2
$\begingroup$

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

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.

$\endgroup$

2 Answers 2

8
$\begingroup$

Assume that $\alpha_1,...,\alpha_n$ are $n$ real roots of the polynomial. Notice that $$\sum_{i<j} \alpha_i \alpha_j = a_2 > a_1^2 \geq 0\implies a_1^2-a_2 = \sum_i \alpha_i^2+\sum_{i<j}\alpha_i \alpha_j>0$$ which is a contradiction with our assumption that $a_1^2-a_2<0$. Hence, these real roots cannot exist at the same time.

$\endgroup$
2
  • $\begingroup$ @Lazy Lee Why $\sum\limits_{i}\alpha^2+\sum\limits_{i<j}\alpha_i\alpha_j>0$? $\endgroup$ Jun 7, 2017 at 15:09
  • 1
    $\begingroup$ 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" $\endgroup$
    – Lazy Lee
    Jun 7, 2017 at 15:19
3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .