How to properly set up partial fractions for repeated denominator factorsI was just trying to solve a problem that had the following item which I needed to split into separate generating functions: x ( 1 − 2 x ) 2 ( 1 − 5 x ) I had assumed I needed to split it into: A 1 − 2 x + B 1 − 2 x + C 1 − 5 x But according to Wolfram Alpha it appears I had to split it into: A 1 − 2 x + B ( 1 − 2 x ) 2 + C 1 − 5 x Can anyone explain the intuition behind this? Is this a general rule that when you have a repeated factor in the denominator, you split it into all powers of that factor?

Question

How to properly set up partial fractions for repeated denominator factorsI was just trying to solve a problem that had the following item which I needed to split into separate generating functions:      x          (      1      −      2      x              )        2            (      1      −      5      x      )      I had assumed I needed to split it into:      A          1      −      2      x        +      B          1      −      2      x        +      C          1      −      5      x      But according to Wolfram Alpha it appears I had to split it into:      A          1      −      2      x        +      B          (      1      −      2      x              )        2              +      C          1      −      5      x      Can anyone explain the intuition behind this? Is this a general rule that when you have a repeated factor in the denominator, you split it into all powers of that factor?

Cristopher Barrera · Accepted Answer

Consider the simplest of cases.      ξ          x      3      Assuming   ξ is some polynomial, one could carry on long division to determine the quotient and the remainder. Thus one could get      ξ          x      3        =  q  (  x  )  +            r      (      x      )              x      3      Now what can we say for certain about   r  (  x  ) ? We can say that it is the remainder thus of smaller degree than       x    3  . The most general possible polynomial of degree 2 or smaller is   A  +  B  x  +  C      x    2  . Then            r      (      x      )              x      3        =            A      +      B      x      +      C              x        2                    x      3      Further more, we can split the right side            r      (      x      )              x      3        =      A          x      3        +            B      x              x      3        +            C              x        2                    x      3      which becomes            r      (      x      )              x      3        =      A          x      3        +      B          x      2        +      C    x  Thus the above would be a sensible what to try to split any proper fraction             r      (      x      )              x      3       Moreover, the similar idea would hold for             r      (      x      )              (      x      +      a              )        3            , meaning a sensible way to split it would be            r      (      x      )              (      x      +      a              )        3              =      A          (      x      +      a              )        3              +      B          (      x      +      a              )        2              +      C          (      x      +      a      )      hope that helps..

Lovellss · Accepted Answer

Yep - if there&#039;s a factor of   (  x  −  a      )    k   in the denominator, you need to include       1          x      −      a        ,      1          (      x      −      a              )        2              ,  .  .  .  ,      1          (      x      −      a              )        k             terms in your expansion.Note that in your attempt, you could combine the first two terms to             A      +      B              1      −      2      x      , making one of those constants redundant.

How to properly set up partial fractions for repeated denominator factors I was just trying to solv

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