How prove that polynomial has only real root. Let this polynomial $f(x)=\displaystyle\sum_{i=1}^{n}a_{i}x^i,\;\;a_{i}\in \mathbb{R} $ have only real roots. Prove:
The polynomial $g(x)=\displaystyle\sum_{i}^{n}C_{n}^{i}a_{i}x^i$ has only real roots where $C_{n}^{i}=\dfrac{n!}{i!(n-i)!}$
My idea:
let $x_{1},x_{2},\cdots,x_{n}$ was $f(x)=0$ real root,and we have 
$$x_{1}+x_{2}+\cdots+x_{n}=-\dfrac{a_{n-1}}{a_{n}}$$
$$x_{1}x_{2}+x_{1}x_{3}+\cdots+x_{1}x_{n}+x_{2}x_{3}+\cdots+x_{n-1}x_{n}=\dfrac{a_{n-2}}{a_{n}}$$
$$\vdots$$
$$x_{1}x_{2}\cdots x_{n}=(-1)^n\dfrac{a_{1}}{a_{n}}$$
and I find a book has this theory:
if Polynomial $f(x)=\displaystyle\sum_{i=0}^{n}a_{i}x^n(a_{0},a_{n}\neq 0),a_{i}\in R$ only have real roots, then we have 
$$ \Delta_{1}=(n-1)a^2_{n-1}-2na_{n-2}a_{n}\ge 0 $$
and
$$\Delta_{2}=(n-1)a^2_{1}-2na_{2}a_{0}\ge 0$$
 A: This is problem 719 in the exercises book by Faddeev and Sominski. Here is
their proof.
Denote by $D$ the set of all real univariate polynomials all of whose roots are
real.
 Lemma 1  If $f\in D$ and $\lambda \in {\mathbb R}$, then
$f'+\lambda f \in D$ also.
 Proof of lemma 1.  The case $\lambda=0$ follows from Rolle’s theorem, so assume $\lambda \neq 0$. Denote by $x_1<x_2< \ldots <x_r$ the distinct roots of
$f$, and denote by $m_i$ the multiplicity of $x_i$ in $f$. Then $\sum_{k=1}^r m_k=n$ where
$n$ is the degree of $f$. Consider the rational fraction $g=\frac{f'}{f}$. Its poles are
the $x_k$, and $g$ is surjective $(x_k,x_{k+1}) \to {\mathbb R}$ for each $k\lt n$. Also, on the border, $g$ is surjective $(-\infty,x_1) \to (-\infty,0)$ and surjective $(x_n,+\infty) \to (0,+\infty)$. We deduce that there is a $y_k\in (x_k,x_{k+1})$ with $g(y_k)=-\lambda$ and that there is an yet another $y$ satisfying $g(y)=-\lambda$ (so $y$ will be $\lt x_1$ if $\lambda$ is negative and $\gt x_n$ if $\lambda$ is positive). Let us now count all the roots we have found for $f'+\lambda f$ : we have $y,y_1,y_2, \ldots y_{r-1}$, plus the $x_k$ with multiplicities $m_k-1$. This makes up a total of $n$ real
roots and concludes the proof.
 Lemma 2  If $f,g\in D$ with $g=\sum_{k=0}^n {\gamma}_k x^k$, then
$h=\sum_{k=0}^n {\gamma}_k f^{(k)} \in D$ also.
 Proof of lemma 2.  Let $\lambda_1, \ldots ,\lambda_n$ denote the roots
of $g(-x)$, so that $g(x)=\prod_{k=1}^n (x+\lambda_k)$. Iterating lemma 1 above, we see
successively that $F_1=f’+\lambda_1f, F_2=F_1'+\lambda_2 F_1, \ldots$, etc, up to
$F_{n}=F_{n-1}'+\lambda_n F_{n-1}=h$, are all in $D$.
 Lemma 3  Let $f\in D$, $f=\sum_{k=0}^n a_k x^k$, and $m$ be a positive integer. Then $\sum_{k=0}^n \frac{m!}{(m-n+k)!} a_k x^k \in D$ also.
 Proof of lemma 3.  Use lemma 2 with $g(x)=x^m$, and multiply by $x^{n-m}$ if needed.
 Main theorem  Let $f\in D$, $f=\sum_{k=0}^n a_k x^k$. Then 
$\sum_{k=0}^n \binom{n}{k} a_k x^k \in D$ also.
 Proof of main theorem.  Using lemma 3 with $m=n$, we see that 
$h_1=\sum_{k=0}^n \frac{n!}{k!} a_k x^k \in D$. So $x^nh_1(\frac{1}{x}) \in D$ also,
which means that $h_2=\sum_{k=0}^n \frac{n!}{(n-k)!} a_k x^k \in D$. Using lemma 3 again
with $h_2$ in place of $f$ and $m=n$, we see that 
$h_3=\sum_{k=0}^n \frac{n!}{k!}  \frac{n!}{(n-k)!} a_{n-k} x^k \in D$. Then, divide by
$n!$.
