<s>[INST] 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.
 [/INST] Given three roots $\alpha, \beta,  \gamma$ of a polynomial, it can generally be written as
$$(x-\alpha)(x-\beta)(x-\gamma)=0$$
Note how each of the roots contribute in making the equality work. On expanding the braces, 
$$(x-\alpha)(x-\beta)(x-\gamma)=0$$
$$[(x^2-x(\beta)-x(\alpha)+(\alpha\beta)](x-\gamma)=0$$
$$(x^3-x^2(\gamma)-x^2(\beta)+x(\beta\gamma)-x^2(\alpha)+x(\alpha\gamma)+x(\alpha\beta)-(\alpha\beta\gamma)=0$$
$$x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\alpha\gamma)x -(\alpha\beta\gamma)=0$$
Note that if any given equation is scaled down in such a way that the coefficient of $x^3$ is $1$ then the coefficient of $x^2$ gives the negative of sum of roots. More precisely, sum of roots of cubic equation = $-$(coefficient of $x^2$)/(coefficient of $ x^3$)
I hope you can take over from here. As pointed out already sum of roots is $-2$
Bonus: It can also be seen that product of roots of any given cubic equation is equal to the negative of the constant term divided by coefficient of $x^3$ and the sum of product of roots taken two at a time is, well, (coefficient of $x$)/(coefficient of $x^3$)
These are enough conditions on the given roots to get you started.
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