Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: −2 2+1 x xx (2) If we are given the form of f (x) in (2), it is a matter of Intermediate Algebra to determine a common denominator to obtain the form of f (x) given in (1). The focus of this section is to develop a method by which we start with f (x) in the form of (1) and ‘resolve it into partial fractions’ to obtain the form in (2). Essentially, we need to reverse the least common denominator process. Starting with the form of f (x) in (1), we begin by factoring the denominator x2 − x − 6 x2 − x − 6 =22 4 + x2 x x (x + 1) We now think about which individual denominators could contribute to obtain x2 x2 + 1 as the least common denominator. Certainly x2 and x2 + 1, but are there any other factors? Since x2 + 1 is an irreducible quadratic1 there are no factors of it that have real coeﬃcients which can contribute to the denominator. The factor x2 , however, is not irreducible, since we can think of it as x2 = xx = (x − 0)(x − 0), a so-called ‘repeated’ linear factor.2 This means it’s pos...
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