Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha^3 = -2\alpha^2+5\alpha -1$. 
And, $(\alpha^2+2\alpha-4)^3+2(\alpha^2+2\alpha-4)^2-5(\alpha^2+2\alpha-4)+1=0$ gives $\alpha^6+6\alpha^5+2\alpha^4-13\alpha^2+54\alpha-11=0$
Additionally, $\alpha^6 = (-2\alpha^2+5\alpha -1)^2 = 4\alpha^4-20\alpha^3+29\alpha^2-10\alpha+1$
I do not know how to move forward from here. I tried setting $f(x)$ equal to the product of the roots and expanding that out to a 14-term polynomial with $\alpha$ and $\beta$ coefficients but that seems unproductive.

Call the roots $\alpha, \alpha', \alpha''$ with $\alpha' = \alpha^2 + 2\alpha - 4$.
$$p(x) = x^3 + 2x^2 - 5x + 1 = (x - \alpha)(x - \alpha')(x - \alpha'') = q(x)(x - \alpha'')$$
with $$q(x) = x^2 + (-\alpha - 3 \alpha + 4) x + (\alpha^3 + 2 \alpha^2 - 4\alpha) = x^2 + (-\alpha - 3 \alpha + 4) x + (\alpha - 1)$$
(replacing $\alpha^2$ with $-2\alpha + 4$)
We can now compute the polynomial long division $p(x)/q(x)$ in $\mathbb Z(\alpha)$ to get $x - \alpha''$:
$$p(x) - x q(x) = (\alpha^2 + 3 \alpha - 2) x^2 + (-\alpha - 4) x + 1$$
$$p(x) - x q(x) - (\alpha^2 + 3 \alpha - 2) q(x) = 0$$
so $$p(x)/q(x) = x - (- \alpha^2 - 3 \alpha + 2).$$
