Finding remaining polynomial after finding complex factors I want to express this polynomial as a product of linear factors:
$x^5 + x^3 + 8x^2 + 8$
I noticed that $\pm$i were roots just looking at it, so two factors must be $(x- i)$ and $(x + i)$, but I'm not sure how I would know what the remaining polynomial would be. For real roots, I would usually just do use long division but it turns out a little messy in this instance (for me at least) and was wondering if there was a simpler method of finding the remaining polynomial. 
Apologies for the basic question!
 A: If you divide $$ x^5 + x^3 + 8x^2 + 8$$ by $$(x-i)(x+i) = x^2+1$$ you will get $$x^3+8$$ which factors as $$(x^3+8) = (x+2)(x^2-2x+4)$$ which has a solution of $x=-2$
Now use quadratic formula to solve $x^2-2x+4=0$ to find other roots and factor if you wish.
A: If you spotted this by looking at it you have good intuition. When roots come in complex pairs you can always combine them to find a quadratic factor with real coefficients. Here $(x+i)(x-i)=x^2+1$
If you could do the first bit with intuition, I am sure you can do that division and complete the factorisation.
A: Yo can also do long division for complex polynomials. But in this case, I would suggest to pull out the two factors at once, i.e., divide by their product, which is $(x^2+1)$.
$(x^5+x^3+8x^2+8):(x^2+1)= x^3+8$. 
You can easily factorize this polynomial. (Hint: $8=2^3$.)
A: Alt. hint:   you might as well notice that $\,x=-2\,$ is a root by the rational root theorem. Then if the remaining two roots are $\,a,b\,$, by Vieta's relations $\,i+(-i)+(-2)+a+b=0 \iff a+b = 2\,$ and $\,i \cdot (-i)\cdot(-2)\cdot a \cdot b = -8 \iff a \cdot b = 4\,$, so $\,a,b\,$ are the roots of $\,x^2 - 2 x + 4 = 0\,$.
