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?
 A: Given $f(x)=x^5-5x^3+4x-1$, differentiate first:
$$ f'(x) = 5x^4 - 15x^2 +4 . $$
Setting the derivative equal to $0$, we have
$$ 0 = 5x^4 - 15x^2 +4 . $$
This is quadratic in $x^2$, so we can use the quadratic formula:
$$ x^2 = \frac{15 \pm \sqrt{145}}{10} . $$
Since $\sqrt{145}<15$, $x^2>0$, so our solutions are
$$ x = \pm \sqrt{\frac{15 \pm \sqrt{145}}{10}} . $$
So, we have four distinct local extrema for $f$. If there are five distinct roots, each must be on one of the following intervals:
$$ \left(-\infty, -\sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15+\sqrt{145}}{10}}, -\sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15+\sqrt{145}}{10}}, \infty\right) . $$
A: 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.
A: If you rewrite the equation as
$$x^4-5x^2+4={1\over x}$$
and then sketch the curves $y=x^4-5x^2+4=(x^2-1)(x^2-4)$ and $y=1/x$ sufficiently carefully, you can see that the curves cross at two points in the interval $(-2,-1)$, two more points in the interval $(0,1)$, and one final point in the interval $(2,\infty)$.
The intersections in $(0,1)$ are the hardest to see (you might miss them if your sketch is too sketchy, as mine was at first), but are easily confirmed by comparing the values of $y$ at $x=1/2$, since the hyperbola clearly lies above the polynomial curve at $x=1$ and as $x\to0^+$. Doing so tells you one root lies in the interval $(0,1/2)$ and the other in the interval $(1/2,1)$. Similarly, comparing the $y$ values at $x=-3/2$ splits $(-2,-1)$ into $(-2,-3/2)$ and $(-3/2,-1)$, each with one root. And finally, it's easy to see that $(3^2-1)(3^2-4)\gt1/3$, so the largest root lies in the interval $(2,3)$.
Note, this curve-sketching approach works as nicely as it does only because of the special form of the equation, in particular the nice factorization of $x^4-5x^2+4$, which makes that curve's general shape easy to draw. It's not an approach you can always rely on to be helpful, but it's worth keeping in mind for occasions when it is.
A: I would start with the rational root theorem, which fails but gets you values at $x=\pm1$.
Because the coefficients are small, the roots are all close to $0$.  Checking values at the integers close to $0$ is a good approach.  Once you go from $-3$ to $+3$ you have found one root in $(2,3)$.  You should be able to convince yourself there are not roots outside $(-3,3)$ because the $x^5$ term dominates.  Now checking at half integers is a good next step and finds the roots.  There was no guessing involved.  The real question comes if you don't find the roots with the half integral evaluations.  Do you trust the problem setter to guarantee there are five roots?  If so, trying quarters is the next logical step.  Or do you worry the question is wrong?  I have no advice here.
