partial fractions help,

Sep 2008
631
2
Express the following improper fractions as partial fractions.

\(\displaystyle \frac{x^{3} - x^{2} - x -3}{x(x-1)} \)

I know I am suppose to dive the numerator by the denominator, but I keep getting the wrong answer,
 

Soroban

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May 2006
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Hello, Tweety!

Express the following improper fraction as partial fractions:

. . \(\displaystyle \frac{x^3 - x^2 - x -3}{x(x-1)} \)

We have: .\(\displaystyle \frac{x^3-x^2-x-3}{x^2-x} \;=\;x -\frac{x+3}{x(x-1)} \)


We will decompose the fraction:

. . \(\displaystyle \frac{x+3}{x(x-1)} \;=\;\frac{A}{x} + \frac{B}{x-1} \quad\Rightarrow\quad x+3 \;=\;A(x-1) + Bx \)


Let \(\displaystyle x = 0\!:\quad0+3 \:=\:A(0 -1) + B(0) \quad\Rightarrow\quad A \:=\:-3\)

Let \(\displaystyle x = 1\!:\quad 1+3 \:=\:A(1-1) + B(1) \quad\Rightarrow\quad B \:=\:4\)


The problem becomes: . \(\displaystyle x - \left[-\frac{3}{x} + \frac{4}{x-1}\right] \;\;=\;\;x + \frac{3}{x} - \frac{4}{x-1}\)

 
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Apr 2010
41
15
India
\(\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}\)
now,
\(\displaystyle \frac{x+3}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}\)
\(\displaystyle x+3=A(x-1)+Bx\)
comparing coeff. of x and constant term we have, A=-3 & B=4
so,
\(\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}=x-[\frac{-3}{x}+\frac{4}{x-1}]\)
 
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Sep 2008
631
2
\(\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}\)
now,
\(\displaystyle \frac{x+3}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}\)
\(\displaystyle x+3=A(x-1)+Bx\)
comparing coeff. of x and constant term we have, A=-3 & B=4
so,
\(\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}=x-[\frac{-3}{x}+\frac{4}{x-1}]\)

I am unsure as to how you got the first line? Could you please explain, how \(\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}\)

thank you
 
Apr 2010
384
153
Canada
I am unsure as to how you got the first line? Could you please explain, how \(\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}\)

thank you
factor the numerator \(\displaystyle (x^3-x^2)-x-3 = x^2 (x-1) - x - 3 \)

So we get,

\(\displaystyle \frac{ x^2 (x-1) - x - 3 } { x(x-1) } = \frac{ x^2 (x-1) } { x(x-1) } - \frac { x + 3 } { x(x-1) } = x - \frac{x+3}{x(x-1)}\)
 
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