If a+b+c=0, then the equation 3 a x^2+2 b x+c=0 has, in the interval…

Consider the polynomial $f(x)$ given by

$f(x) =a x^3+b x^2+c x, \quad x \in[0,1]$

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

Clearly, $f(x)$, being a polynomial, is continuous on $[0,1]$ and differentiable on $(0,1)$.

Also, $f(0)=0$ and $f(1)=a+b+c=0$          [Given]

Thus, 0 and 1 are two roots of $f(x)$.

Therefore, by the algebraic interpretation of Rolle's theorem, $f^{\prime}(x)=0$ i.e. $3 a x^2+2 b x+c=0$ has at least one root between 0 and 1 .