I'm on my last problem of a matrix theory project.
I need to describe the basis of P3 that partial fractions asserts for the polynomial
$$q(x)=(x^2+2x+2)(x^2-x+2)$$ with degree 4
Now, I understand how to describe the basis once I figure out what the polynomial of p(x) looks like. The problem is I do not understand how to find p(x) by using partial fractions, that is I do not know how to arrive at the following:
$$\frac{p(x)}{q(x)}=\frac{b_1}{?}+\frac{b_2}{?}+\frac{b_3}{?}+\frac{b_4}{?}$$
because q(x) is an irreducible quadratic.
I was thinking I need to complete the square but that leaves me confused.
For instance, $$(x^2+2x+2)$$ is
$$(x+1)^2+1$$
and $$(x^2-x+2)$$ is $$(x-\frac{1}{2})^2+\frac{7}{4}$$
So does this mean the equation should look like
$$\frac{p(x)}{q(x)}=\frac{b_1}{(x+1)^2+1}+\frac{b_2}{(x+1)+1}+\frac{b_3}{(x-\frac{1}{2})^2+\frac{7}{4}}+\frac{b_4}{(x-\frac{1}{2})+\frac{7}{4}}$$
There are two ways to do this: leave the quadratics unfactored, or play with complex numbers.
With the first, your expansion will look like this: $$ \frac{p(x)}{q(x)} = \frac{b_1 x + b_2}{x^2 + 2x + 2} + \frac{b_3 x + b_4}{x^2 - x + 2} $$
Alternatively, you can just factor them, and have: $$ \frac{p(x)}{q(x)} = \frac{z_1}{x + (-1 + i)} + \frac{z_2}{x + (-1 - i)} + \frac{z_3}{\frac{1}{2} (1 + \sqrt{7}i)} + \frac{z_4}{\frac{1}{2} (1 - \sqrt{7}i)} $$
Frankly, I think having quadratics in the denominator is a small price compared to these unpleasant roots. Plus, the $z$s probably aren't nice either.