Let  $f\left(x\right)=a_5{x}^5+a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x$  where   $a_i’s$  are real and $f\left(x\right)=0$ has a positive root ${\alpha }_0.$ Then

$f\left(x\right)=a_5{x}^5+a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x$ Also, given that $f\left(x\right)=0$ has positive root $a_0$ Thus, the equation must have at least three real roots (as complex root occurs in conjugate pair). Thus $f\text{'}\left(x\right)=0$  has at lest two real roots as between two roots of $f\left(x\right)=0$, there lies at least one roots of $f\text{'}\left(x\right)=0$  Similary, we can say that $f"\left(x\right)=0$has at least one real root. Futher, $f\text{'}\left(x\right)=0$has one rootbetween roots $x=0$ and $x=a_0$  of $f\left(x\right)=0$