 So we put together all of the techniques we've learned about factoring large polynomials and use them to help find factorizations here So consider this degree 5 polynomial that might seem overwhelming, but we have all the tools we need So combining synthetic division with the rational roots theorem We can find the list of potential rationales and we can test each and every one of the sea Which ones are roots to speed up the process we can use things like Descartes variation or Descartes rule of variation of signs And we can use the upper and lower bound theorem Now I like to take a quick look Descartes using Descartes rule right because sometimes the following observation can be very helpful Sometimes maybe not I mean it depends Descartes rule can sometimes be very helpful and sometimes it can be completely useless Let's take a quick look if we look at variation signs it goes from a positive to a negative That's one variation Goes from a negative to a positive then a net positive to a negative then a negative to a positive then a positive to a negative You'll notice in this situation it varied signs each and every time positive negative positive negative positive negative Did I say the right number there? It switched variations every single time So this tells us that the variation of science is going to be five That tells us that our polynomial will either have five three or one Positive root here now what's critical here is that if you look at since it varies each and every time If you swat if you swap this to be f of negative x you switch the signs You're gonna get negative x to the fifth minus five x to the fourth minus 12x cube minus 24x squared minus 32x minus 16 because The variation turned out to be five the negative variation is necessarily gonna have to be zero What you see right here and this observation here is really critical because this tells us that if the negative variation is Zero here then there's gonna be no negative roots That is a super time saver if we start looking for negative roots. We are wasting our time So don't do that We we'd be better off looking for positive roots So then if we apply the rational roots theorem to that context the rational roots theorem is we're gonna be taking Devisors of negative 16 and divide them by divisors of one which is just one there So we're gonna get as our divisors six. We'll start with the small number one two four eight and Sixteen I didn't do plus or minus here because I knew by Descartes rule of signs that I don't need any negative I don't need any negative roots there So we just want to work on the positive ones and so What it was what do we choose like one two four eight sixteen? I might try something in the middle just kind of see how it goes if you try four You're gonna get something like the following. Let's make sure we don't write down the coefficients one minus five 12 negative 24 32 minus 16 if I tried four We could bring down the one one times four is Well, of course gonna be four minus Four minus five is gonna be a negative one four times negative one is going to be here negative four plus twelve is an eight times four is Gonna be thirty two thirty two minus twenty four is Eight right there again, so we try this again thirty two plus thirty two is sixty four and then sixty four minus sixteen That should be a forty eight right here So when you look at this thing right here, you look at the coefficients of the bottom The denominator is getting bigger or sorry the bottom row is just gonna be bigger bigger bigger bigger bigger Now there is a negative sign right here So the lower the upper bound theorem doesn't automatically apply but noticing the coefficients are just getting bigger bigger bigger This suggests to me that well one we know for a fact that four doesn't work for didn't work at all But this suggests to me that maybe I should try something smaller Because four was barely barely had a negative in here You had a negative one if I tried something like eight this would be even exacerbated This would be much much worse right if you tried an eight right here You're gonna get a three and then these numbers gonna be bigger bigger bigger So the fact that we almost had all positives on the bottom suggests to me I should probably try something smaller. I should try one or two and Whichever one you prefer it really doesn't matter I'm gonna try just one just in this situation here now one thing you can do is if you want to you Can erase the numbers you have before you don't have to keep every failed attempt here So I'm just gonna kind of erase this to save just a little bit of time a little bit of space on the screen Right here. So I'm gonna try one this time So you bring down the one one times one is one minus five is negative four times one is negative four plus 12 is eight times one is eight minus 24 is a negative 16 times one is negative 16 Plus that 32 you get 16 one times 16 is 16 plus 16 you get zero so we see that X equals one one in fact work my my guess of trying something something smaller actually paid off So now we can factor f whenever you find a route stop and make a factorization You're gonna get x minus one because one turned out to be successful right here And then look at the depressed polynomial of quotient We're gonna get x to the fourth we go down a power minus 4x cubed Plus 8x squared minus 16x plus 16 Like so if we were to run the rational roots theorem again, you get the exact same numbers One two four eight sixteen four doesn't work anymore. So we're not gonna try that one But one thing I should mention is that just because one worked doesn't mean it won't work again It could be that one is a repeated route And so it is worth trying again, but don't try it with the big with the original polynomial You have to try it with the quotient because if you try with the original polynomial again You're gonna see that one is a route one could be a repeated route So we need to try it with this depressed polynomial. So if we start the process over again Using the coefficients from the previous one one negative four eight negative sixteen sixteen Like so if we try one on the quotient We bring down the one one times one is one minus four is negative three times one is negative three plus eight is a five five times one is gonna be a five right here Add that to sixteen We're gonna get negative eleven and then one times negative eleven is negative eleven you add sixteen to that we end up with a positive five And we get We get the one didn't work the second time so that just means one's not a repeated route So if we come and record our potential rational roots what we can say is the following So we have one four. Sorry one two four eight Sixteen So four didn't work and now we've exhausted the possibility of one now when we checked for remember We didn't we didn't have the proof that four was definitely a upper bound, but it seemed really close So I'm very suspicious of sixteen and eight and so my likelihood would try two next That's what I feel like trying because one no longer works. So if we try this again with a two Right now when you look at this one right here, you might think that oh, this is the lower bound Right one negative three five negative eleven five. It's alternating again The lower bound theorem only applies to negative numbers the upper bound theorem only applies to Positive numbers, but it is somewhat suggestive to me that I should try something bigger. It's not a guarantee Because one is not a negative number, but it does seem to suggest that's the direction you want to go in here So erasing the numbers that we had I'm gonna try this again with two So if we try two here bring down the one one times two is two Minus four is negative two times two is negative four plus eight is positive four times two is eight Minus sixteen is a negative eight times two is a negative sixteen which gives us a zero so two in fact did work That's good. So now let's write the factorization. We have this time So we have f of x equals x minus one times x minus two and then what's left over is gonna be the cubic polynomial x cube minus 2x squared Plus 4x minus eight and so then by the rational roots theorem We could continue to look for roots where two would be a thing to check we could try to again Right and if you did that If you tried to again you get one negative two four negative eight trying to here You're going to get bring down the one one times two is two that negative two plus two is zero times two is zero Bring down the four two times Two times four is eight which then turns out oh two is in fact a repeated zero and this gives you the quotient as well But I'm actually gonna try this from a different perspective Synthetic division worked out great here, but when I look at this cubic polynomial, this makes me think I have four terms I think I could factor this by groups right if you do factored my group should take x cube minus 2x squared And then you'd have a 4x minus 8 if you factor out an x squared from the first group that leaves x minus 2 behind Factor out a 4 from the second group that leaves x minus 2 behind and so then you see that this factored as x square plus 4 and x minus 2 which is the exact same things synthetic division gave you and So then the factorization of f of x turns out to be x minus 1 x minus 2 turned out twice so it's a repeated root and then you get x squared plus 4 Which that would the roots of that thing are gonna be plus or minus 2i So there's no real roots of that polynomial right there. This is the real factorization of the polynomial and so The thing I want you to get from this one is you know combine things like the upper lower bound theorems Decards rule of science to help you speed up the process of synthetic division So you don't have to try every single number you can really you know kind of zero in on the roots of the polynomial if you choose your guesses strategically Another thing to notice here is that there can be repeated roots It's perfectly fine And when the polynomial gets smart small enough your elementary techniques of factoring might come into play And you might choose to use those instead of synthetic division because the factoring groups works out really nicely here And so I would suggest using that keep that on your radar whenever possible