# Thread: Partial fractions : I don't understand my notes

1. ## Partial fractions : I don't understand my notes

My teachers shown an example of partial fraction in order to solve an integral. The problem is that I don't understand everything.
Say we have $\deg P(x)< \deg Q(x)$ and that $Q(x)=(x+1)(x-2)^3(x^2-4x+13)(x^2+3x+8)^2$.
Then my notes say $\frac{P(x)}{Q(x)}=\frac{a}{x+1}+\frac{b}{x-2}+\frac{c}{(x-2)^2}+\frac{d}{(x-2)^3}+\frac{f\cdot x+g}{x^2-4x+13}+\frac{h\cdot x+i}{x^2+3x+8}+$ $\frac{j\cdot x+k}{(x^2+3x+8)^2}$.
What I don't understand : why the numerators are not all constants, like a, b, c, etc. but appear to be lineal functions in the 3 last fractions?

2. Originally Posted by arbolis
My teachers shown an example of partial fraction in order to solve an integral. The problem is that I don't understand everything.
Say we have $\deg P(x)< \deg Q(x)$ and that $Q(x)=(x+1)(x-2)^3(x^2-4x+13)(x^2+3x+8)^2$.
Then my notes say $\frac{P(x)}{Q(x)}=\frac{a}{x+1}+\frac{b}{x-2}+\frac{c}{(x-2)^2}+\frac{d}{(x-2)^3}+\frac{f\cdot x+g}{x^2-4x+13}+\frac{h\cdot x+i}{x^2+3x+8}+$ $\frac{j\cdot x+k}{(x^2+3x+8)^2}$.
What I don't understand : why the numerators are not all constants, like a, b, c, etc. but appear to be lineal functions in the 3 last fractions?
They are linear factors when the denominators are quadratic.

3. I understood it better now. But I have a question :
They are linear factors when the denominators are quadratic.
What about the example $\frac{x^3}{(-6x^3+2x^2-5)^2}$, can it be "reduced" to fractions with a quadratic function as numerator? I'm asking this because as you said, they are linear factors when the denominators are quadratic, so I bet that there are quadratic factors when the denominators are cubic. Am I right to bet so?

4. Originally Posted by arbolis
I understood it better now. But I have a question : What about the example $\frac{x^3}{(-6x^3+2x^2-5)^2}$, can it be "reduced" to fractions with a quadratic function as numerator? I'm asking this because as you said, they are linear factors when the denominators are quadratic, so I bet that there are quadratic factors when the denominators are cubic. Am I right to bet so?
The cubic can be factorised into a linear factor and an irreducible quadratic factor. So the denominator consists of a repeated linear factor and a repeated irreducible quadratic factor.

The cubic is difficult to factorise. And once factorised, the algebra in getting the partial factions is grinding and tedious.

So I'm afraid that you have two very unpleasant jobs if you want to express this rational function in partial fraction form.

5. Originally Posted by arbolis
I understood it better now. But I have a question : What about the example $\frac{x^3}{(-6x^3+2x^2-5)^2}$, can it be "reduced" to fractions with a quadratic function as numerator? I'm asking this because as you said, they are linear factors when the denominators are quadratic, so I bet that there are quadratic factors when the denominators are cubic. Am I right to bet so?
Yes..

$\frac{x^3}{(-6x^3+2x^2-5)^2}=\frac{Ax^2+Bx+C}{-6x^3+2x^2-5}+\frac{Dx^2+Ex+F}{(-6x^3+2x^2-5)^2}$ is one possibility, but not the most advanced decomposition, like Mr Fastastic stated.

6. Thank you both. I never saw such an example; the one I posted directly came from my fingers, so I hope I won't find a similar one in my exam!