# Thread: Partial Fraction Decomposition Problem

1. ## Partial Fraction Decomposition Problem

I'm trying to figure out this problem:

At the moment, I've got it down to this:
$3x-4=A(x-1)+B(x-1)^2$

To find $B$, I let $x=1$, which also gets rid of $A$. Any ideas?

2. Originally Posted by BeSweeet
I'm trying to figure out this problem:

At the moment, I've got it down to this:
$3x-4=A(x-1)+B(x-1)^2$

To find $B$, I let $x=1$, which also gets rid of $A$. Any ideas?
Hi BeSweet,

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

There must be a partial fraction present for each power of $(x - c)^n$ less than or equal to n. We can find B, but not A using Heaviside's method. Only the constant corresponding to $(x-c)^n$ can be found using this method.

First, multiply the original proper fraction by $(x-c)^2$ and evaluate the result at x = c to get the value of the constant.

$\frac{3x-4}{(x-1)^2}(x-1)^2=3x-4 \Rightarrow B=3(1)-4=-1$

$\boxed{B=-1}$

$\frac{3x-4}{(x-1)^2}=\frac{A}{x-1}+\frac{-1}{(x-1)^2}$

To find the other constant, multiply both sides by the least common denominator to clear the fractions:

$3x-4=A(x-1)-1$

$3x-4=Ax + (-1-A)$

The polynomial on the left of the equal sign must have the same coefficient and constant as the polynomial on the right of the equal sign.

$\boxed{A=3}$

You could also say $-4=(-1-A)$

You still get $\boxed{A=3}$

The decomposition is:

$\boxed{\frac{3x-4}{(x-1)^2}=\frac{3}{x-1}+\frac{-1}{(x-1)^2}}$