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Current time:0:00Total duration:4:54

CCSS.Math:

let's see what if we can figure out what X plus 3 times X minus 3 is and I encourage you to pause the video and see if you can work this out well one way to tackle it is the way that we've always tackled when we multiply binomials is just apply the distributive property twice so first we could take this entire yellow X plus 3 and multiply it times each of these two terms so first we can multiply it times this X so that's going to be x times X plus 3 and then we are going to multiply it times we could say this is negative 3 so we could write minus 3 times now that's going to be multiplied by X plus 3 again X plus 3 and then we apply the distributive property one more time where we take this magenta X and we distribute it across this X plus 3 so x times X is x squared x times 3 is 3x and then we do it on this side negative 3 times X is negative 3x and negative 3 times 3 is negative 9 and what does this simplify to well we're going to get x squared now we have 3x minus 3x so these two characters cancel out and we are just left with x squared minus 9 and you might see a little pattern here notice I added 3 and then I subtracted 3 and I got this I got the x squared and then if you take 3 and multiply it by negative 3 you are going to get a negative 9 and notice the middle terms canceled out and one thing you might ask is well will that always be the case if we add a number and then we subtract that same number like that and we could try it out let's let's talk in general terms so for you instead of doing X plus 3 times X minus 3 we could write the same thing as instead of 3 let's just say you have X plus X plus a times X minus a times X minus a and I encourage you to pause this video and work it all out just to assume a is some number like three or some other number and apply the distributive property twice and see what you get well let's work through it so first we can distribute this yellow X plus a onto the X and the negative a so X plus a times X or we could say x times X plus a so it's going to be that's going to be x times X plus a and then we're going to have minus a or this negative a times X plus a so minus then we're going to have this minus a times X plus a times X plus a times X plus a notice all I did is I distributed this yellow this I just distributed this big chunk of this expression I just distribute it onto the X and on to this negative a I'm multiplying it times the X and I'm multiplying it by the negative a and now we can apply the distributive property again x times X is x squared x times a is ax and then we get negative a times X is negative a X and the negative a times a is negative a squared and notice regardless of my choice of a I'm going to have a X and then minus a X so this is always going to cancel out it didn't just work for the case when a was 3 for any a if I have a times X and then I subtract a times X that's just going to cancel out so this is just going to cancel out and what are we going to be left with we are going to be left with x squared minus a squared x squared minus a squared and you could view this as a special case when you have something X plus something times X minus that same something it's going to be x squared minus that something squared and this is a good one to know in general this is a good one to know in general and we could use it to quickly figure out the products of other binomials that fit this pattern here so if I were to say quick what is X plus 10 times X minus 10 well you could say all right this is just going this fits the pattern it's X plus a times X minus a so it's going to be x squared minus a squared if a is 10 if a is 10 a squared is going to be 100 so you can do it really quick once you recognize the pattern