1. ## partial fractions help,

Express the following improper fractions as partial fractions.

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

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

2. Hello, Tweety!

Express the following improper fraction as partial fractions:

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

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

We will decompose the fraction:

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

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

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

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

3. $\frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}$
now,
$\frac{x+3}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$
$x+3=A(x-1)+Bx$
comparing coeff. of x and constant term we have, A=-3 & B=4
so,
$\frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}=x-[\frac{-3}{x}+\frac{4}{x-1}]$

4. Originally Posted by slovakiamaths
$\frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}$
now,
$\frac{x+3}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$
$x+3=A(x-1)+Bx$
comparing coeff. of x and constant term we have, A=-3 & B=4
so,
$\frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}=x-[\frac{-3}{x}+\frac{4}{x-1}]$

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

thank you

5. Originally Posted by Tweety
I am unsure as to how you got the first line? Could you please explain, how $\frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}$

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

So we get,

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