Stitz-Zeager_College_Algebra_e-book

1 sequences 553 3 from 2n 1 we have that an 2n 1

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Dx + E =22 =+ +2 4 + x2 2 x x (x + 1) x x x +1 1 2 Recall this means it has no real zeros; see Section 3.4. Recall this means x = 0 is a zero of multiplicity 2. 8.6 Partial Fraction Decomposition 523 C C However, if we look more closely at the term Bx+C , we see that Bx+C = Bx + x2 = B + x2 . The x x2 x2 x2 B A term x has the same form as the term x which means it contributes nothing new to our expansion. Hence, we drop it and, after re-labeling, we find ourselves with our new guess: x2 − x − 6 x2 − x − 6 A B Cx + D =22 = + 2+ 2 4 + x2 x x (x + 1) x x x +1 Our next task is to determine the values of our unknowns. Clearing denominators gives x2 − x − 6 = Ax x2 + 1 + B x2 + 1 + (Cx + D)x2 Gathering the like powers of x we have x2 − x − 6 = (A + C )x3 + (B + D)x2 + Ax + B In order for this to hold for all values of x in the domain of f , we equate the coefficients of corresponding powers of x on each side of the equation3 and obtain the system of linear equations (E 1) A + C (E 2) B...
View Full Document

Ask a homework question - tutors are online