Stitz-Zeager_College_Algebra_e-book

n1 computing the rst few dierences we nd a2 a1 2

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Unformatted text preview: licity two. We have x3 3 A 3 B C =+ = + 2+x 2 − 2x x(x − 1) x x − 1 (x − 1)2 Clearing denominators, we get 3 = A(x − 1)2 + Bx(x − 1) + Cx, which, after gathering up the like terms becomes 3 = (A + B )x2 + (−2A − B + C )x + A. Our system is A+B = 0 −2A − B + C = 0 A=3 Substituting A = 3 into A + B = 0 gives B = −3, and substituting both for A and B in −2A − B + C = 0 gives C = 3. Our final answer is x3 3 3 3 3 =− + 2+x − 2x x x − 1 (x − 1)2 3. The denominator factors as x x2 − x + 1 . We see immediately that x = 0 is a zero of multiplicity one, but the zeros of x2 − x + 1 aren’t as easy to discern. The quadratic doesn’t factor easily, so we check the discriminant and find it to be (−1)2 − 4(1)(1) = −3 < 0. We find its zeros are not real so it is an irreducible quadratic. The form of the partial fraction decomposition is then x3 3 3 A Bx + C = = +2 2+x 2 − x + 1) −x x (x x x −x+1 Proceeding as usual, we clear denominators and get 3 = A x2 ...
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