If $4a+2b+c=0,$ then the equation $3a{x}^2+2bx+c=0$ has at least one real root lying in the interval 

Solution:

Let $f\left(x\right)=a{x}^3+b{x}^2+cx.$Then, $f\left(0\right)=0$ and 

$f\left(2\right)=2(4a+2b+c)=0$. Therefore, 0 and 2 are roots of $f\left(x\right).$ Therefore, $f^{\mathrm{\prime }}\left(x\right)=0$ i.e. $3a{x}^2+2bx+c=0$ has a root between 0 and 2.