# Partial fractions

1. Mar 21, 2015

### MohammedRady

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 $b$?

2. Mar 21, 2015

### Staff: Mentor

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 $\frac{x^3}{(x - a)(x - b)}$.
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.

Last edited: Mar 21, 2015
3. Mar 21, 2015

### MohammedRady

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 $x = 2$.

4. Mar 21, 2015

### Staff: Mentor

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.
Yes, of course.

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