If $a, b, c$ are real numbers such that $\frac{3a+2b}{c+d}+\frac{3}{2}=0$ then the equation $a{x}^3+b{x}^2+cx+d=0$ has

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

$\Rightarrow f\left(x\right)=\frac{a{x}^4}{4}+\frac{b{x}^3}{3}+\frac{c{x}^2}{2}+dx,f(+2)=0=f\left(0\right)$

$a{x}^3+b{x}^2+cx+d=0$ has at least one root in $\left[0.2\right]$