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

Mar 12, 2017

$p \left(x\right) = - \left(x + 2\right) \left(x - 1\right) \left(x - 4\right) = - {x}^{3} + 3 {x}^{2} + 6 x - 8$

#### Explanation:

Define:

$p \left(x\right) = - \left(x + 2\right) \left(x - 1\right) \left(x - 4\right) = - {x}^{3} + 3 {x}^{2} + 6 x - 8$

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 $p \left(x\right)$.

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

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

$\left\{\left(q \left(x\right) > 0 \text{ for all " x in (-oo, -2) uu (-2, 1) uu (1, 4) uu (4, oo)), (q(x) >= 0 " for all } x \in \left\{- 2 , 1 , 4\right\}\right)\right.$

For example, we can multiply $p \left(x\right)$ by $q \left(x\right) = {\left(x - 1\right)}^{2} \left({x}^{2} + 2\right)$ to get a septic polynomial that is non-positive in $\left[- 2 , 1\right] \cup \left[4 , \infty\right)$.

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