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Unformatted text preview: Solutions to Quiz 2B
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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This note was uploaded on 07/12/2011 for the course MAC 1140 taught by Professor Williamson during the Fall '08 term at University of Florida.
 Fall '08
 WILLIAMSON
 Math, Calculus, Algebra

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