Hi guys, today I will discuss partial fractions of irreducible quadratic factors. I already talked about the partial fraction of an expression where the quadratic factor can be factorised.

You can also check out related posts like how to determine partial fractions of higher degree numerators. Or, the way to get the partial fractions of expressions where both the numerator and the denominator have the same degree.

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

Today I will discuss those cases.

Let’s start then.

### Partial fractions of irreducible quadratic factors with example

##### Example

Write the following expression into partial fraction form:

.

##### Solution

Here the given expression is: .

I can rewrite the expression as

Here 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 .

This is the solution to the given example.

Dear friends, this is the end of my today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

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