Positive and negative intervals of polynomials | Polynomial graphs | Algebra 2 | Khan Academy

Let's say that we have the polynomial p of x, and when expressed in factored form, it is (x + 2)(2x - 3)(x - 4). What we're going to do in this video is use our knowledge of the roots of this polynomial to think about intervals where this polynomial would be positive or negative. The key realization is that the sign of a polynomial stays the same between consecutive zeros.

Let me just draw an arbitrary graph of a polynomial here to make you appreciate why that is true. So, x-axis, y-axis, and if I were to draw some arbitrary polynomial like that, you can see that between consecutive zeros, the sign is the same. Between this zero and this zero, the polynomial is positive. Between this zero and this zero, the polynomial is negative. That's almost intuitively true because if the sign did not stay the same, that means you would have to cross the x-axis, so you would have a zero. But we're saying between consecutive zeros; so between this zero and this zero, it is positive again. Then, after that zero, it stays negative once again. The only way it wouldn't stay negative is if there were another zero.

So now let's go back to this example here and let me delete this because this is not the graph of p of x which I have just written down. Let's first think about its zeros. The zeros are the x values that would either make (x + 2) = 0, (2x - 3) = 0, or (x - 4) = 0.

So first, we can think about well what x values would make (x + 2) = 0. Well, that of course would be x = -2. What x values would make (2x - 3) = 0? (2x - 3) = 0. Add three to both sides, you get (2x = 3). Divide both sides by two, you get x = 3/2. And then, last but not least, what x values would make (x - 4) = 0? Add four to both sides, you get x = 4.

So if we were to plot this, it would look something like this: this is x = -2, x = -1, this is 0, this is 1, 2, 3, and 4. Let me draw the y-axis here, so the y-axis would look something like this. X and y, we have a zero at x = -2, so our graph will intersect the x-axis there. We have a zero at x = 3/2, which is 1.5, which is right over there. And we have a zero at x = 4, which is right over there.

And so we have several candidate intervals, and actually let me write this down in a table so the intervals over which—and this is really between consecutive zeros— intervals to consider. So, I'll draw a little table here.

So you have x < -2. That's one interval. x is between -2 and 3/2, so -2 < x < 3/2; that would be this interval right over here. You have the interval, I'm trying to use all my colors, between 3/2 and 4; this interval here. So that would be 3/2 < x < 4. And then, last but not least, you have the interval where x > 4; that interval right over there. So, x > 4.

Now all we—there’s a couple of ways of thinking about whether over that interval our function is positive or negative. One method is to just evaluate our function at a point in the interval, and if it's positive, well that means that that whole interval is positive. If it's negative, that means that the whole interval is negative. And once again, it’s intuitive because if for whatever reason it were to switch, we would have another zero. I know I keep saying that, but another way to think about it is over that interval, what is the behavior of (x + 2), (2x - 3), and (x - 4)? Think about whether they're positive or negative and use our knowledge of multiplying positives and negatives together to figure out whether we’re dealing with a positive or negative.

So let's do it. We can do it both ways. So let’s think of this as our sample x value, and then let’s see what we can intuit about or deduce about whether over that interval we are positive or negative. So for x < -2, maybe an easy one or an obvious one to use—it could be any value where x is less than -2—but let’s try x = -3. So you could try to evaluate p of -3. You could just evaluate that. Actually, let’s just do that. So that’s going to be equal to (-1)(2)(-3) = -6 - 3 = -9. So -9 times (-3 - 4) = -7.

So if you were to multiply all of this out, this would give you -63, which is clearly negative. So over this interval right over here, our polynomial is going to be negative. Then we can move on to the next one. An interesting thing is we didn’t have to figure out the -63 part. We can just see that there’s a negative times a negative times a negative, which is going to be negative.

So let’s just do that going forward. Let’s just think about whether each of these are going to be positive or negative and what would happen when you multiply those positive and negatives together. Now in this second interval, between -2 and 3/2, what is going to happen? Well, we could do a sample point. Let’s say x = 0. That might be pretty straightforward. Well, when x = 0, we’re going to be dealing with a positive times a negative times a negative—a positive times a negative times a negative. And the reason why I did that just in my head, I said okay that’s going to be (2)(-3)(-4). So a positive times a negative times a negative.

Well, a negative times a negative is a positive, and a positive times a positive is a positive, so we are positive over that interval. If you were to evaluate p of 0, you will get a positive value.

Now what about this next interval? What about this next interval here between 3/2 and 4? We could try x = 2. When x = 2, we are going to get a positive times a positive times (2 - 4), which is negative times a negative. So this is going to be negative over that interval.

And then, last but not least, when x > 4, we could try x = 5. We are going to have a positive times a positive times a positive times a positive; so we are going to have a positive. As I mentioned, you could also do it without the sample points. You could say okay, when x > 4, you can say okay for any x greater than 4, if you add 2 to it, that for sure is going to be positive. For any x greater than 4, if you multiply it by 2 and subtract 3, well that’s still going to be positive because 2 times something greater than 4 is definitely greater than 3. And for any x greater than 4, if you subtract 4 from it, you’re still going to have a positive value.

So that’s another way to think about it even if you don’t use a sample point. But there you have it—we’ve figured out the intervals over which the function is negative or positive. We don’t know exactly what the function looks like but, generally speaking, it’s negative over this first interval, so it might look something like this. It’s positive over the next interval, and then it’s negative over that third interval, and then it’s positive over that last interval. So it would have a general shape like this. We don’t know without trying out more points exactly how high or low it would actually go.