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^2+5x+4>0

Accepted Answer

Hello. Today we're going to be solving the given polynomial inequality. So what we are given is X squared plus seven X plus 10. Greater than zero. Now, in order to solve this inequality, we want to first completely factor the given polynomial X squared plus seven. X plus 10 is a fact herbal try no meal as this is going to factor into the quantities X plus five times X plus two. And now that the fact the polynomial is completely factored, we want to go ahead and identify the zeros for X in order to identify the zeros for X. We're going to take each factor of X. Set it equal to zero and solve. So this is going to give us X plus five is equal to zero and X plus two is equal to zero. On the left hand side. We're going to subtract five to both sides of the equation. And that's going to give us X is equal to negative five as our first zero. And on the right hand side we're going to subtract two to both sides and that's going to give us X is equal to negative two as our second zero. Now that we've identified the zeros for X. We're gonna go ahead and plot them on a number line and use that to figure out what regions of the number line satisfied the given inequality. So this number line is going to start a negative infinity, go to our first zero which is negative five, then from negative 52 negative two. And finally ending with negative two going to positive infinity. And keep in mind that the original inequality is using the greater than sign. Since we're 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 that we have the number line, we want to go ahead and pick a testing point from each region on the number line, plug it into our original inequality and whichever one satisfies the given inequality is going to be included in our solution set. So let's go ahead and test the region to the left of negative five. And a testing point from this region can be when X is equal to negative six. When x is equal to negative six, we get negative six squared plus seven times negative six plus 10 greater than zero. Negative six squared is going to give us positive 36 7 times negative six is going to give us negative 42 plus 10, greater than zero. And adding the values together is going to give us four greater than zero, which is a true inequality. So since this testing point satisfies the given inequality, we're going to be including the region to the left of -5. Now let's pick a testing point between negative five and negative two and that can be when x is equal to negative three. When x is equal to negative three, we get negative three squared plus seven times negative three plus 10 greater than zero. Native three squared is going to give us positive 97 times negative three is going to give us negative 21 plus 10 is greater than zero. And adding the three numbers together is going to give us negative two greater than zero, which is a false inequality. Since this didn't satisfy the inequality, we're not going to be including any values in the region between -5 and -2. Now let's go ahead and test the region to the right of -2. And that can be when X is equal to negative one. When X is equal to negative one, I'm sorry. We can make it easier. And by picking X is equal to zero because zero is to the right of negative two. So when X is equal to zero, we get zero squared plus seven times zero plus 10 is greater than zero. Anything times zero is just going to give us zero. So what we're left with is zero plus zero plus 10 greater than zero. And that's going to give us 10 is greater than zero, which is true. So since the region to the right satisfies, we're going to include the region to the right of negative two. Now let's go ahead and write our answer in interval notation. So we've identified that the first region that satisfies the inequality starts at negative infinity and ends at negative five. So we're going to start this region with negative infinity and assigning a parentheses. Because negative infinity doesn't have a finite value. And this region is going to end with negative five. And since negative five is not included, we give it a parentheses E. Then we're going to union the next region, which is going to start a negative two and negative two is not included. So we give it a parentheses E and ended at positive infinity, which also gets a parentheses. And this is going to be the 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^2+5x+4>0

Key Concepts

Polynomial Inequalities

Interval Notation

Graphing on a Number Line

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