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Calculate a closed interval of $\mathbb{R}$ that contains the roots of $$ x^5 + x^4 + x^3 + x^2 + 1 $$

Should I use the Intermediate Value Theorem and try to guess multiple points in which I know the function crosses the x-axis? I am pretty sure there is a smarter way of doing this, but I am not quite getting there.

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    If $f(x) = x^5+x^4+x^3+x^2+1$, $f(-2) = -19, f(-1) = 1$. There is a root between $-2$ and 1. There are no other real roots. This can be seen as follows: When $x>0$, all terms are positive and hence no roots. When $x < -2$, $x^5$ dominates and hence no zeros. – 2017-01-10

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Intuitively, as $x $ grows much larger the function goes to $\infty $; as $x $ gets "very" negative, the function goes to $-\infty $. All it takes is to try and measure that "very large $x $" and that "'very' negative $x $". One can do that by differentiating and finding the points from which the derivative will always be positive, if you increase $x $ or if you decrease $x $. Call them $a, b $ with $b > a $.

If that polynomial were to be negative at $a $, then it is negative to the left of $a $. Similarly for being positive to the right of $b $. If the polynomial does not have the sign you wish, just stretch the bounds a bit. For example, if your polynomial is positive at $a $, try at $a - 1$ or $a - 10$ or something you feel that will do. Call $a'$ to the point with $a' \leq a $ for which the polynomial is always negative to its left. Call $b' \geq b $ to the point for which the polynomial is always positive to its right. Then $[a', b'] $ is a closed interval containing all roots of the polynomial.

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Multiply your polynomial with $x-1$ to get $x^6-1$. Now you can see that all roots of your original polynomial are roots of unity. In particular you have for a root $\xi$ of $x^5+\ldots+1$, that $\vert\xi\vert=1$. What will your intervall be?