Find the PFD of the folowing:
$$\frac{x^6-x^5+48x^3-53x^2+99x+48}{x^5-2x^4+2x^3-4x^2+x-2}$$ Initially I used long division and got $$x+1+\frac{50(x^3-x^2+2x+1)}{x^5-2x^4+2x^3-4x^2+x-2}$$ I then used trial and error to factor the denominator as: $$x+1+\frac{50(x^3-x^2+2x+1)}{x^5-2x^4+2x^3-4x^2+x-2}=x+1+\frac{50(x^3-x^2+2x+1)}{(x-2)(x^4+2x^2+1)}$$ How do I factor the second part of the denominator?
The second part of the denominator factors as $$ (x^4 + 2x^2 + 1) = ((x^2)^2 + 2(x^2) + 1) = (x^2 + 1)^2 $$ Therefore, your partial fraction decomposition is of the form $$ \frac{A}{x-2} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2} $$