quiz 2A sol - Solutions to Quiz 2A...

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Unformatted text preview: Solutions to Quiz 2A www.math.ufl.edu/˜harringt September 8, 2007 1. Factor Completely. (a) x2 + 7x + 5 Here we will use the AC -Method. The AC = 5 ∗ 1 = 5. However, there are no two numbers a, b such that a + b = 7 and ab = 5. Hence, the polynomial is prime. (b) 2x2 + 7xy − 4y 2 AC -Method: Here AC = −4 ∗ 2 = −8. Futhermore, AC = −8 = 8 ∗ (−1) and 8 − 1 = 7 (The middle term.) Thus, 2x2 + 7xy − 4y 2 = 2x2 + 8xy − xy − 4y 2 = 2x(x + 4y ) − y (x + 4y ) = (2x − y )(x + 4y ) (c) (y − 1)3 − x3 Recall: A3 − B 3 = (A − B )(A2 + AB + B 2 ). So in our case: A = y − 1 and B = x. So we have: (y − 1)3 − x3 = (y − 1 − x)((y − 1)2 + (y − 1)(x) + x2 ) = (y − 1 − x)(y 2 − 2y + 1 + yx − x + x2 ) 1 ...
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