<s>[INST] Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots 
Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.

By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 roots.
But I wonder that there is some logical ways we can find intervals that contain roots without guessing?
 [/INST] We may apply Sturm's theorem https://en.wikipedia.org/wiki/Sturm%27s_theorem.
Let $\mathrm{rem}(p(x),q(x))$ denote the remainder of $p(x)$ divided by $q(x)$ for two polynomials $p(x), q(x)$.
The Sturm sequence is given by:
$p_0(x) = x^5-5x^3+4x-1$
$p_1(x) = p_0'(x) = 5x^4 - 15x^2 + 4$
$p_2(x) = -\mathrm{rem}(p_0, p_1) = 2x^3 - \frac{16}{5}x + 1$
$p_3(x) = -\mathrm{rem}(p_1, p_2) = 7x^2 + \frac{5}{2}x - 4$
$p_4(x) = -\mathrm{rem}(p2, p_3) = \frac{883}{490}x - \frac{29}{49}$
$p_5(x) = -\mathrm{rem}(p3, p4) = 1889881/779689$
As this is a constant, this finishes the computation of the Sturm sequence.
We have $V(-\infty) - V(+\infty) = 5 - 0 = 5$.
So there are five real roots.
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