Today we will discuss again on partial fractions. We will focus on the expressions with higher degree numerators. Our aim will be to reduce its degree first. Then we will try to get partial fractions of the expression. By now, we all know how to find out the degrees of the numerator and the denominator. So let’s start with a few examples.
Partial fractions of higher degree numerators with examples
Write the following expression into the partial fraction form:
Here the degree of the numerator is 3. The denominator’s degree is 2. First of all, we divide the numerator with the denominator.
So we can rewrite the expression as
Now is reduced to two parts:
one is and
the other is .
The first part cannot be reduced any more. But we can still reduce the second part. So let’s start then.
Like in previous posts, here also we start with the factorisation. Then we use the assumptions.
Now we have the same denominator on both left hand and right hand sides. Therefore we can say,
We can compare the coefficients of on both sides. Then we get
We can also compare constants on both sides. From that we have
We substitute this value of in equation (1). Then we have
So the partial fraction of is .
Therefore we can say that the partial fraction of is . This is the solution of the problem.
In my next post I will discuss more on the partial fractions of expression where repeated roots come in the denominator. So bye for today and enjoy reading!!