<s>[INST] Rolles theorem used for solving equation $ax^3+bx^2+cx+d=0$ If a,b,c,d are Real number such that 
$\frac{3a+2b}{c+d}+\frac{3}{2}=0$.   Then the equation $ax^3+bx^2+cx+d=0$ has 
(1) at least one root in [-2,0]
(2) at least one root in [0,2]
(3) at least two root in [-2,2]
(4) no root in [-2,2]
I am doing hit and trial method by using $f'(x)=0$, put x=1, we get $3a+2b+c=0$, putting $3a+2b=-c$ in 
$\frac{3a+2b}{c+d}+\frac{3}{2}=0$, i get relation between c & d, but not able to proceed.
 [/INST] Let $$f(x)=\frac{ax^4}{4}+\frac{bx^3}{3}+\frac{cx^2}{2}+dx.$$
Then $f0)=0$ and $f(2)=\frac{16a}{4}+\frac{8b}{3}+\frac{4c}{2}+2d=\frac{48a+32b+24c+24d}{12}=\frac{16(3a+2b)+24(c+d)}{12}=0$ (from the condition given).
Now apply Rolle's theorem. There exists a point in $[0,2]$, where the derivative of $f$ must be zero.
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