Stitz-Zeager_College_Algebra_e-book

556 sequences and the binomial theorem equation 91

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Unformatted text preview: x + 1 + (Bx + C )x or 3 = (A + B )x2 + (−A + C )x + A. We get A+B = 0 −A + C = 0 A=3 From A = 3 and A + B = 0, we get B = −3. From −A + C = 0, we get C = A = 3. We get x3 4. Since 4x3 x2 − 2 3 3 3 − 3x = +2 2+x −x x x −x+1 isn’t proper, we use long division and we get a quotient of 4x with a remainder of 3 8x 8x 8x. That is, x4x 2 = 4x + x2 −2 so we now work on resolving x2 −2 into partial fractions. The 2− quadratic x2 − 2, though it doesn’t factor nicely, is, nevertheless, reducible. Solving x2 − 2 = 0 8.6 Partial Fraction Decomposition 527 √ be gives us x = ± 2, and each of these zeros must√ of multiplicity one since Theorem 3.14 √ enables us to now factor x2 − 2 = x − 2 x + 2 . Hence, 8x 8x A B √ √= √+ √ = x2 − 2 x− 2 x+ 2 x− 2 x+ 2 √ √ √ Clearing fractions, we get 8x = A x + 2 + B x − 2 or 8x = (A + B )x + (A − B ) 2. We get the system A+B = 8 √ (A − B ) 2 = 0 √ From (A − B ) 2 = 0, we get A = B , which, when subst...
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