Let $a,b,c,d$ are non-zero real numbers such that $6a+4b+3c+3d=0$,then the equation $ax^3+bx^2+cx+d=0$ has 
Let $a,b,c,d$ are non-zero real numbers such that $6a+4b+3c+3d=0$. Then the equation $ax^3+bx^2+cx+d=0$ has:
(A) At least one root in $[-2,0]$
(B) At least one root in $[0,2]$
(C) At least two roots in $[-2,2]$
(D) No root in $[-2,2]$

Let $f(x)=ax^3+bx^2+cx+d$
$f(x)$ has at least one root in [-2,0] if $f(-2)f(0)<0$:
$$(-8a+4b-2c+d)d<0$$
$f(x)$ has at least one root in [0,2] if $f(2)f(0)<0$:
$$(8a+4b+2c+d)d<0$$
$f(x)$ has at least two roots in [-2,2] if $f(2)f(0)>0$:
$$(-8a+4b-2c+d)(8a+4b+2c+d)>0$$
Am I right uptil here? I am stuck from hereon.
 A: We show that (B) must be true. We are given $d\ne0$, so switching the signs of all of $a,b,c,d$ if necessary (which does not affect the existence of roots or the relation $6a+4b+3c+3d=0$) we can assume $d>0$ and hence $f(0)>0$. We show that $f(1)>0$ and $f(2)>0$ leads to a contradiction. We have:
$d>0$ (1); $a+b+c+d>0$ (2); $8a+4b+2c+d>0$ (3); and $6a+4b+3c+3d=0$ (4)
(4)-3(2): $3a+b<0$
(1)+(2)+(3)-(4): $3a+b>0$
Contradiction. So we must have either $f(1)<0$ or $f(2)<0$ and either is sufficient to give (B) true.
It is not hard to construct examples where (A), (C), (D) are false.
A: This question can be solved by using Rolle's theorem.
Let's consider a function
$$f(x)=\frac{ax^4}{4}+\frac{bx^3}{3}+\frac{cx^2}{2}+dx=0$$
[Considered so that on differentiating it, we would get our orignal equation ]
$$f(0)=0$$
$$f(2)=4a+\frac{8b}{3}+2c+2c$$
$$    =\frac{2}{3}(6a+4b+3c+3d)$$
$$=0$$
Thus according to Rolle' theorem ,
$$f'(x)=0$$for atleast one value of x between (0,2)
$$f'(x)=ax^3+bx^2+cx+d=0$$will have atleast one real solution between (0,2)
Option (b)
