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+3)(x−5)>0

Accepted Answer

Hello today we're gonna be solving the given polynomial inequality. So what we are given is the quantity X plus six times the quantity X minus nine. And this is going to be greater than zero. So since we are given a simplified version of a polynomial, what we want to go ahead and do is figure out the zeros for X. To plot on the number line in order to get the zeros for X. We're going to take each factor of X set equal to zero and solve. So this is going to give us X plus six is equal to zero and x minus nine is equal to zero. On the left hand side. We're going to subtract six to both sides of the equation and this is going to give us X is equal to negative six. And on the right hand side we're going to add nine to both sides of the equation. And that's going to give us X is equal to nine. Now that we've identified the zeros for X. We're gonna go ahead and plot it on a number line and use this to figure out the regions that satisfy the given inequality. So the number line is going to start from negative infinity, going to pot negative six, then from negative six to positive nine. And then from positive nine to positive infinity. And since the original inequality is using the greater than sign, we're not going to include the zeros in our solution set and we're going to represent that with open circles on the number line. Now that we have the regions that we want to test, we're going to pick a testing point from each region, plug it into our original inequality and whichever region satisfies the given inequality is going to be included in our solution set. So let's pick a testing point to the left of -6. And this can be when X is equal to negative seven. When x is equal to negative seven, we get negative seven plus six times negative seven minus nine. And that's going to be greater than zero. Negative seven plus six is going to give us negative one and negative seven minus nine is going to give us negative 16 negative one times negative 16 is going to give us positive 16 greater than zero, which is a true statement. Since negative seven satisfies the inequality. We're going to be including the region to the left of -6. Now, let's pick a testing point between negative six and nine. And this can be when X is equal to zero. When x is equal to zero, we get zero plus six which is going to give us six and zero minus nine, which is going to give us negative nine And six times negative nine is going to give us -54, greater than zero, which is a false statement. And since this value didn't satisfy the inequality, we're not going to include any values between negative six and positive nine. Now let's pick a testing point to the right of positive nine. And this can be when X is equal to positive 10. When x is equal to positive 10, we get 10 plus six times 10 minus nine. Greater than zero. 10 plus six is going to give us 16 and 10 minus nine is going to give us positive one. 16 times one is going to give us 16, which is greater than zero. And that's going to be a true inequality. And since this testing point satisfies the inequality, we're going to be including the region to the right of positive nine. Now let's go ahead and write a solution and interval notation. So the first region that satisfies the inequality starts at negative infinity and ends at -6. So we're gonna write negative infinity as our starting point. And exciting the parentheses since it doesn't have a finite value, then we're going to end this first region with negative six. And since negative six is not being included, we're going to give it a parentheses. Then we're going to union the next region which is going to be from nine to positive infinity. So the next region starts at nine. And since nine is not being included, we're going to assign it a parentheses E and then end that region with positive infinity. And since positive infinity doesn't have a finite value, we're going to give it a parentheses. E 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 be 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+3)(x−5)>0

Key Concepts

Polynomial Inequalities

Interval Notation

Graphing Solution Sets

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