If 27 a+9 b+3 c+d=0, then the equation 4 a x^3+3 b x^2+2 c x+d=0 has…

Consider the polynomial $f(x)$ given by

$f(x)=a x^4+b x^3+c x^2+d x$

$\Rightarrow f^{\prime}(x)=4 a x^3+3 b x^2+2 c d+d$

We have,

$f(0)=0$

and, $f(3)=81 a+27 b+9 c+3 d=3(27 a+9 b+3 c+d)=0$         [Given]

Therefore, 0 and 3 are roots of $f(x)=0$.

Consequently, by Rolle's theorem $f^{\prime}(x)=0$ i.e. $4 a x^3+3 b x^2+2 c x+d=0$ has at least one real root between 0 and 3.