# Thread: partial fractions (quardratic factor in denominator)

1. ## partial fractions (quardratic factor in denominator)

hi am having difficulty figuring this one out can u help?

6/(x+1)(x-1)(x-4)^2

so far this is what i got but i'm not sure its correct....

A/(x+1) + B/(x-1) + (Cx+D)/(x-4)^2
considering just the frc (Cx+D)/(x-4)^2
and letting D= -4C+E
then

frc(Cx+D)/(x-4)^2 = (Cx - 4C + E)/(x-4)^2

= [C/(x-4)] + E/(x-4)^2

therefore

6/(x+1)(x-1)(x-4)^2 = A/(x+1) + B/(x-1) + C/(x-4) + E/(x-4)^2

so am i going good so far? cuz when i substitute values of x am getting weird numbers.

2. Originally Posted by helloworld101
hi am having difficulty figuring this one out can u help?

6/(x+1)(x-1)(x-4)^2
you have a repeated linear factor ...

$\frac{6}{(x+1)(x-1)(x-4)^2} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{C}{x-4} + \frac{D}{(x-4)^2}$

$6 = A(x-1)(x-4)^2 + B(x+1)(x-4)^2 + C(x+1)(x-1)(x-4) + D(x+1)(x-1)$

let $x = -1$ ...

$6 = -50A$ ... $A = -\frac{3}{25}$

let $x = 1$ ...

$6 = 18B$ ... $B = \frac{1}{3}$

let $x = 4$ ...

$6 = 15D$ ... $D = \frac{2}{5}$

let $x = 2$ ...

$6 = 4A + 12B - 6C + 3D = -\frac{12}{25} + 4 - 6C + \frac{6}{5}$

$C = -\frac{16}{75}$

$\frac{6}{(x+1)(x-1)(x-4)^2} = -\frac{3}{25(x+1)} + \frac{1}{3(x-1)} - \frac{16}{75(x-4)} + \frac{2}{5(x-4)^2}$

3. $\frac{6}{(x+1)(x-1)(x-4)^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x-4}+\frac{D}{(x-4)^2}$

$6=A(x-1)(x-4)^2+B(x+1)(x-4)^2+C(x+1)(x-1)(x-4)+D(x+1)(x-1)$

$x=1\Rightarrow 6=18B\Rightarrow B=\frac{1}{3}$

$x=-1\Rightarrow 6=-18A\Rightarrow A=-\frac{1}{3}$

$x=4\Rightarrow 6=15D\Rightarrow D=\frac{2}{5}$

Now differentiate both members:

$0=A(x-4)^2+2A(x-1)(x-4)+B(x-4)^2+2B(x+1)(x-4)+$

$+C(x-1)(x-4)+C(x+1)(x-4)+C(x+1)(x-1)+D(x-1)+D(x+1)$

$x=4\Rightarrow 0=15C+8D\Rightarrow C=-\frac{16}{75}$