Use the partial fractions technique when you're asked to evaluate a rational function that

• has a lower degree in the numerator than in the denominator, and
    
• has a denominator that can be factored into distinct linear factors.

Sample Problem

We can use the method of partial fractions on

because the numerator has degree 0, the denominator has degree 2, and the denominator factors into

x² – 2x – 3 = (x – 3)(x + 1).

Sample Problem

We wouldn't use the method of partial fractions on

because the denominator factors into

x² + 2x + 1 = (x + 1)(x + 1).

These are not distinct linear factors.

Actually, it is possible to use the method of partial fractions on this example, but you don't need to know it for the AP exam or for most calculus classes.