MHB How do we solve a quadratic inequality with multiple factors?

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To solve the quadratic inequality (x^4)(x - 2)(x - 16) ≥ 0, critical points are identified by setting each factor to zero, resulting in x = 0, x = 2, and x = 16. The root x = 0 has an even multiplicity, meaning the sign of the expression remains unchanged across this point, while the others have odd multiplicities, indicating a sign change. Testing intervals around the critical points reveals that the solution set is (-infinity, 2] U [16, infinity). The approach of plotting the critical points on a number line and testing intervals is confirmed as correct. This method effectively identifies where the inequality holds true.
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This is the last quadratic inequality problem (for now) before moving on to Chapter 3, Section 3.1 THE DEFINITION OF A FUNCTION.

Section 2.6
Question 30

Solve the quadratic inequality.

(x^4)(x - 2)(x - 16) ≥ 0

Do I set each factor to 0 and solve for x? The values of x are then plotted on the number line for testing.

Correct?
 
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RTCNTC said:
This is the last quadratic inequality problem (for now) before moving on to Chapter 3, Section 3.1 THE DEFINITION OF A FUNCTION.

Section 2.6
Question 30

Solve the quadratic inequality.

(x^4)(x - 2)(x - 16) ≥ 0

Do I set each factor to 0 and solve for x? The values of x are then plotted on the number line for testing.

Correct?

Yes, observe that the root $x=0$ is of even multiplicity (4), and so the sign of the expression will not change across this root. The others are of odd multiplicity (1) and so the sign will change across those roots.
 
I will work on this tomorrow. Working right now.
 
(x^4)(x - 2)(x - 16) ≥ 0

Our critical points are x = 0, x = 2 and x = 16.

I can see that our critical points are also included.

<-------(0)--------(2)--------(16)------>

For (-infinity, 0), let x = -1. True statement.

For (0, 2), let x = 1. True statement.

For (2, 16), let x = 3. False statement.

For (16, infinity), let x = 4. True statement.

Solution:

(-infinity, 2] U [16, infinity)

Correct?
 
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