution Grouping x 3 + 6x 2 – 25x – 150 = (x 3 + 6x 2 ) – (25x + 150) = x 2 (x + 6) – 25(x + 6) = (x + 6)(x 2 – 25) = (x + 6)(x + 5)(x – 5) Must change the sign of each term. Example Factor: x 2 + 8x + 16 – y 2. Solution x 2 + 8x + 16 – y 2 = (x 2 + 8x + 16) – y 2 = (x + 4) 2 – y 2 = (x + 4 + y)(x + 4 – y)

9. Example Solve: x 3 + 6x 2 = 25x + 150. Solution—Algebraic x 3 + 6x 2 = 25x + 150 x 3 + 6x 2 – 25x – 150 = 0 (x 3 + 6x 2 ) – (25x + 150) = 0 x 2 (x + 6) – 25(x + 6) = 0 (x + 6)(x 2 – 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 x = –6, –5, or 5. Solving Equations We can now solve polynomial equations involving differences of squares and perfect-square trinomials.

10. Graphical Solution The solutions are x = –6, –5, or 5.