Hi guys, this is my last post on partial fraction.

Today we will discuss about partial fractions of irreducible quadratic factors. We already know about the partial fraction of an expression where the quadratic factor can be factorised.

But there are also some cases where the denominator has an irreducible quadratic factor. That factor cannot be factorised into two simple factors.

Today we will discuss those cases.

Let’s start then.

#### Partial fractions of irreducible quadratic factors with example

Write the following expression into partial fraction form:

.

We can rewrite the expression as

where and are constants.

Now we simplify the right-hand side of the equation.

Now the denominator of both left-hand side and right-hand side are the same. So we can compare numerators of both sides.

.

We can compare the coefficients of on both sides.

(1)

Now we compare the coefficients of on both sides.

(2)

Finally, we compare the constants on both sides.

(3)

Therefore we have three linear equations (1), (2) and (3) with three unknowns and . Now our task is to solve them to get the values of and . I have already discussed before how to solve linear equations.

Now we substitute the value of equation (1) in equation (2).

Now we substitute this value of in equation (3).

At the end, we substitute in equation (1).

Therefore we can say is the partial fraction form of .

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