How do you define a polynomial p(x) that is zero or negative whenever x in [-2, 1] uu [4, oo) ?

1 Answer
Mar 12, 2017

p(x) = -(x+2)(x-1)(x-4) = -x^3+3x^2+6x-8

Explanation:

Define:

p(x) = -(x+2)(x-1)(x-4) = -x^3+3x^2+6x-8

graph{ -x^3+3x^2+6x-8 [-10, 10, -15, 15]}

This cubic has a leading negative coefficient -1 and zeros at x=-2, x=1 and x=4.

As a result, it satisfies the required conditions.

Any polynomial in x with these zeros will be a multiple (scalar or polynomial) of p(x).

Any positive scalar multiple of p(x) will also be a cubic satisfying the conditions.

If p(x) is multiplied by any polynomial q(x) satisfying the following conditions, then the resulting polynomial also satisfies the conditions:

{ (q(x) > 0 " for all " x in (-oo, -2) uu (-2, 1) uu (1, 4) uu (4, oo)), (q(x) >= 0 " for all " x in { -2, 1, 4 }) :}

For example, we can multiply p(x) by q(x) = (x-1)^2(x^2+2) to get a septic polynomial that is non-positive in [-2, 1] uu [4, oo).

graph{(-x^3+3x^2+6x-8)(x-1)^2(x^2+2) [-10, 10, -240, 700]}