Division algorithm for polynomials
- [Instructor] 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, and the reason why it says interactive graph is this is a screenshot from this type of exercise on Khan Academy, and on Khan Academy, you'd be able to click on points here and it'd put little dots and you can either delete them and put 'em someplace else, so it'd 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 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 would 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, I could think about whether factoring by grouping could work here, and 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 would say, is there anything I could factor out of these first two terms that would, or 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, and then it 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 and these first two terms become x squared times x plus one, and then for these second two terms, I can factor out a negative nine, so I could rewrite it as negative nine times x plus one. Now, that all worked out quite nicely, because now we see, if we view, if we view this as our now our first term and this as our second term, we can see that x plus one is a factor of both of them. And so we can factor that out. We can factor out the x plus one, and I'll do that in this light blue color, actually let me do it with slightly darker blue color. And so if you factor out the x plus one, you're left with x plus one times x squared, x squared, minus nine. Minus nine. And that is going to be equal to zero. Now, we are not done factoring yet, because now we have a difference of squares. X squared minus nine, this is going to be equal to, and let me just write it all out, so I have this x plus one here, so I have x plus one, and then the x squared minus nine, I can write as x plus three times x minus three. 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, and 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, any, 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 gonna do that in 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, and 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 gonna draw on it. So we have x equals negative one, which is right over there, x equals negative three, which is right over there, and x equals three, which is right over there. And 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.