What is the solution set to x3+x2−4x≥4?

2 Answers
Noah G · Tony B
Jun 30, 2017

Solutions are x∈[−2,−1] and x∈[2,∞).

Explanation:

I'm going to assume that you mean to say

x3+x2−4x≥4

We can solve this by factoring.

x3+x2−4x−4≥0

I would write as an equation.

x3+x2−4x−4=0

x2(x+1)−4(x+1)=0

(x2−4)(x+1)=0

(x+2)(x−2)(x+1)=0

x=−2,2and−1

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

Test point 1: x=0

03+02−4(0)−4≥?0

−4≥0

This is obviously false, therefore, (−1,2) obviously is not part of the solution set.

Test point 2: x=3

33+32−4(3)−4≥?0

27+9−12−4≥?0

20>0

So [2,∞) is a solution.

Since (−1,2) is not a solution, by the alternating signs of the function, [−2,−1], is a solution. On the other hand (−∞,−2) is not.

Our solutions are: x∈[−2,−1] and x∈[2,∞).

Hopefully this helps!
Tony BTony B

Nghi N
Jun 30, 2017

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

Explanation:

Solving by graphing.
First, graph the function f(x)=x3+x2−4x−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(x)≥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]}

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