Is there a unique partial fraction decomposition for every rational polynomial

Question

Consider this

$$({2x-3})/({x^2-1})(2x+3)$$

Here can i do decomposition as

$$Ax+B/(x^2-1)+C/(2x+3)$$ Instead of $$A/(x-1)+B/(x+1)+C/(2x+3)$$

And if not then why?

Answer 1

The partial decomposition is unique providing that the factorization of the denominator $g$ is performed in terms of irreducible polynomials

$$g(x) = \prod_{i=1}^n p_i^{n_i}(x).$$

$p(x)=x^2-1$ is not irreducible. This explains why you got the two partial decompositions.

Answer 2

Note tha in your case there are three unknowns in RHS. If you simplify yowill get $A(x+1)(2x+3)+B(x-1)(2x+3)+C(x^2-1)=0x^2+2x-3$. Simplify this and equate the cpefficients of the same powers of $x$ on both sides, you get three eqns. to find $A,B,C$.

The process of partial fractions is interesting and requires some rules and it yields the unknown. $F(x)=\frac{P_m(x)}{P_n(x)}, m<n$. $P_n(s)$ should contain linear factors , yheir powers or quadratic factors. For a repeated linear factor, we write $$\frac{A}{(x-a)}+\frac{B}{(x-a)^2}+\frac{C}{(x-a)^3}$$ for $$\frac{1}{(x-a)^{3}}$$ For $\frac{1}{ax^2+bx+c}$, we write $\frac{Ex+G}{ax^2+bx+c}.$

For $\frac{1}{(ax^2+bx+)^2}$, we write $\frac{Ex+G}{ax^2+bx+c}+\frac{Ux+V}{(ax^2+bx+c)^2}.$ For a good example: $$\frac{1}{(x-1)(x-2)^2(x^2+1)^2(x^2+3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}+\frac{Dx+E}{x^2+3}+\frac{Gx+H}{x^2+1}+\frac{Jx+K}{(x^2+1)^2}.$$ Thre are 9 unknowns and then there will be 9 equations. Some time one may also go for $x^2+1=(x+i)(x-i).$ Normally, partial fractions have linear or quadratic factors.