total number of positive and negative roots of the equation $ax^3+bx^2+cx+d=0$ Suppose $a,b,c,d$ are non zero real numbers and $ab>0,$ and 
$\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx = \int^{2}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
Then total number of positive and negative roots of the equation $ax^3+bx^2+cx+d=0.$
what i have try $\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx = \int^{2}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
$\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=\int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx +\int^{2}_{1}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
Am I not able to go further. Could someone help me with this? Thanks

We know $a,b$ have the same sign, so the number of sign changes in the cubic is either $0, 1$ or $2$. Thus by Descartes rule of signs, at most it can have only $2$ positive roots. 
From the integrals given, it is clear there must be sign changes in both the intervals $(0,1)$ and $(1,2)$. So there are two positive roots, then the remaining root has to be real and negative. 
