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# Factoring higher-degree polynomials

CCSS.Math:

## Video transcript

plot the real zeroes of the given polynomial on the graph below and they give us P of X is equal to 2x to the fifth plus X to the fourth minus 2x minus one when they say plotted they give us this little widget here where if we click at any point on this we get our point and we get as many points as we would like and we can drag these points around or if we don't want these points anymore we can just dump them in this little trash bin at the bottom right so let's think about what the zeros of this polynomial actually are to do that I'll take my scratch pad out and this is a little daunting at first this is a fifth degree this is a fifth degree polynomial here factoring fifth degree polynomials is really something of an art you're really going to have to sit and look for patterns if they're actually expecting you to find the zeros here without the help of a computer without the help of a calculator then there must be some type of pattern that you can pick out here so let me just rewrite P of X so P of X is equal to 2x to the fifth plus X to the fourth minus 2x minus one and one way that's typically seen when you're trying to factor this type of polynomial is to try to essentially undo the distributive property a few times and if you want to relate it to techniques for factoring quadratics it's essentially factoring by grouping so for example you see a 2x you see a 2x minus 1 or something that looks like a 2x minus 1 right over here and over here you have a 2x to the fifth plus X to the fourth so you have a 2x of a higher degree term plus a 1 X of a 1 degree lower so there it seems to be some type of a pattern 2 times X of a higher degree this is the first degree term minus 1 times you could view this as X to the 0 of a lower degree term and so let's think about it a little bit what happens if we essentially try to group these two terms and we group these two terms right over here and we try to factor out anything to essentially clean it up a little bit to see if we can make sense of it well these two terms the greatest common factor is X to the 4 we could write this as X to the fourth times 2x plus 1 and this should get us excited because this looks pretty close to that especially if we were to factor out a negative 1 here so we could factor out a negative 1 and then this is going to be 2x plus 1 and it's exciting because now we can factor out a 2x plus 1 from each of these each of these terms so you have a 2x plus 1 we're going to factor both of these we're going to factor both of these out to get 2x plus 1 which we just factored it out and if you factored it out of this term right over this term right over here you're left with X to the fourth and you factor out this term you're left with just the minus 1 minus 1 and now this is exciting because this is much to X plus 1 this is pretty easy to figure out when does this thing equal 0 and we'll do that in a little bit and this is pretty easy to factor this is a difference of squares this right over here can be re-written as can be re-written as x squared plus 1 times x squared minus times x squared minus 1 and of course we still have this 2x plus 1 out front 2x plus 1 and once again we have another difference of squares we have another difference of squares right over here that's the same thing as X plus 1 times X minus 1 and let me just write all the other parts of this expression x squared plus 1 and you have 2x plus 1 2x plus 1 and I think I factored P of X about as much as could be reasonably except reasonably expected so P of X is equal to all of this business here remember the whole reason why I wanted to factor it is I wanted to figure out when desisting equal 0 so if P of X can be expressed as the product of a bunch of these expressions it's going to be 0 whenever at least one of these expressions is equal to 0 if any of these is equal to 0 then that's just going to make this whole expression equal to 0 so when does 2x plus 1 equals 0 so 2x plus 1 is equal to zero well you could probably do this in your head what we do it we can do it systematically as well subtract one from both sides you get two x equals negative one divide both sides by 2 you get X is equal to negative 1/2 so when x equals negative 1/2 or you one way to think about P of negative 1/2 is 0 so P of negative 1/2 is 0 so this right over here is a point on the graph and it is one of the real zeroes now we could try to solve this x squared plus 1 equals 0 I'll just write it down just to show you if we try to isolate the X term on the left subtract 1 from both sides you get x squared is equal to negative 1 now if we were to go if we start thinking about imaginary numbers we could think about what what X could be but they want us to find the real zeros the real zeros so there is no real number where that number squared is equal to negative 1 so we're not going to get any zeros by setting this real zeros by setting this thing equal to 0 in the for real for there's no real number x where x squared plus 1 is going to be equal to 0 now let's think about when X plus 1 could be equal to 0 we'll subtract 1 from both sides you get X is equal to negative 1 so P of negative 1 is going to be 0 so that's another one of our zeros right there and then finally we have let's think about when X minus 1 is equal to 0 well add 1 to both sides X is equal to 1 so we have another 0 we have another real 0 right over there and so we could plot them let me so it's negative 1 negative 1/2 and 1 so it's negative 1 negative 1/2 and 1 and we can check our answer and we got it right now one thing that might be saying you know bugging you is like hey you know Sal you just happen to group this in exactly the right way what if I try to group in a different way what if I what if I try to and actually let's try to do that that could be interesting just just to show you this isn't voodoo and actually there's there's several ways to get there there's several ways to get there so what if instead of writing it like this we're you're writing it kind of in the highest-degree term then the next highest degree and so on and so forth you were to write it like this P of X is equal to 2 X to the fifth minus 2x plus X to the fourth minus 1 well actually even this way you could do a fairly interesting grouping if you group these two together you see that they have the common factor 2x you factor 2x out you get 2x times X to the 4th minus 1 and I think you see what's going on and then this can be re-written as plus 1 times X to the 4th my X to the 4th minus 1 minus 1 and now you could factor out an X to the 4th minus 1 and you're left with just in a neutral color X to the 4th minus 1 times 2x plus 1 which is much easier to factor now difference of squares exactly what we did the last time around so there's several ways that you could have reasonably grouped this and reasonably undone the distributive property but I do admit it is something of an art you really just have to play around and see let's group the first two terms let's see if there's a common factor here let's group the second two terms see if there's a common factor here hey once we factor out those common factors it looks like both of these two terms have this common expression as a factor and then you can start to factor that out