Partial Fractions: Decomposing a Rational Function

[PFuser1232]

Suppose we have a rational function $P$ defined by:
$$P(x) = \frac{f(x)}{(x-a)(x-b)}$$
This is defined for all $x$, except $x = a$ and $x = b$.
To decompose this function into partial fractions we do the following:
$$\frac{f(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$
Multiplying both sides by $(x-a)(x-b)$:
$$f(x) = A(x-b) + B(x-a)$$
Which again holds whenever $x$ does not equal $a$ or $b$.
To find $A$ or $B$, we set $x$ equal to $a$ or $b$, which is confusing. Didn't we state earlier that the rational function, and all steps that took us from the first equation to the last equation, are only valid when $x$ does not equal $a$ or $b$?

 

[Mark44]

Suppose we have a rational function $P$ defined by:
$$P(x) = \frac{f(x)}{(x-a)(x-b)}$$
This is defined for all $x$, except $x = a$ and $x = b$.
To decompose this function into partial fractions we do the following:
$$\frac{f(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$

In general, no. If f(x) happens to be a constant or a linear polynomial, the above is valid, but if f(x) is a higher degree polynomial or other function, it doesn't work. For example, it doesn't work if P(x) is $\frac{x^3}{(x - a)(x - b)}$.

Multiplying both sides by $(x-a)(x-b)$:
$$f(x) = A(x-b) + B(x-a)$$
Which again holds whenever $x$ does not equal $a$ or $b$.
To find $A$ or $B$, we set $x$ equal to $a$ or $b$, which is confusing. Didn't we state earlier that the rational function, and all steps that took us from the first equation to the last equation, are only valid when $x$ does not equal $a$ or $b$?

The equation f(x) = A(x - b) + B(x - a) is valid for all values of x, including a and b, so there is no problem setting x to either of these values.

Consider the equation $\frac{x^2 - 4}{x - 2} = x + 2$. The left side is not defined if x = 2. If we multiply both sides by x - 2, we get x² - 4 = (x + 2)(x - 2). Here, both sides are defined for all values of x. Since we're no longer dividing by an expression that could be zero, there are no longer any restrictions on x.

 

[PFuser1232]

In your example, when both sides of the equation are multiplied by $x-2$, isn't the fact that $x ≠ 2$ implied?
The value of $\frac{x-2}{x-2}$ is indeterminate if $x = 2$.

 

[Mark44]

In your example, when both sides of the equation are multiplied by $x-2$, isn't the fact that $x ≠ 2$ implied?

Yes. However, in the new equation, there are no restrictions on x.
The two expressions...
$\frac{x^2-4}{x-2}$ and x + 2
have identical values except when x = 2.
The expression on the left is undefined when x = 2, but the expression on the right is defined, and has a value of 4.

When you multiply both sides of the equation $\frac{f(x)}{(x - a)(x - b)} = \frac A {x - a} + \frac B {x - b}$, you get a new equation that does not involve division, and so is defined for all x, including x = a and x = b. So there is no problem in substituting either of these values to find your constants A and B.

The value of $\frac{x-2}{x-2}$ is indeterminate if $x = 2$.

Yes, of course.