If $2a+3b+6c=0,$ the at least one root of the equation $a{x}^2+bx+c=0$ lies in the interval 

Solution:

Consider the function 

$f\left(x\right)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx$

We find that $f\left(0\right)=0$ and,

$f\left(1\right)=\frac{a}{3}+\frac{b}{2}+c=\frac{2a+3b+6c}{6}=\frac{0}{6}=0[\because 2a+3b+6c=0]$

Therefore, 0 and 1 are roots of the polynomial $f \left(x\right).$

Consequently, there exists at least one root of the polynomial

$f^{\mathrm{\prime }}\left(x\right)=a{x}^2+bx+c$ lying between 0 and 1.