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Example

Example. Determine whether each of the following is a perfect-square trinomial. a) x 2 + 8 x + 16 b) t 2  9 t  36 c) 25 x 2 + 4  20 x Solution a) x 2 + 8 x + 16 1. Two terms, x 2 and 16 , are squares. 2. Neither x 2 or 16 is being subtracted.

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Example

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  1. 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. = (x + 4)2 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 5xand 2, is 20x, the opposite of the remaining term, 20x. Thus 25x2  20x + 4 is a perfect-square trinomial. = (5x – 2)2

  2. Example Factor: 16a2 24ab + 9b2 Solution 16a2 24ab + 9b2= (4a  3b)2 Example Factor: 12a3108a2 + 243a Solution Always look for the greatest common factor. This time there is one. We factor out 3a. 12a3108a2 + 243a = 3a(4a2 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.

  3. Differences of Squares • An expression, like 25x2 36, 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.

  4. 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) c) 25  36a12 = (5 + 6a6)(5  6a6) d) 98x2  8x8 = 2x2(49  4x6) Greatest Common Factor = 2x2(7 + 2x3)(7  2x3)

  5. A2 + B2 is Prime!!

  6. Must change the sign of each term. Sometimes when factoring a polynomial with four terms, we may be able to factor further. More Factoring by Grouping Example Factor: x3 + 6x2 – 25x – 150. Solution Grouping x3 + 6x2 – 25x – 150 = (x3 + 6x2) – (25x + 150) = x2(x + 6) – 25(x + 6) = (x + 6)(x2 – 25) = (x + 6)(x + 5)(x – 5) Factor: x2 + 8x + 16 – y2. Solution x2 + 8x + 16 – y2 = (x2 + 8x + 16) – y2 = (x + 4)2 – y2 = (x + 4 + y)(x + 4 – y) Example

  7. Solving Equations Example We can now solve polynomial equations involving differences of squares and perfect-square trinomials. • Solve: x3 + 6x2 = 25x + 150. • Solution—Algebraic • x3 + 6x2 = 25x + 150 • x3 + 6x2 – 25x – 150 = 0 • (x3 + 6x2) – (25x + 150) = 0 • x2(x + 6) – 25(x + 6) = 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 x = –6, –5, or 5.

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

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