Partial Fractions: Explained

[FS98]

im a bit confused about partial fractions

If we have something like x/((x+1)(x+2)) we could decompose it into a/(x+1) +b/(x+2)

If we had something like x/(x+1)^2 we could decompose it into a/(x+1) + b/(x+1)^2

We use a different procedure when there is a square in part of the polynomial in the denominator.

What I don’t understand is why in the second case it wouldn’t be split into a/(x+1) + b/(x+1).

x/(x+1)^2 can be written as x/(x+1)(x+1). Then couldn’t we just use the procedure used in the first case?

 

[fresh_42]

In the decomposition to partial fractions, we use the fact that factors are coprime. For coprime elements $n,m$ we can find other elements $a,b$ such that $1=an+bm$ by the Euclidean algorithm and the $1$ can be extended to what we need.

Say we want to solve $\dfrac{z}{n\cdot m} = \dfrac{q}{n}+\dfrac{p}{m}$ with coprime $n,m$ and we found $a,b$ such that $1=an+bm$. Then we have $z= qm+pn=z\cdot 1 = zan+zbm$ and thus $q=zb$ and $p=za$ or $\dfrac{z}{n\cdot m}=\dfrac{zb}{n}+\dfrac{za}{m}$. All this because of Bézout's identity, resp. the fact that $n,m$ are coprime.

 

[Mark44]

What I don’t understand is why in the second case it wouldn’t be split into a/(x+1) + b/(x+1).

The simple answer is because it doesn't work. Notice that $\frac a {x + 1} + \frac b {x + 1}$ is really the same as $\frac c {x + 1}$, so you won't get that second term with the squared factor in the denominator.