# What is the solution set to x^3 + x^2 - 4x ≥ 4?

Jun 30, 2017

Solutions are $x \in \left[- 2 , - 1\right]$ and $x \in \left[2 , \infty\right)$.

#### Explanation:

I'm going to assume that you mean to say

${x}^{3} + {x}^{2} - 4 x \ge 4$

We can solve this by factoring.

${x}^{3} + {x}^{2} - 4 x - 4 \ge 0$

I would write as an equation.

${x}^{3} + {x}^{2} - 4 x - 4 = 0$

${x}^{2} \left(x + 1\right) - 4 \left(x + 1\right) = 0$

$\left({x}^{2} - 4\right) \left(x + 1\right) = 0$

$\left(x + 2\right) \left(x - 2\right) \left(x + 1\right) = 0$

$x = - 2 , 2 \mathmr{and} - 1$

Now select test points between these values of $x$ to see whether they satisfy the initial inequality.

Test point 1: $x = 0$

0^3 + 0^2 - 4(0) - 4>=^? 0

$- 4 \ge 0$

This is obviously false, therefore, $\left(- 1 , 2\right)$ obviously is not part of the solution set.

Test point 2: $x = 3$

3^3 + 3^2 - 4(3) - 4 >=^?0

27 + 9 - 12 - 4 >=^? 0

$20 > 0$

So $\left[2 , \infty\right)$ is a solution.

Since $\left(- 1 , 2\right)$ is not a solution, by the alternating signs of the function, $\left[- 2 , - 1\right]$, is a solution. On the other hand $\left(- \infty , - 2\right)$ is not.

Our solutions are: $x \in \left[- 2 , - 1\right]$ and $x \in \left[2 , \infty\right)$.

Hopefully this helps!

Jun 30, 2017

[-2, -1]
[2, + infinity)

#### Explanation:

Solving by graphing.
First, graph the function $f \left(x\right) = {x}^{3} + {x}^{2} - 4 x - 4$
The 3 x-intercepts are: - 2, - 1, and 2
Find parts of the graph that stay above the x-axis , meaning f(x) > 0.
The solution set, where $f \left(x\right) \ge 0$, are:
Closed interval [-2, - 1],
and half closed interval [2, + infinity)
graph{x^3 + x^2 - 4x - 4 [-5, 5, -2.5, 2.5]}