Let $x \in \Bbb R $. Prove that if $x^5+4x^4+3x^3-x^2+3x-4 \ge 0$, then $x \ge 0$

Question

Let $x \in \Bbb R $. Prove that if $x^5+4x^4+3x^3-x^2+3x-4 \ge 0$, then $x \ge 0$

My try:

I tried to factor the polynomial, but i couldn't do it. I don't know how to proceed in this problem.

It's an exercise from a class in my university. The last exercise and is the only that i can't do and i'm looking for some hints.

Answer 1

Because $$0\leq x^5+4x^4+3x^3-x^2+3x-4<x^5+4x^4+3x^3-x^2+3x=$$ $$=x(x^4+4x^3+3x^2-x+3)=x\left(\left(x^2+2x-\frac{2}{3}\right)^2+\frac{1}{9}(3x^2+15x+23)\right).$$

Answer 2

If $x<0$, then $$ x^5+4x^4+3x^3-x^2+3x-4 <x^5+3x^3+3x-4 =\frac{(x^2+1)^3-1}{x}-4<0$$