Find the minimal polynomial of $t^2+t$ over $\mathbb{Q}$ where t satisfies $x^3-3x^2-3$ Find the minimal polynomial of $t^2+t$ over $\mathbb{Q}$ where t satisfies $t^3-3t^2-3=0$.
Okay, so I was working on this one for awhile today with my buddy and we couldn't figure it out, haha. We got creative with this and tried a lot of stuff but couldn't figure it out, so i'm betting somebody here makes it look really easy like you always do.
One thing we did was try plugging $t^2+t$ into $x^3-3t^2-3$ and trying to look for clues or even make it zero. Another route I took was dividing $\frac{x^3-3x^2-3}{x-t}$ and yeah of course it divides without remainder but yeah I don't know haha I need a new perspective
 A: A more conceptual method:
If $t$ is a root of $t^3-3t^2-3=0$, then $t$ is an eigenvalue of the matrix
$$A = \begin{pmatrix}0 & 0 & 3 \\ 1 & 0 & 0 \\ 0& 1 & 3\end{pmatrix}$$
hence $t^2+2t$ is an eigenvalue of the matrix $$A^2+2A = \left(
\begin{array}{ccc}
 0 & 3 & 15 \\
 2 & 0 & 3 \\
 1 & 5 & 15 \\
\end{array}
\right)$$ computing characteristic polynomial shows $t^2+2t$ satisfies $x^3-15 x^2-36 x-69 = 0$. It is indeed the minimal polynomial because $[\mathbb{Q}(t^2+2t):\mathbb{Q}] = 3$.
However, in a computational point of view, this method is quite expensive when dimension of matrix is large.
A: Let $\alpha=t^2+t$.
Then $$\begin{align}\alpha^2&=t^4+2t^3+t\\&=(t+2)t^3+t\\&=(t+2)(3t^2+3)+t\\&=3t^3+6t^2+4t\\&=3\cdot(3t^2+3)+6t^2+4t\\&=15t^2+4t+9\end{align}$$
and
$$\begin{align}\alpha\cdot(\alpha^2-15\alpha)&=(t^2+t)\cdot(-11t+9)\\
&=-11t^3-2t^2+9t\\&=-11(3t^2+3)-2t^2+9t\\&=-35t^2+9t-33\end{align} $$
so that
$$\alpha^3= \alpha\cdot(\alpha^2-15\alpha)+15\alpha^2=190t^2+69t+102.$$
Now find a linear dependency between
$$\begin{matrix}\alpha^3=&190t^2&+69t&+102\\
\alpha^2=&15t^2&+4t&+9\\
\alpha^1=&t^2&+t\\
\alpha^0=&&&1\end{matrix} $$
