Let $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ be a polynomial of degree $n \ge 3$, Knowing that $a_{n-1}=- {n \choose 1}$ and $a_{n-2}={n \choose 2}$, and that all roots are real, find the remaining coefficients.
$n$ is obviously even. Now the product of its roots is $a_0$ and the sum is $n$.I cannot do anything else. Please help me.
Please don't use Calculus.
On
Hint: if $\,P(x)\,$ has all roots real, then the $\,k^{th}\,$ derivatives $\,P^{(k)}(x)\,$ have all roots real for $1 \le k \lt n\,$.
In the case here $\,P^{(n-2)}(x) = \frac{n!}{2!}(x-1)^2\,$, and therefore $\,P^{(n-3)}(x)=\frac{n!}{3!}(x-1)^3 + C\,$ for some constant $\,C \in \Bbb R\,$. The roots of $\,P^{(n-3)}(x)=0\,$ are $\,x=1 + \frac{3!}{n!} \cdot \sqrt[3]{-C}\cdot \omega^k \mid k = 0,1,2\,$ where $\,\omega\,$ is a primitive cube root of unity. For all the roots to be real, it is necessary and sufficient that $\,C=0\,$, so $\,P^{(n-3)}(x)=\frac{n!}{3!}(x-1)^3\,$.
Repeat the same argument to show that in the end $\,P(x)=\frac{n!}{n!}(x-1)^n=(x-1)^n\,$.
Alt. hint: for a purely algebraic solution, consider the equality case of the RMS-AM inequality, which must hold since all roots $\,x_k\,$ are real:
$$ 1 = \frac{\sum_k x_k}{n} \le \sqrt{\frac{\sum_k x_k^2}{n}} = \sqrt{\frac{\left(\sum_k x_k\right)^2 - 2 \sum_{i \lt j} x_i x_j}{n}} = \sqrt{\frac{n^2 - n(n-1)}{n}} = 1 $$