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

Question

Solve each polynomial inequality. Give the solution set in interval notation. -(x - 3)(x - 4)² (x - 5) > 0

Accepted Answer

Hey, everyone for this problem for the given inequality we're asked to solve and to write this solution set in interval notation. Now, the inequality we're given is negative X minus seven multiplied by X minus 11 squared, multiplied by X minus 13 is greater than zero. We're given four answer choices. Option A is the interval from negative infinity to infinity. Option B is a union of the intervals from negative infinity to seven and from 13 to infinity. Option C is the interval from 7 to 13. And option D is the union of the intervals from 7 to 11 and from 11 to 13, I'm gonna start by rewriting what we were given. So we have negative X minus seven multiplied by X minus 11 squared, multiplied by X minus 13 is greater than zero. What you'll notice is that the left hand side is already factored. OK. So this is great. A lot of the work's been done for us. This is factored. We can go ahead and identify our points of interest. OK. So what we wanna do at this stage to identify our points of interest is look for anywhere where this left-hand side is equal to zero. OK? So our points of interest in this case, we can look at the first factor we have X minus seven. If that is equal to zero, then X is equal to seven, we move to the next factor X minus 11 squared that is equal to zero and X must be equal to 11. And finally, X minus 13 setting that equal to zero, we get that X is equal to 13. And so we have three points of interest. X equals 7-Eleven or 13. So what we're gonna do is we're gonna go ahead and draw number and we're gonna conclude these points of interest. So we have seven, we have 11 and we have 13. So these three points break our number line into four intervals. We can see these four segments, we're gonna go through each one of those and figure out what's going on and whether that interval is going to be included in our solution. Now, starting from the left, OK, we're on the left of X equals seven. That means the X values are less than seven. Now, we can pick a test point, say X is equal to zero. OK. That's in this interval. And we were gonna look at each individual factor. So we know that X minus seven, X is less than seven. It's gonna be negative, it's less than zero, X minus 11 squared is always gonna be positive. Maybe because it's squared, whether you have a negative or a positive you square that value, it's gonna give you a positive and X minus is also going to be negative. So if you look at the left hand side of our function, OK, which we're gonna write as L H S, we have negative X minus seven multiplied by X minus 11 squared multiplied by X minus 13. So we have a negative that negative out in front multiplied by a negative from the first factor multiplied by a positive multiplied by a negative. That's gonna give us that. The left hand side of this inequality is less than zero. OK. We have three negative values and one positive value being multiplied. But we want the left hand side to be greater than zero. Hey, that's our inequality. So if the left hand side is less than zero, this will not satisfy our inequality. And so this is not included in our solution set. We're gonna draw a red X in this interval to indicate that we do not include it in our solution set. Now we're moving to the right. And the next thing we hit on our number line is the point X equals seven. Now at X equals seven, we get that the left hand side is equal to zero. So inequality is that zero is greater than zero, but that's not true. OK? Because we have a strict inequality, we have strictly greater than zero. We don't have greater than or equal to zero. The left-hand side equal and zero will not satisfy this equation. It is not a solution. And so we're gonna draw an open circle at X equals seven to indicate that we do not include this point. OK. And we're just drawing this number line out to help us get a sense of what's going on. It helps us visualize what's going on and then it'll be easier to convert this into interval notation at the end. So moving to our next inter we're X values between seven and 11. So we can choose a test point, say X equals 10, we have that X minus seven will be positive X minus 11 squared again is gonna be positive as it always will be and X minus 13 will still be negative or less than zero. So we have a negative multiplied by a positive multiplied by a positive multiplied by a negative two positives, two negatives. This is gonna give us that the left-hand side is greater than zero. And so this interval gets a green check because that satisfies our inequality. Moving to the right again, we have the point X equals 11. And we're going to treat that in the same way as X equals seven, we substitute X equals 11 into the left hand side. We forget that it's equal to zero. So we have zero is greater than zero, which is not true. The inequality is not satisfied and therefore, this is not gonna be included in our solution. So we're gonna use a round circle to indicate that we do not include the point X equals 11 moving to the right. Once again, we're in a new interval from X equals 11 to X equals 13, we can choose a test point, say X equals 12. In that case, X minus seven is going to be greater than zero, X minus 11 squared, greater than zero like it always is and X minus 13 will be less than zero. So the left hand side of inequality is going to be a negative multiplied by a positive multiplied by a positive multiplied by a negative. OK. And that is gonna give us a value that is greater than 02 negatives, two positives all multiplied together. And this satisfies our inequality. OK? Just like the last interval, we want the left hand side greater than zero. And that's exactly what we have. So we're gonna give this a green check mark and we're gonna include this interval in our solution set. We move to the right, we get X equals 13. And again, the left hand side will be zero, which does not satisfy our inequality. So we use an open circle to indicate that we exclude that point. And our final interval is when X is greater than 13. That's the case, then X minus seven is greater than zero, X minus 11 squared is greater than zero X minus 13 is also greater than zero. So all of our factors are positive, but we still have this negative multiplied out in front. And so the left hand side is going to end up being negative or less than zero, which does not satisfy our inequality. And so we have a red X on this interval, it is not part of our solution. So, so we found that we have two intervals that satisfy our solution set. We have the interval from 7 to 11. We have the interval from 11 to 13. We have to be a little bit careful here. OK. It can be easy to say, well, these intervals are side by side. So we can just write this as the interval from 7 to 13. But that wouldn't be correct because that would be including the point X equals 11, which we've indicated we need to exclude. So we need to write this as two intervals. We have the interval from 7 to 11 and we're using round brackets to indicate that we do not include the end points. Let me take the union of that with the interval from 11 to again with round brackets. OK. So writing it like this of these two separate intervals allows us to exclude that point X equals 11. And that is the answer to this problem. That is the solution that we were looking for. And if we compare this to our answer choices, we can see that this corresponds with answer choice D 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. -(x - 3)(x - 4)² (x - 5) > 0

Key Concepts

Polynomial Inequalities

Critical Points and Sign Analysis

Multiplicity of Roots

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Solve each polynomial inequality. Give the solution set in interval notation. -(x - 3)(x - 4)2 (x - 5) > 0

Key Concepts

Polynomial Inequalities

Critical Points and Sign Analysis

Multiplicity of Roots

Watch next

Solve each polynomial inequality. Give the solution set in interval notation. -(x - 3)(x - 4)² (x - 5) > 0