Alg_Complete

# Notice as well that weve now got a linear factor to a

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: he denominator as completely as possible. Then for each factor in the denominator we can use the following table to determine the term(s) we pick up in the partial fraction decomposition. Factor in denominator Term in partial fraction decomposition ax + b A ax + b ( ax + b ) Ak A1 A2 + +L + 2 k ax + b ( ax + b ) ( ax + b ) k ax 2 + bx + c ( ax 2 + bx + c ) © 2007 Paul Dawkins Ax + B ax + bx + c Ak x + Bk A1 x + B1 A2 x + B2 + +L + 2 k 2 ax + bx + c ( ax 2 + bx + c ) ( ax2 + bx + c ) 2 k 271 http://tutorial.math.lamar.edu/terms.aspx College Algebra Notice that the first and third cases are really special cases of the second and fourth cases respectively if we let k = 1 . Also, it will always be possible to factor any polynomial down into a product of linear factors ( ax + b ) and quadratic factors ( ax 2 + bx + c ) some of which may be raised to a power. There are several methods for determining the coefficients for each term and we will go over each of those as we work the examples. S...
View Full Document

## This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

Ask a homework question - tutors are online