Prove that $x^6$ + $x^4$ +$x^2$+ $x$ + $3$ has no positive real roots.
My attempt: I couldn't really think of a way to do this. I have looked at a similar problem but it had a cubic equation where you assign variables to the roots and use them to prove, but that method becomes tedious if you have to do it for this equation.
If $x$ is positive then $x^2,x^4,x^6$ are also positive so $$x^6 + x^4 + x^2 + x + 3 \ge 3.$$