Problem

Solve each inequality, and graph the solution set…

07:38

Problem 27 Medium Difficulty

Solve each inequality, and graph the solution set. See Example 4.
$$
(x-1)(x-2)(x-4)<0
$$

Answer

$$
(-\infty, 1) \cup(2,4)
$$

Related Courses

Precalculus

Algebra

Beginning and Intermediate Algebra

Chapter 11

Quadratic Equations, Inequalities, and Functions

Section 8

Polynomial and Rational Inequalities

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Video Transcript

right, So we have polynomial X minus one times X minus two times X minus four, which is all less than zero. So this product is less than zero Normally, the first thing we would do is factor, but that's already been done for us. So what we're gonna do first here is we're gonna take each factor and let it equals zero. So that way we can figure out what our roots are. So if x minus one is equal to zero at one of both sides, you get X equals one that's a root. If X minus two is equal to zero and tunable signs you get X equals two. And if x minus four is equal to zero, then we have another one at X equals four. Now the's air where this polynomial would be equal to zero. But that's not what I'm looking for. I'm looking for where it is less than zero. So the intervals where this thing is less than zero are gonna be around these endpoints. So what ends up happening is we have to test. We need to test points to determine where are intervals are, right. If we were to think about a number line here, one to for and so on. We would have points at 12 and four. What we don't know is which way our solutions at what intervals air solution sets around. So I have to test a point that's on this interval. The left of one. I have to test a point between one and two. A point between two and four and something to the right of four. So there's four places where I need to test a value. Okay, so to the left of one, it would make sense the plug in zero. So if I plug zero in for axe, I'm gonna get zero minus one times zero minus two times zero minus four. And if this values on this interval or in the solution set, then this should check out. This will give us negative one times negative, too. Times negative. Four. Well, that's going to be naked. Once presented to his two terms, thinking of four is gonna be negative. Eight, which is in fact, less than zero. So that indicates to me that the values to the left of one are part of our solution set. So now I'm gonna pick a value between one and two three halves or 1.5 seems to be the most convenient. That's going to give us 1.5 minus one times 1.5 minus two times 1.5 minus four. Well, this is going to be 0.5 times negative 0.5 times negative, 2.5. But right here doesn't even matter what this answer is. Positive terms of negative times a negative is going to give me a positive number, which is not less than zero, so that it point doesn't check out for us. That indicates to me that the values between one and two are not part of our solution set. So next I'm gonna check for X Equal Street. That'll give us three minus one times three minutes to times three minus four. And that's going to be too times one terms negative one, which will be negative two, which is less than zero. So that indicates that the values between two and four are also part of our solution set. So now we gotta check this interval to the rate of four. So I'm gonna check five. This will give us five minutes, One terms, five minutes to times five minutes for all of these were going to be positive values, right? We're gonna have four times, three times one. So that can't possibly be less than zero, because it's a positive number. So that interval is not part of our solution set. So what is our solution set, then? Well, it's everything to the left of one and everything between two and four. So when we write this out, it's going to be the interval from Negative Infinity 21 or the interval from two before, and that is going to be our answer.

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