Don't solve it!

Let $p(x) = x^4 + x^3 + x^2 - x - 1 = 0 \Rightarrow$ $p(-1) = 1 > 0$ and $p(0) = - 1 < 0 \Rightarrow$ $p(x)$ has a negative real solution $x \in (-1,0)$ due to Intermediate value theorem . Furthemore, $p'(x) = 4x^3 + 3x^2 + 2x - 1$ and $p''(x) = 12x^2 + 6x + 2 > 0, \space \forall x \in \mathbb{R}$ since its discriminant is less than 0. This implies that $p'(x)$ is a strictly increasing function with $\displaystyle \lim_{x \to -\infty} p'(x) = -\infty \wedge \lim_{x\to\infty} p'(x) = \infty$ (this means $p(x)$ is strictly decreasing in an interval $(-\infty, a)$ and strictly increasing in an interval $(a,\infty)$ )and hence $p(x)$ has at most two real roots, and now applying again the Intermediate value theorem and due to $p(0) = -1$ and $p(1) = 1$ we can deduce the solution, i.e, the polynomial above only has a negative real root, a positive real root and 2 complex roots (conjugated)