# How do you solve the polynomial inequality and state the answer in interval notation given x^4<=16+4x-x^3?

Nov 25, 2017

The solution is $\left[- 2 , 2\right]$

#### Explanation:

Let us use a simplified version of 'sign chart' method.

${x}^{4} \le 16 + 4 x - {x}^{3}$ can be written as ${x}^{4} + {x}^{3} - 4 x - 16 \le 0$

As $2$ and $- 2$ are the zeros

${x}^{4} + {x}^{3} - 4 x - 16 = \left(x - 2\right) \left(x + 2\right) \left({x}^{2} + x + 4\right)$

Observe that ${x}^{2} + x + 4$ is always positive (note that ${x}^{2} + x + 4 = {\left(x + \frac{1}{2}\right)}^{2} + \frac{15}{4}$ - sum of two positive numbers) and hence

whether ${x}^{4} + {x}^{3} - 4 x - 16 \le 0$ depends on three intervals

$x \le - 2$, $- 2 \le x \le 2$ and $x \ge 2$

Let us pick up three values each in one of the range,

say $x = - 4$, $x = 0$ and $x = 4$

as only $x = 0$ satisfies the inequality,

solution is $- 2 \le x \le 2$ and in interval notation $\left[- 2 , 2\right]$

graph{x^4+x^3-4x-16 [-10, 10, -5, 5]}