Stitz-Zeager_College_Algebra_e-book

# 1 an 2 bk 5n1 n1 3n 1k 2k 1 k0 3 2n 1 n1 4 1

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Unformatted text preview: sible that a term with a denominator of just x contributed to the expression as well. What about something like x x2 + 1 ? This, too, could contribute, but we would then wish to break down that denominator into x and x2 + 1 , so we leave out a term of that form. At this stage, we have guessed x2 − x − 6 x2 − x − 6 ? ? ? =22 = + 2+ 2 4 + x2 x x (x + 1) xx x +1 Our next task is to determine what form the unknown numerators take. It stands to reason that 2 − x− since the expression xx4 +x26 is ‘proper’ in the sense that the degree of the numerator is less than the degree of the denominator, we are safe to make the ansatz that all of the partial fraction resolvents are also. This means that the numerator of the fraction with x as its denominator is just a constant and the numerators on the terms involving the denominators x2 and x2 + 1 are at most linear polynomials. That is, we guess that there are real numbers A, B , C , D and E so that x2 − x − 6 x2 − x − 6 A Bx + C...
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