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Stitz-Zeager_College_Algebra_e-book

# Finally f3 3 f2 3 2 6 some remarks about example 911

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Unformatted text preview: + D A (E 3) (E 4) B = 0 = 1 = −1 = −6 From From From From equating equating equating equating coeﬃcients of x3 coeﬃcients of x2 coeﬃcients of x the constant terms To solve this system of equations, we could use any of the methods presented in Sections 8.1 through 8.5, but none of these methods are as eﬃcient as the good old-fashioned substitution you learned in Intermediate Algebra. From E 3, we have A = −1 and we substitute this into E 1 to get C = 1. Similarly, since E 4 gives us B = −6, we have from E 2 that D = 7. Our ﬁnal answer is x2 − x − 6 1 6 x+7 x2 − x − 6 =22 =− − 2 + 2 4 + x2 x x (x + 1) xx x +1 which matches the formula given in (2). As we have seen in this opening example, resolving a rational function into partial fractions takes two steps: ﬁrst, we need to determine the form of the decomposition, and then we need to determine the unknown coeﬃcients which appear in said form. Theorem 3.16 guarantees that any polynomial with real coeﬃcients can be factored over the real numbers as a product of linear fac...
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