Find the galois group of the polynomial when a root is given

If $\alpha$ is a root of a polynomial $f(x)=x^3 +x^2-4x+1$ then show that $2 - 2\alpha - \alpha^2$ is also a root of $f(x)$. Use this fact to compute the Galois group of the splitting field of $f(x)$ over $\mathbb{Q}$.

I am just introduced to field theory. I can determine the Galois group of the splitting field of the given polynomial by computing its discriminant.

But i have no idea how can we find the Galois group by the use of given roots.

Any hints will be highly appreciated. Thank you.

To show that $2-2\alpha-\alpha^2$ is a root, we can just use brute force: $$\begin{eqnarray*}f(2-2\alpha-\alpha^2)&=&\left(-\alpha ^2-2 \alpha +2\right)^3+\left(-\alpha ^2-2 \alpha +2\right)^2-4 \left(-\alpha ^2-2 \alpha +2\right)+1\\ &=&-\alpha ^6-6 \alpha ^5-5 \alpha ^4+20 \alpha ^3+16 \alpha ^2-24 \alpha +5 \\ &=&-\left(\alpha ^3+\alpha ^2-4 \alpha +1\right) \left(\alpha ^3+5 \alpha ^2+4 \alpha -5\right)\\ &=&-0\times \left(\alpha ^3+5 \alpha ^2+4 \alpha -5\right)\\ &=&0 \end{eqnarray*}$$ Now that that's verified, we know that $\alpha \mapsto 2-2\alpha-\alpha^2$ is an element of the Galois group. So let's plug $2-2\alpha-\alpha^2$ into $\alpha \mapsto 2-2\alpha-\alpha^2$ to get $$-\alpha ^4-4 \alpha ^3+2 \alpha ^2+12 \alpha -6\equiv \alpha ^2+\alpha -3\pmod{\alpha ^3+\alpha ^2-4 \alpha +1}$$ And of course then plugging $\alpha ^2+\alpha -3$ into $\alpha\mapsto -\alpha ^2-2 \alpha +2$ yields $$-\alpha ^4-2 \alpha ^3+3 \alpha ^2+4 \alpha -1\equiv \alpha \pmod{\alpha ^3+\alpha ^2-4 \alpha +1}$$ That means $\alpha \mapsto 2-2\alpha-\alpha^2$ is a three cycle.
I think that if you note that the sum of the roots of $f(x)$ is $1$ then you can see that all the roots of $f$ are linear combinations of $\alpha$. So all the roots of $f$ are contained in $\mathbb{Q}(\alpha)$. Any splitting field contains $\mathbb{Q}$ and $\alpha$ so this is the splitting field of $f$ over $\mathbb{Q}$.