Finding the real zeros of $m(t)=t^7-3t^6+4t^3-t-1$

Question

Finding the real zeros of $m(t)=t^7-3t^6+4t^3-t-1$.

Using the rational theorem, I have found that $m(t)$ has $1$ as a zero with multiplicity of $2$. So $$m(t)=(t-1)^2(t^5-t^4-3t^3-5t^2-3t-1).$$ However, I don't know or can't see a trick to factor the second polynomial.

Answer 1

$$m(t)=t^7-3t^6+4t^3-t-1$$.

Let's replace $t^3$ with 1 wherever we see it. we get

$t - 3 +4 - t -1 = 0 $ For any $t$ given $ t^3=1$.

Hence $(t^3-1) | m(t)$.

So $$m(t) = (t^3-1)(t^4 -3t^3+t+1)$$ after some Euclidean division.

The second polynomial factor has root 1, so $$ m(t) = (t^3-1)(t-1)(t^3-2t^2-2t-1) $$ after more Euclidean division.

We know that the first polynomial factor has one real root, namely 1. The second one has one real root, again 1. The third one has a real root, that is horrible to express, where did you get the polynomial? Because the root would be cumbersome to express by hand.

In more generality, if we have a polynomial $p(t)$ with a root $\sqrt[m]{q}$. And we replace $t^m$ with $q$ in a similar fashion as above, and get a similar result, i.e we get that $p(t) = 0$, for any $t^m=q$, so we must have $(t^m-q) | f(t)$.

Aside: I worked this out during my course on Galois Theory, so I would appreciate any literature on the subject.

Answer 2

Trick

One of the tricks I use to factor school problems like this one where the coefficients kind of add to zero, (and these usually work a third of the times), is to put $\color{blue}{t = \omega}$. Then you get $t^7 - 3t^6 + 4t^3 - t - 1 = \omega - 3 + 4 - \omega - 1 = \color{blue}{0}$.

Hence $\color{blue}{t^2 + t + 1}$ is also a factor. The rest is division.

So finally you have $(t-1)^2(t^2 + t + 1)(t^3 - 2t^2 - 2t - 1)$ as the final factors.