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Current time:0:00Total duration:4:55

Video transcript

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 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 since you 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 9x minus 9 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'd say is there anything I could factor out of these first two terms that would that would or what's the most that I could factor over these first two terms and then what's the most that I could factor out these last two terms and then would it 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 9 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 if we view this is our now our first term and this is our second term we can see that X plus 1 is a fat is 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 in a 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're 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 we do that 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 3 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 any of these x values will be equal to zero so we can plot it here on this interactive graph well I'm just going to 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 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