If 3(a+2c) = 4(b+3d), then the equation ax³+bx²+cx+d = 0 will have

Let $\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{ax}}^4}{4}+\frac{{\mathrm{bx}}^3}{3}\frac{{\mathrm{cx}}^2}{2}+\mathrm{dx}$
Which is continuous and differentiable.
f(0) = 0
$\mathrm{f}(-1)=\frac{\mathrm{a}}{4}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}-\mathrm{d}=\frac{1}{4}(\mathrm{a}+2\mathrm{c})-\frac{1}{3}(\mathrm{b}+3\mathrm{d})=0$
So, according to Rolle’s theorem, there exists at least one root of f'(x) = 0 in (-1,0).