TMUA: What's So Special About This Polynomial?

Added: 2026-05-25

[00:00:00]Today I'm going to be sharing with you one of the many problems that I have that I give to my students who are preparing for the TMUA. And one of the reasons I create these problems is A, to give them a bit of a stretch because if you do some of the TMUA papers, certainly the earlier years, these questions can be a lot easier than the actual TMUA questions that you'll get nowadays. So, if you do the earlier 2016, 2017 papers, you might get a higher score than you actually are kind of at and give you that false sense of security. But also, like it's important to do questions that you find challenging. That's my general rule of thumb is when you are preparing for Oxford and Cambridge level maths, you need to be doing questions that are out of your comfort zone. That is the only way you're going to get better. This one is slightly more challenging than the average TMUA question, but this this problem could very easily be in a TMUA and I'm going to be showing you a couple of ways to think about this problem. So, we've got the polynomial P(x) which is x^4 - 6x^3 + 11x^2 - 6x + 1 has four real roots R1, R2, R3, R4. What is the value of 1/R1 + 1/R2 + 1/R3 + 1/R4 + R1 + R2 + R3 + R4? Do pause the video and give this problem a go for yourself. I'm going to dive right into a solution here. So, if you've seen stuff like this before, you're probably immediately thinking of Vieta's formulas or what's informally called roots of polynomials at A-level. And this allows me to encourage what I normally say to students here, which I will still say, is that if you are preparing for the TMUA and you genuinely want to get a grade nine, the very first thing you should do is teach yourself all of maths and further maths. Yes, it is a big ask, but B, if you are someone who wants to study maths at university, it shouldn't feel like that difficult of an objective, right? You should want to learn more maths. And so, definitely, definitely, definitely go out of your way to learn all of maths and further maths. It genuinely should only take a couple of weeks and that sounds a bit like, "What? How am I doing two years of content in just two weeks?" Well, you've already done half of it and you'd be surprised with that if you lock in how much you can learn. Anyway, let's let's actually get into solving this. So, the sum of the roots here and then the sum of the reciprocals of the roots. So, the sum of the roots, if we quote Vieta's formulae, is going to be the negative of this guy divided by this guy, which is one. So, this thing here is just going to be minus minus six over one, which is just six. Great. How do we work out the reciprocals of the roots, or the sum of the reciprocals? Well, what we're going to do is write this this guy here in a slightly different way. We can think of this as R2 R3 R4 + R1 R3 R4 + R1 R2 R4 + R1 R2 R3. Oops.

[00:02:31]over R1 R2 R3 R4. And all we need to know now is the product of the four roots and the sum of the what I call tri products of the roots. And so, the product of the roots, again just using Vieta's formulae here, that's just going to be one.

[00:02:47]I'll put that on the bottom one, and the uh so, doing this divided by one. So, one divided by one is one. And then the sum of the tri products is going to be the negative six divided by one. So, negative negative six over one, which is six. So, the answer here is six plus six, which is 12. Amazing. Lovely. And that would be the solution to this problem. But, I actually wanted to add something else here because there's something rather nice about this polynomial.

[00:03:13]And that is that this polynomial here, x to the four minus six x cubed plus 11 x squared minus six x plus one, is symmetric in the sense all the coefficients are symmetric. It goes one minus six 11 minus six one. And then when you see a polynomial like this, we we should jump for joy because something very very nice is happening. What I'm going to do here is factor out an x squared, which initially might seem a bit bit odd.

[00:03:35]But, just just trust me on this. Going to factor out an x squared here. So, we get x squared minus six x plus 11 minus six over x plus one over x squared. And the reason it's useful to write it like this is this thing now we can kind of write quite nicely. Uh And I'll show you how.

[00:03:56]We can write this as x squared and then this we can think of as like x plus one over x squared. Now, if you imagine expanding x plus one over x squared, you're going to get an x squared, you're going to get a one over x squared, and then you're also going to get a plus two. And so, what we're left with is like plus nine, and then we've also got minus six lots of x plus one over x like so. Great. Awesome.

[00:04:25]Well, let's continue this. Let me zoom out a little bit.

[00:04:28]And we've got x squared. And then this thing here, if I just rewrite this, this is x plus one over x squared minus six times x plus one over x plus nine. And I might look at this for a second and go, "Well, this factorizes nicely. This is just x squared times x plus one over x minus three all squared. Like, if you think of a minus three all squared, that's just going to equal a squared minus six a plus nine. And our a here is just x plus one over x.

[00:04:57]And that's very nice cuz we're just solving the equation of course it yeah, p of x equals zero to get the roots. So, we just want that to be zero. And we know that x can't be zero because of the fact well, if you plug in x is zero here, p of x is not zero, it's one.

[00:05:12]So, we can ignore those solutions we get from that. And then we just get that x plus one over x must equal three. And so, our uh solutions here are going to be the solutions of this guy. So, what will what does it take for x plus one over x to equal three? Well, we can just solve this and and and and work out the exact roots themselves. And that's another way to think about uh attacking this problem. So, really really nice way of thinking about this problem. So, the two morals of the story here is firstly, learn Vieta's formulae. Learn all of A-level further maths uh and single maths for that matter. But also, B, if you have a polynomial with symmetric roots, uh sorry, symmetric coefficients, um you can use this very nice little trick of basically making it look more symmetric, and then what you're left with in in the middle will be a function of like X plus one over X, and you can use that to your advantage to perhaps factorize the thing.