# Prove that a real polynomial $x^n+ a_1x^{n-1}+ \cdots +a_n$ cannot be completely resolved into linear factors if $a_1^2<a_2$.

Prove that a real polynomial $$x^n+ a_1x^{n-1}+ \cdots +a_n$$ cannot be completely resolved into linear factors if $$a_1^2.

Here's what I've got. Let $$\alpha_1, \dots, \alpha_n$$ be the roots of the polynomial. Then $$\displaystyle{a_1^2-a_2=(\alpha_1+ \cdots +\alpha_n)^2 - \sum_{k,j \leq n; k \neq j}\alpha_{j}\alpha_{k} = \alpha_1^2+ \cdots +\alpha_n^2 + \sum_{k,j \leq n; k \neq j}\alpha_{j}\alpha_{k}}$$, but I have no idea how to prove this nonnegative. Please help.

Assume that $$\alpha_1,...,\alpha_n$$ are $$n$$ real roots of the polynomial. Notice that $$\sum_{i a_1^2 \geq 0\implies a_1^2-a_2 = \sum_i \alpha_i^2+\sum_{i0$$ which is a contradiction with our assumption that $$a_1^2-a_2<0$$. Hence, these real roots cannot exist at the same time.
• @Lazy Lee Why $\sum\limits_{i}\alpha^2+\sum\limits_{i<j}\alpha_i\alpha_j>0$? Jun 7, 2017 at 15:09
• It's because $\sum_{i<j} \alpha_i \alpha_j = a_2 > a_1^2 \geq 0$, where the $>$ is from the "if a_2>a_1^2" Jun 7, 2017 at 15:19
Let $$f(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+\dots+a_n$$ has $$n$$ real roots.
Hence, $$f^{(n-2)}$$ has two real roots, which says $$\frac{n!}{2}x^2+(n-1)!a_1x+(n-2)!a_2=0$$ or $$\frac{n(n-1)}{2}x^2+(n-1)a_1x+a_2=0$$ has two real roots.
Thus, $$(n-1)^2a_1^2-2n(n-1)a_2\geq0,$$ which gives $$(n-1)a_1^2\geq2na_2>2na_1^2,$$ which is contradiction.