Factors of a polynomial over $\mathbb{C}$

Question

For one of the problems on my algebra homework I am asked to find the zeros of $p(t)=t^5+t+1$ over $\mathbb{C}$. I have factored it into $$ p(t) = (t^2+t+1)(t^3-t^2+1).$$ We can compute the zeros of the quadratic term easily from the quadratic formula, however I am having trouble with the cubic factor. Using Cardano's formula we can find the only real zero of the cubic factor, but I am stuck after this. Normally, I'd use synthetic division to find the quotient (quadratic) polynomial and then factor from there, but the real zero is irrational, which makes the above method intractable. Is there an efficient way of determining the remaining complex zeros of the cubic factor without using too much machinery from Galois theory (because we have not covered much of it in lecture)?

Answer 1

$q(t)=t^3-t^2+1$ has only one real root in the interval $(-1,0)$; let $\xi$ be such a root and $a+bi,a-bi$ the other two complex roots. By Viète's formulas we know that: $$ \xi + (a-bi) + (a+bi) = 1, $$ hence $a=\frac{1-\xi}{2}$, and we know also: $$ -1=\xi(a-bi)(a+bi)=\xi(a^2+b^2), $$ from which we can compute $b$ in terms of $\xi$.