<s>[INST] 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.
 [/INST] 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 $b=-1$, $c=1$.
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