Partial Fractions Examples

Write the terms for partial fraction decomposition but don't solve for the constants.

Since the denominator can be factored, we will rewrite as:

Our partial fraction decomposition will now look like this:

Take the previous example and solve for the constants in the partial fractions.

First, we will clear the fractions by multiplying each side by (x – 4)(x + 1).

5x + 1 = A(x + 1) + B(x – 4)

Try x = -1:

Try x = 4:

So...

Find the partial fraction decomposition of:

Factor the bottom and setup constants:

Clearing fractions by multiplying each side by (2x – 3)(x + 1):

7x – 4 = A(x + 1) + B(2x – 3)

Now let's try to solve by choosing values of x that will cancel out terms.

Try x = -1:

Try .

Why? Because we want 2x – 3 to be zero so we solved for 2x – 3 = 0 which gave us an x value of :

Yes! Our partial fraction decomposition is: