Zeros of polynomials (with factoring): grouping | Polynomial graphs | Algebra 2 | Khan Academy

So we're told that p of x is equal to this expression here, and it says plot all the zeros or x intercepts of the polynomial in the interactive graph. The reason why it says interactive graph is this is a screenshot from this type of exercise on Khan Academy. On Khan Academy, you'd be able to click on points here, and they'd put little dots, and you can either delete them and put them someplace else. So it would be an interactive graph, but this is just a screenshot, so I'm just going to draw on top of this.

But the main goal is: what are the zeros of this polynomial? And then you just have to plot it on this graph. So pause this video and have a go at it.

All right, now to figure out the zeros of a polynomial, you essentially have to figure out the x values that would make the polynomial equal to zero. Or another way to think about it is the x values that would make this equation true: x to the third plus x squared minus nine x minus nine is equal to zero.

Now, the best way to do that is to try to factor this expression. Now, this is a third-degree polynomial, which isn't always so easy to factor. So let's see how we might approach it. The first thing I look for is: are there any common factors to all of these terms? And it doesn't look like there is.

The next thing I could look for is whether factoring by grouping could work here. When I think about factoring by grouping, I would look at the first two terms, and I would look at the last two terms, and I’d say is there anything I could factor out of these first two terms? What’s the most that I could factor out of these first two terms? And then what’s the most that I could factor out of these last two terms? That would leave something similar.

Once I've done that factoring—now what I mean is for these first two, we have a common factor of x squared, so let's factor out an x squared. These first two terms become x squared times x plus one. Then for these second two terms, I can factor out a negative nine, so I could rewrite it as negative 9 times x plus 1.

Now that all worked out quite nicely because now we see if we view this as our first term and this is our second term, we can see that x plus one is a factor of both of them. So we can factor that out. I’ll do that in this light blue color—actually, I’m doing a slightly darker blue color.

So if you factor out the x plus one, you're left with x plus one times x squared minus nine, and that is going to be equal to 0. Now we’re not done factoring yet because now we have a difference of squares: x squared minus 9.

This is going to be equal to—and let me just write it all out—so I have this x plus 1 here. So I have x plus 1, and then the x squared minus 9 I can write as x plus 3 times x minus 3.

If any of what I’m doing feels unfamiliar to you, if that first factoring feels unfamiliar, I encourage you to review factoring by grouping. If what I just did looks unfamiliar, I encourage you to look at factoring differences of squares. But anyway, all of that would be equal to zero.

Now if I have the product of several things equaling zero, if any one of those things is equal to zero, that would make the whole expression equal to zero. So we have a situation where one solution would be the solution that makes x plus one equal to zero.

And once again, I’m doing a darker color: x plus one equal zero, and that of course is x is equal to negative one. Another solution is what would make x plus three equal to zero, and that of course is x is equal to negative three—subtract three from both sides.

Then another solution is going to be whatever x value makes x minus three equal to zero—add three to both sides, you get x is equal to three. So there you have it. We have our three zeros: our polynomial evaluated at any of these x values will be equal to zero.

So we can plot it here on this interactive graph. I’m just going to draw on it, so we have x equals negative 1, which is right over there, x equals negative 3, which is right over there, and x equals 3, which is right over there.

The reason why you might want to do this type of thing—this exercise just asks us to do this and we’re done—but the reason why this is useful is this can help inform what the graph looks like. This tells us where our graph intersects the x-axis. So our graph might do something like this, or it might do something like this, and we would have to look at other information to think about what that might be. But I’ll leave you there.