Solve each polynomial inequality in Exercises 1–42 and graph the ...

Question

Solve each polynomial inequality in Exercises 1–42 and graph the solution set on a real number line. Express each solution set in interval notation. (x−4)(x+2)>0

Accepted Answer

Hello. Today we're going to be solving the given polynomial inequality. So what we are given is x minus eight times the quantity X plus three and that's going to be greater than zero. Since this polynomial inequality is already factored for us. We need to go ahead and identify the zeros for X. In order to do that. We're going to take each factor of x set equal to zero and solve for X. So this is going to give us x minus eight is equal to zero and X plus three is equal to zero on the left hand side. If we add eight to both sides, we get X is equal to eight which is our first zero. And on the right hand side we're going to subtract three to both sides. And that's going to give us x minus three or X is equal to negative three as the next zero. So now that we have both of the zeros for X, we can go ahead and plot them on the number line and then use that to test for the regions on the number line. So this number line is going to start negative infinity, Go to negative three. And then from negative three to positive eight. And then from positive eight ending with positive infinity. And keep in mind that the original polynomial is using the greater than sign. Since it's using the greater than sign, we're not going to be including the zeros in our solution set and we're going to represent that with open circles on the number line. Now what we want to go ahead and do is take a testing point from each region. On the number line, plug it into the original inequality and whichever one satisfies is going to be included in our solution set. So let's pick a testing point to the left of -3. And this can be when X is equal to negative four. When X is equal to negative four, we get negative four minus eight times negative four plus three is greater than zero. Negative four minus eight is going to give us negative 12 and negative four plus three is going to give us negative one And negative 12 times negative one is going to give us positive 12 greater than zero, which is a true statement. Since the testing point satisfies the inequality we're going to be including the region to the left of -3. Now, let's pick a test point between native three and eight and this can be when X is equal to zero. When X is equal to zero, we get zero minus eight, which is going to give us negative eight times zero plus three, which is going to give us positive three and negative eight times three is going to give us negative 24 greater than zero, which is a false statement. Since this testing point doesn't satisfy, we're not going to include any values between -3 and eight. Now, let's pick a testing point to the right of eight and this can be when X is equal to nine. When x is equal to nine, we get nine minus eight times nine plus three greater than zero, nine minus eight is going to give us one and nine plus three is going to give us 12 times one is going to give us 12 which is greater than zero. And that's going to be a true inequality. And that means that the region to the right of eight is going to satisfy the original inequality. Now let's go ahead and write a solution. And interval notation. The first region that satisfies the inequality starts at negative infinity and ends at -3. So we're gonna start this with negative infinity and assigning the parentheses because it doesn't have a finite value. Then the first region ends at negative three. Since negative three is not being included, we're going to assign it to parentheses. E then we're going to union the next region which starts at positive eight. And since positive it's not being included, we assign it a parentheses. And this region is going to end with positive infinity which gets a parentheses because it's not a finite value. And this is going to be the solution to the given inequality. And with that being said, the answer to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video

Solve each polynomial inequality in Exercises 1–42 and graph the solution set on a real number line. Express each solution set in interval notation. (x−4)(x+2)>0

Key Concepts

Polynomial Inequalities

Interval Notation

Graphing Solution Sets

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