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.

Here is another, slightly more sophisticated, take:
The other root is $\beta= \gamma ^2+2\gamma-4$, where $\gamma = \alpha^2+2\alpha-4$.
Expanding $\beta$ in terms of $\alpha$ and reducing it mod $\alpha^3+2\alpha^2-5\alpha+1$ gives $\beta=-\alpha^2 - 3 \alpha + 2$. Note that $\beta=g(\gamma)=g(g(\alpha))$, where $g(x)=x^2+2x-4$.
(This is because $\mathbb Q(\alpha)$ must be the splitting field of $x^3+2x^2-5x+1$ since it already contains two roots and so must contain the third. The Galois group is cyclic of order $3$ and so the roots are $\alpha$, $g(\alpha)$, $g^2(\alpha)$.)
