Roots and point of inflections Let $b$ and $c$ be the roots of a four degree polynomial. Also $x=b$ and $x=c$ are the real points of inflection of this four degree polynomial. If the other two roots of the polynomial be $a$ and $d$ where $a<b<c<d$ then prove that $$\int_{a}^{d}f(x)dx=0$$
My Attempt
I framed the equation $f(x)=k\left(x^4-10x^3+24x^2+5x-20\right)=k\left(x^2-5x+4\right)\left(x^2-5x-5\right)$ where $1$ and $4$ are the points of inflection as well as the roots and on integrating $f(x)$ I obtained $$\int_{\frac{5-3\sqrt{5}}{2}}^{\frac{5+3\sqrt{5}}{2}}f(x)dx=0$$But there must be a general way out.
One observation for the given polynomial is that $b+c=a+d$. Beyond this I am not able to generalize beyond this.
 A: Shift the coordinate system so that the new origin is at $\left(\frac{b+c}{2},0\right)$; this obviously does not change the integral. We can then rephrase the problem as: suppose $\pm B = \pm\frac{b+c}{2}$ are two roots of a quartic and that these are also the $x$-coordinates of the points of inflection. If $A, D$ are the other two roots, show that $\int_A^D f(x)\,dx=0$.
We have $f''(x) = (x-B)(x+B) = x^2-B^2$, so that $f'(x) = \frac{1}{3}x^3-B^2x+K_1$ and then $f(x) = \frac{1}{12}x^4 - \frac{1}{2}B^2x^2 +K_1x +K_2$. After multiplying through by $12$ and renaming constants, we may assume $f(x) = x^4-6B^2x^2+K_1x+K_2$. Solving $f(B) = f(-B) = 0$ for $K_1$ and $K_2$ gives $K_1 = 0$ and $K_2 = 5B^4$, so that
$$f(x) = x^4 - 6B^2x^2 + 5B^4.$$
This factors as $(x^2-B^2)(x^2-5B^2)$, so that the other two roots are $x = \pm B\sqrt{5}$. Then a computation shows that
$$\int_{-B\sqrt{5}}^{B\sqrt{5}}f(x)\,dx = 0.$$
A: If $f$ has degree $4$, $f''$ has degree $2$, so if $b$ and $c$ are inflection points (and thus  roots of $f''$), we must have $f''(x) = k (x-b)(x-c)$ for some constant $k$.  By integrating twice, we get $f$ as a polynomial of degree $4$ with coefficients of $x^1$ and $x^0$ arbitrary.  But those coefficients will be determined by the requirement that $f(b)=f(c)=0$.  So, given $b$ and $c$, $f$ is determined up to the constant factor $k$.
Now if $f(x)$ satisfies the conditions with inflection points $b$ and $c$, $f(A x + B)$ will satisfy them with inflection points $(b-B)/A$ and $(c-B)/A$.  The other roots are simlarly transformed, and
$$\int_{(a-B)/A}^{(d-B)/A} f(Ax+B)\; dx = \frac{1}{A} \int_a^d f(t)\; dt$$
The conclusion is that if it's true for one particular example, it's true in general.  You might as well take something conveniently symmetric such as $b=-1$, $c=1$.
