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.
 A: Note 
$$\alpha (\alpha^2+2\alpha -4)\beta =-1$$
Thus,
$$\beta =- \frac1{\alpha (\alpha^2+2\alpha -4)}
 =- \frac1{\alpha^3+2\alpha^2-4\alpha}
 =- \frac1{5\alpha -1 -4\alpha} =\frac1{1-\alpha}
$$
A: Since by Vieta's formulas the sum of the roots is $-2$, there is not much to do here: given that two roots are $\alpha$ and $\alpha^2+2\alpha-4$, the third one has to be $-\alpha^2-3\alpha+2$.

The interesting question, however, is: how do we realize that there are two 
   roots $\alpha,\beta$ fulfilling $\beta=\alpha^2+2\alpha-4$ ?

Well, the discriminant of the polynomial is $361=19^2$, so all the roots are real  and the Galois group over $\mathbb{Q}$ is not $S_3$ but something simpler. By considering the depressed cubic we have
$$ \frac{27}{38\sqrt{19}}\,\underbrace{p\left(\frac{2}{3}(x\sqrt{19}-1)\right)}_{q(x)}=4x^3-3x+\frac{7}{2\sqrt{19}} $$
so by trigonometry we have that a root is given by 
$$ \zeta = -\frac{2}{3}+\frac{2}{3}\sqrt{19}\cos\Big(\underbrace{\frac{1}{3}\arccos\left(\frac{-7}{2\sqrt{19}}\right)}_{\theta}\Big) $$
and the other roots are given by
$$ -\frac{2}{3}+\frac{2}{3}\sqrt{19}\cos(\theta+2\pi/3)\qquad\text{and}\qquad -\frac{2}{3}+\frac{2}{3}\sqrt{19}\cos(\theta+4\pi/3) $$
i.e. by 
$$ -\frac{2}{3}-\frac{1}{3}\sqrt{19}\cos(\theta)-\frac{1}{\sqrt{3}}\sqrt{19}\sin(\theta)\qquad\text{and}\qquad -\frac{2}{3}-\frac{1}{3}\sqrt{19}\cos(\theta)+\frac{1}{\sqrt{3}}\sqrt{19}\sin(\theta) $$
which are clearly related (to each other, and to $\zeta$) via the Pythagorean theorem $\sin^2\theta+\cos^2\theta=1$.
A: 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 $\alpha$, $g(\alpha)$, $g^2(\alpha)$.)
A: 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 $$p(x)/q(x) = x - (- \alpha^2 - 3 \alpha + 2).$$
A: 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.
