Let $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}_5{\mathrm{x}}^5+{\mathrm{a}}_4{\mathrm{x}}^4+{\mathrm{a}}_3{\mathrm{x}}^3+{\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{a}}_1\mathrm{x},$ where a_i's are  real and f (x) = 0 has a positive root ${\alpha }_0$. Then

Clearly, f (0) = 0. So, f(x) = 0 has two real roots 0, ${\mathrm{\alpha }}_0\left(>0\right)$.

Therefore, $\mathrm{f}\text{'}\left(\mathrm{x}\right)=0$ has a real root  ${\mathrm{\alpha }}_1$ lying between 0 and ${\mathrm{\alpha }}_0$

So, $0<{\mathrm{\alpha }}_1<{\mathrm{\alpha }}_0$

Again, f '(x) = 0 is a fourth-degree equation. As imaginary  roots occur in conjugate pairs, f '(x) = 0 will have another real root ${\mathrm{\alpha }}_2$. Therefore,  f"(x) = 0 will have a real root lying between ${\mathrm{\alpha }}_1 \mathrm{and} {\mathrm{\alpha }}_2$.  As f(x) = 0 is an equation of the fifth degree, it will have at least three real roots and, so f '(x) will have at least two real roots.