Solve each polynomial inequality. Give the solution set in interv...

Question

Solve each polynomial inequality. Give the solution set in interval notation. See Examples 2 and 3. (x - 3)(x - 4)(x - 5)^2 ≤ 0

Accepted Answer

Hey, everyone. Welcome back in this problem for the given inequality we're asked to solve and to write the solution set in interval notation. The inequality we're given is X minus seven multiplied by X minus 11, multiplied by X minus 13 squared is less than or equal to zero. We have four answer choices here. Options A and C of the union of two intervals and option D is that there is no solution. Now, we're gonna start by rewriting what we were given. So our inequality is X minus seven multiplied by X minus 11, multiplied by X minus 13 squared is less than or equal to zero. Now, this is already factored. OK? We have zero on the right hand side, on the left hand side, we have this function that's already factored. So that's great. That's the form that we want it in and that saves us a little bit of extra work here. Now, when it's in this form, what we want to do is identify our points of interest and our points of interest are gonna come whenever this function on the left hand side is equal to zero in this case, and we don't have a rational function, we don't have a denominator. So we're just looking at when this function is equal to zero and we have three factors multiplied together. So if any one of these factors is equal to zero, the entire equation will be equal to zero. So we're gonna take them one by one. If we have X minus seven equals to zero, we get that X is equal to seven. If X minus 11 is equal to zero, we get X is equal to 11. And if X minus 13 squared is equal to zero, we get that X is equal to 13. We have these three X values 7-Eleven and 13 that are points of interest for this inequality. And what we're gonna do is we're gonna take a number line and we're gonna add these points on it and these points are gonna divide that number line into four different intervals. OK? And then we're gonna evaluate each one of those one by one to see whether it's satisfies or inequality or not. All right. So we draw our number line, we're gonna add these three points. So we have seven, we have 11 and we have and maybe not drawn quite to scale. It gives us an idea. Now, the reason we use these points of interest is because this is where that function is gonna switch from being positive to being negative potentially. OK? When, when one of those factors is zero. Then on either side, it may be positive or it may be negative. That's where it's gonna change science. And that's when we're gonna figure out if it satisfies the inequality or not. So, starting on the far left, we have X values less than seven. If X is less than seven, then X minus seven is going to be less than zero. It's going to be negative X minus 11 is also going to be less than zero and negative X minus whoops 13 squared is going to be greater than zero. OK? And this is gonna be true all the time. We're taking a value and we're squaring it whether that value is negative or positive, this is always going to end up being positive. And so the left hand side of our inequality, which we're gonna call L H S, OK, X minus seven multiplied by X minus 11, multiplied by X minus 13 square will be a negative multiplied by a negative multiplied by a positive. So that's gonna give us a positive value. The left hand side is going to be greater than zero. What that tells us is that this interval is not part of our solution and we want less than or equal to zero, not greater than, and so we're gonna give a red X on this interval. We are not including that in our solution set. Now as we move further, right, we hit this point at X equals seven when X equals seven. The left-hand side of our function or inequality is going to be equal to zero. OK? And that satisfies our inequality. OK? Because we have less than or equal to zero. OK? We don't have a strict inequality. We have less than or equal to zero. So zero does satisfy that. So the 00.7 is going to be included in our interval. And I'm gonna draw it as a solid circle to indicate that we're including that point. OK. And this is just gonna help us when we go to write this in interval notation to remember to include that point, moving to the next interval we have from X equals seven to X equals 11. Now you can choose a test point in this interval if you want. So you could say choose X equals 10. If X is equal to 10, then X minus seven is greater than zero, X minus 11 is less than zero, it's negative and X minus 13 squared again is going to be positive like it always will be. Now we have a positive multiplied by a negative multiplied by a positive value. That's gonna give us that the left hand side is less than zero. And that's exactly what we want that satisfies our inequality. And so this interval is going to be included in our solution set. We're gonna give it a green check moving further to the right X is equal to 11. And the exact same thing is going to occur when X is equal to 11. OK. As it did with X equals seven, the left hand side will be zero that satisfies our inequality. So we do a solid circle there to indicate that we're including this point. Keep moving to the right. We have the interval X equals ele 11 to X equals 13. Hm. So we can choose a test point X equals 12, X minus seven is going to be greater than zero X minus 11 is also going to be greater than zero and X minus 13 squared is going to be greater than zero. Like it always will. Now we have three positive values being multiplied together. So the left hand side is going to be positive which does not satisfy our inequality. So red X on this interval as well and then we move to the right again, we hit our point X equals 13, same thing as the other two points. The left hand side will be 01, X is equal to 13 that satisfies our inequality. So this point is included. We draw a full circle. Finally, our last interval when X is greater than 13, OK, we can choose a test point, say X equals 15 X minus seven is going to be positive X minus 11 is going to be positive and X minus squared will also be positive. So just like the previous interval, we have three positive values multiplied together gives us that the left hand side is greater than zero. And so this is not a solution to our inequality. We exclude this interval. All right. So we can eliminate option D we know there is a solution. OK? What does that solution look like? Well, we know we don't include this far left interval that comes from negative infinity. We can eliminate option A as well because we're not including that negative infinity. Let's try to write this in interval notation. So the first interval we had, that was part of our solution was from 7 to 11. And we have closed circles indicating that we're including the end points. So what we're gonna do is use square brackets. We have the interval from 7 to 11. We're writing it with square brackets to indicate that we include those end points. And you can see that option B and C both have the same first interval here. OK? So either of those could be correct based on this first interval. You might be thinking, well, we only have one interval. None of the other intervals are solutions. And that is true. But you have to remember this point at X equals 13. OK. We said that that is included in our solution. And so we have to take the union of this set containing just that point X equals 13. OK. So always remember to go back and look at all of your intervals and all of your points because it can be really easy to miss ones like these. So that is our solution. We have the unit of the interval from 7 to 11 including the end points with the 110.13. And this corresponds with answer choice. C Thanks everyone for watching. I hope this video helped see you in the next one.

Solve each polynomial inequality. Give the solution set in interval notation. See Examples 2 and 3. (x - 3)(x - 4)(x - 5)^2 ≤ 0

Key Concepts

Polynomial Inequalities

Interval Notation

Sign Analysis

Watch next