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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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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