Factorize the polynomial $P(z) = z^4 - 2z^3-z^2+2z+10$, into linear and/or quadratic factors with real coefficients $2+i$ is given to be one of the roots of the polynomial.
I am doing this as a practice for exam prep.
Since $2+i$, is a root, then $(z-2-i)$ is a factor?
So I have:
$(z-2-i)(z^3-Az^2-Bz+C) = z^4-2z^3-z^2+2z+10$
Then, I group the $z$ terms of the same degrees after I multiply out?
Am I on the right track? Because it seems like this will take quite a while in exam conditions? Thanks for any direction!

Since the question has the "calculus" tag, letus use a slightly different method. We start with $$(z^2-4z+5)(z^2+bz+c)= z^4-2z^3-z^2+2z+10.$$ as smanoos does. When we plug zero int both sides, we get $5c=10$, so $c=2$. This gives us the equation, 
$$(z^2-4z+5)(z^2+bz+2)= z^4-2z^3-z^2+2z+10.$$. Now letus take the derivative of both sides to get,
$$(2z-4)(z^2+bz+2)+(z^2-4z+5)(2z+b)=4z^3-6z^2-2z+2$$. Setting $z=0$ gives,
$$-8+5b=2$$, thus $b=2$.
So we factored the polynomial as $(z^2-4z+5)(z^2+2z+2)$, but then we know how $(z^2+2z+2)$ factors, and we must factor $(z^2+2z+2)$, but for that we will refer to the earlier answers.
