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
 A: 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. 
A: Observe that all the functions involved in the integrand are continuous, furthermore the function $(1+e^{x^2})>0$ for all $x$.
From the first condition
$$\int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0.$$
This means the polynomial $ax^3+bx^2+cx+d$ will have at least one root in the interval $[0,1]$. So at least one positive root.
From the last step you get
$$\int^{2}_{1}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0.$$
This means the polynomial $ax^3+bx^2+cx+d$ will have at least one root in the interval $[1,2]$. So at least two positive root. 
By Viete's relations: if $\alpha, \beta, \gamma$ are the roots, then
$$\alpha+\beta+\gamma = -\frac{b}{a}.$$
Since $ab>0$, both $a$ and $b$ are of the same sign. This means at least one root of the cubic must be negative (because we have already established that two roots in $[0,1]$ and $[1,2]$. 
