A more rigorous way to show that $x^5 - 3x = 1$ has at least $3$ roots in $\Bbb R$

Let us use Descartes' sign Rule

$f(x)=x^5-3x-1=0$, the number of sign changes in $f(x)$ is one and number of sigm changes in $f(-x)$ are two. So as per Descarte's rule this equation can have atmost one real positive and atmost two real negative roots. So Descarte's rules permits at most three real root which can be located as : $f(-\infty)<0, f(-1)=+1, f(0)=-1$ means there is at least one real root in $(-\infty, -1)$ and at least one in $(-1,0)$. Next $f(1)=-3, f(2)=+25$. So one real positive roots is in $(1,2)$. In all this equation has three real roots two negative and one positive.

Graphical method can help. $$x^5-3x=1 \stackrel{x\ne 0}{\iff} x^4-3=\frac1x$$ Sketch:

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Note: You can quickly jump to the intervals of interest and compare the values and increase/decrease of the functions.

(1)$f(x)=x^5-3x-1$ has one sign change in its coefficients so at most $1$ positive root. Also, $f(1).f(2)<0$ assures exactly one positive root lying in $(1,2)$.

(2) $f(-x)=-x^5+3x-1$ has two sign changes in its coefficients so at most $2$ negative roots OR more precisely $0$ or $2$ negative roots (possibility of exactly $1$ negative root is discarded due to pairwise occurrence of remaining complex conjugate roots). Now $f(-1).f(0)<0$ is sufficient to assure the existence of exactly $2$ negative roots of $f(x)$.