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.
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).
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.