Unformatted text preview: x) = 2x2 3
x3 − 2x2 + x 4. R(x) = x3 3
− x2 + x 4x3
x2 − 2 5. R(x) = x3 + 5 x − 1
x4 + 6x2 + 9 6. R(x) = 8x2
x4 + 16 Solution.
1. We begin by factoring the denominator to ﬁnd 2x2 − x − 1 = (2x + 1)(x − 1). We get x = − 1
2
and x = 1 are both zeros of multiplicity one and thus we know
x+5
x+5
A
B
=
=
+
−x−1
(2x + 1)(x − 1)
2x + 1 x − 1 2x2 Clearing denominators, we get x +5 = A(x − 1)+ B (2x +1) so that x +5 = (A +2B )x + B − A.
Equating coeﬃcients, we get the system
A + 2B = 1
−A + B = 5
This system is readily handled using the Addition Method from Section 8.1, and after adding
both equations, we get 3B = 6 so B = 2. Using back substitution, we ﬁnd A = −3. Our
answer is easily checked by getting a common denominator and adding the fractions.
x+5
2
3
=
−
−x−1
x − 1 2x + 1 2x2 526 Systems of Equations and Matrices 2. Factoring the denominator gives x3 − 2x2 + x = x x2 − 2x + 1 = x(x − 1)2 which gives x = 0
as a zero of multiplicity one and x = 1 as a zero of multip...
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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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