Quintic Polynomial

Question

I have a polynomial of degree five as ($x > 0$)

$y = A x^5 + B x^4 + C x^3 + D x^2 + E x$,

where $A$ is positive.

I would like to find some sufficient conditions (inequalities) on coefficients in order $y$ becomes negative for some values of $x$. It can be done for example in this way:

In general, a quintic function function looks like as

In order in some way it becomes negative, one can force the derivative of the function to have four real roots (which in principle there are some conditions in order a four degree polynomial has four real roots), then, to demand the value of $y$ at its rightmost minimum becomes negative. So we obtain some inequalities on coefficients, which leads to a negative $y$ near its rightmost minimum.

However, it's somewhat clumsy and cumbersome. Can you think of something simpler?

Example for some illustration:

If we have ($x > 0$)

$y = A x^2 + B x + c$,

where $A > 0$.

One possible way that $y$ can become negative:

One can obtain the minimum of this function, and force $y$ becomes negative at its minimum. So we can obtain some conditions on coefficients in order $y$ becomes negative (for $x$'s near the minimum). This is indeed one possible way to have negative $y$ for some values of $x$.

Answer 1

With $A>0$ : For $x\ne 0$ we have $$y=A x^5(1+(1/A )(B/ x+C/x^2+D/x^3+E/x^4)).$$ So if $$|x|>W=\max (1,|B|/8,|C|/8,|D|/8,|E|/8)$$ then $$y=A x^5 (1+z) \quad \text {where } |z|<1/2.$$ So $x>W\implies y>Ax^5/2$ and $x<-W\implies y<-A|x|^5/2.$ Therefore the range of $y$ is unbounded above and unbounded below.Since $y$ is a continuous function of $x$, its range therefore is all of $R$.