# Quintic Polynomial

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 $x$.

• But if $A \neq 0$, there will be always some $x$ such that $y(x)<0$. – mrprottolo Jan 16 '16 at 11:22
• Are you asking whether y<0 for some x, or are you looking for negative values of y within some limited range of x? The first is easy. – DanielWainfleet Jan 16 '16 at 11:25
• @mrprottolo Yes, I want to obtain some conditions for one possible situation. – user305563 Jan 16 '16 at 11:27
• Like the others hint, $\lim_{x \to -\infty} y(x) = -\infty$, so there is $y$ is always negative for some values of $x$. Do you mean, for example, conditions on the coefficients such that there is some $a$ larger than the smallest root of $y$ such that $y(a) < 0$*, i.e., that $y$ is negative on two intervals and nonnegative between those two interval? – Travis Willse Jan 16 '16 at 11:29
• @user254665 For example if I force the function to behave as above, I am certainly sure that near global minimum it's negative. – user305563 Jan 16 '16 at 11:29

## 1 Answer

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 $R$.

• Do you mean that $y$ can become negative, for some negative $x$? But in my question $x$ is positive. – user305563 Jan 16 '16 at 12:00