I'm trying to figure out this problem:
At the moment, I've got it down to this:
$\displaystyle 3x-4=A(x-1)+B(x-1)^2$
To find $\displaystyle B$, I let $\displaystyle x=1$, which also gets rid of $\displaystyle A$. Any ideas?
I'm trying to figure out this problem:
At the moment, I've got it down to this:
$\displaystyle 3x-4=A(x-1)+B(x-1)^2$
To find $\displaystyle B$, I let $\displaystyle x=1$, which also gets rid of $\displaystyle A$. Any ideas?
Hi BeSweet,
$\displaystyle \frac{3x-4}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}$
There must be a partial fraction present for each power of $\displaystyle (x - c)^n$ less than or equal to n. We can find B, but not A using Heaviside's method. Only the constant corresponding to $\displaystyle (x-c)^n$ can be found using this method.
First, multiply the original proper fraction by $\displaystyle (x-c)^2$ and evaluate the result at x = c to get the value of the constant.
$\displaystyle \frac{3x-4}{(x-1)^2}(x-1)^2=3x-4 \Rightarrow B=3(1)-4=-1$
$\displaystyle \boxed{B=-1}$
$\displaystyle \frac{3x-4}{(x-1)^2}=\frac{A}{x-1}+\frac{-1}{(x-1)^2}$
To find the other constant, multiply both sides by the least common denominator to clear the fractions:
$\displaystyle 3x-4=A(x-1)-1$
This leads to:
$\displaystyle 3x-4=Ax + (-1-A)$
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.
$\displaystyle \boxed{A=3}$
You could also say $\displaystyle -4=(-1-A)$
You still get $\displaystyle \boxed{A=3}$
The decomposition is:
$\displaystyle \boxed{\frac{3x-4}{(x-1)^2}=\frac{3}{x-1}+\frac{-1}{(x-1)^2}}$