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

2. 5.6
Equations Containing Perfect-Square Trinomials and Differences of Squares
■ Perfect-Square Trinomials
■ Differences of Squares
■ More Factoring by Grouping
■ Solving Equations
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3. 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.
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4. 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.
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5. 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.
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6. Factoring a Perfect-Square Trinomial
A2 + 2AB + B2 = (A + B)2;
A2 – 2AB + B2 = (A – B)2
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7. 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!
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8. 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.
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9. Example Factor: 12a3 108a2 + 243a Solution
Always look for a common factor. This time there is one. We factor out 3a.
12a3 108a a = 3a(4a2  36a + 81)
= 3a(2a  9)2
The factorization is 3a(2a  9)2.
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10. 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.
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11. Factoring a Difference of Two Squares
A2 – B2 = (A + B)(A – B)
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12. 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)
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13. 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)
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14. More Factoring by Grouping
Sometimes when factoring a polynomial with four terms, we may be able to factor further.
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15. 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)
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16. Example Factor: x2 + 8x + 16 – y2. Solution
x2 + 8x + 16 – y2 = (x2 + 8x + 16) – y2
= (x + 4)2 – y2
= (x y)(x + 4 – y)
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17. Solving Equations
We can now solve polynomial equations involving differences of squares and perfect-square trinomials.
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18. 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 = or x + 5 = or x – 5 = 0
x = –6 or x = –5 or x = 5
The solutions are –6, –5, and 5.
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19. 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.
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