# Values tried for partial fraction decomposition

I'll explain my question with the following example from wikipedia. Suppose, we have a function: $$f(x)=\frac{1}{x^2+2x-3}$$ Here, the denominator splits into two distinct linear factors: $$q(x)=x^2+2x-3 = (x+3)(x-1)$$ so we have the partial fraction decomposition $$f(x)=\frac{1}{x^2+2x-3} =\frac{A}{x+3}+\frac{B}{x-1}$$ Multiplying through by $x^2 + 2x − 3$, we have the polynomial identity $$1=A(x-1)+B(x+3)$$ Substituting $x = −3$ into this equation gives $A = −1/4$, and substituting $x = 1$ gives $B = 1/4$, so that $$f(x) =\frac{1}{x^2+2x-3} =\frac{1}{4}\left(\frac{-1}{x+3}+\frac{1}{x-1}\right)$$ My doubt: For $x = -3$ and $x = 1$, our function $f(x)$ is undefined. So, is it valid to substitute these values for $x$? If it is, then why? And if it isn't, then why do these values work?

• Just to make my point more clear, please note that when we multiply through $x^2+2x-3$, we'll be multiplying both sides by 0 if $x=-3$ or $x=1$. I really do not see how substituting these particular values for $x$ could be a valid operation. Commented May 11, 2015 at 4:22
• Don't worry ! Your derivation of the parameters $A$ and $B$ is entirely valid and the result is correct. Even if $x = -3$ or $x = 1$ were noy valid points, then still the LHS and RHS should be equal for $x$ very close to these values. In other words, it is impossible that some other solution exists. And your solution is perfect. Commented May 11, 2015 at 5:08

Here is another way of looking at eliminating $A$, which may help. Set $x=1+t$ and $x=1-t$ to obtain:
$$1=At+B(4+t)$$ $$1=-At+B(4-t)$$
Now add these two equations and divide by $2$ to obtain $$1=4B$$
This is exactly what you would have got by putting $x=1$ in the original equation.
• Cool way of eliminating A. How did you come up with $x=1+t$ and $x=1-t$? Commented May 11, 2015 at 15:15
• @shobhu Well they are symmetric around the "problem" point, so the first order term is going to vanish. The $1$ in $1\pm t$ makes the constant term in $A$ vanish and the symmetry deals with the first order term. This is true even in a more complicated case where quadratic terms appear. Equating coefficients of the equivalent of $t$ will eliminate the equivalent of $A$. This process should remind you of differentiation - derivatives do come into the theory when there are double roots in the denominator. It is usually more practical to spot short cuts than to learn a full theoretical solution. Commented May 11, 2015 at 15:27