inequalities and polynomials

Question

a) A polynomial function that has no real zeros will always be exclusively positive or negative (i.e. it will always be either all above or all below the x-axis – eg. f(x) = x2 + 2 has no real roots and it is always positive).

If a polynomial has real zeros, then if it is positive over any given interval, does it necessarily follow that it will be negative over the next sequential interval (i.e. if it is positive between the first and second root, does it necessarily become negative between the second and third root)? Explain using proper mathematical terminology.

b)provide a graph of a polynomial that illustrates your answer

Answer 1

To understand why this is so, we should make a distinction between a "change-of-sign" root and a "tangent" root.

If a root is of odd multiplicity, then the function's values (ie y-values) will change signs, either from negative to positive or from positive to negative. On the graph, this will look like an x-axis crossing. For example, p(x) = (x - 1)(x - 2)³(x - 5)⁴ has change of sign roots at x = 1 and at x = 2. (The graph at x = 2 will look different than at x = 1 because that root of multiplicity 3 will result in a flat point of inflection there also.)

Contrarily, roots of even multiplicity will result in tangent roots, where the function's values do not change signs. Instead, the graph reaches the x-axis, touches it at exactly one point (thus, tangent to the x-axis), and then returns in the direction whence it came. p(x) above has a tangent root at x = 5. A simple and familiar example of this is our favorite parabola, y = x², which has a tangent root at the origin. Its graph "bounces off" the x-axis, returning to positive y-values for x > 0.

What does all of this have to do with solving polynomial inequalities?

Well, let's look at the solution set for the inequality p(x) ≤ 0 for p(x) above. I.e., (x - 1)(x - 2)³(x - 5)⁴ ≤ 0 :

Because the polynomial is even degree (degree = 8), it starts positive. It changes signs at x = 1, dipping below the x-axis, then again at x = 2. However, it does not have a change of sign at x = 5. A sign chart for p(x) is

p(x): +++++++++++ 0 ---------------- 0 +++++++++++ 0 +++++++++++

x: <------------------- | ---------------- | -------------------- | ------------------->

1 2 5

So our solution set is 1 ≤ x ≤ 2 OR x = 5. In set notation, x ∈ [ 1 , 2 ] U { 5 } .

Interestingly, when we start to graph rational functions and solve rational inequalities, we have an analogous situation with zeros in the denominator of a function. These roots of the denominator can produce vertical asymptotes in the graph of a rational function, but depending on the multiplicity of those roots, odd or even, the function's values will either change signs or not. So the graph will head off toward ∞ , for example, and then either disappear for an instant to reappear at negative ∞ (a pretty nifty trick), or it will "bounce off" ∞ and come back down.

Happy solving!