Oxford's Quartic Interview Problem

[@despacitosalad_144]

I would have gone with y=mx+c being the straight line which intersects the polynomial y= 2x^4+7x^3+3x-5, Equating them gives 2x^4+7x^3+3x-5=mx+c 2x^4+7x^3+(3-m)x-(5+c)=0 From vietas formula, the sum of the roots is -b/a so is -7/2 but the mean value is this divided by 4 since the mean is the sum of all the roots divided by 4 Meaning that the mean x value of the roots =-7/8

[@falcongjon7815]

could you try do alot more quetsions in the next 2 days please ive got a cambridge interview on thursday,if you can ,could you go over the number theory and combi qs.Thank you

[@JPiMaths]

Will try to! Good luck with your interviews!

[@r-prime]

I would note that the method here I am relatively sure is actually the same kind of process which defines the group of an elliptic curve, over the prohective plane. So the question itself is not so deep, but certainly there are a whole lot of questions which could branch off of this one quite easily!

[@jamiewalker329]

Hmmm, this implication would be vacuously true for a cubic graph. So I think there is some merit in analysing the graph or asking a follow up on whether this implication is only vacuously true. The quartic above only has one turning point, and there are some quartics for which this there is no way of getting this configuration. In reality we just need to sketch y = 2x^4 + 7x^3 which has a triple repeated root at x = 0 and another at -7/2. Adding a term "ax" onto this for negative a, would split the repeated root into three distinct roots, thus making the premise of the question possible.

[@jamiewalker329]

Just thought, a generalisation could be that n intersections of a fixed degree n polynomial with ANY polynomial of at most degree n-2 will have the same average x value. Just a consequence of the coefficient of x^(n-1) not being altered - and I never thought about this before and it seems both simultaneously trivial and mind-blowing!

[@JPiMaths]

@@jamiewalker329 I agree, there is definitely a lot of exploration to this problem, which in an Oxbridge interview could do you well! And yes, this is a pretty cool generalisation

[@abhirajjoshi1]

Nice

[@gerryiles3925]

I don't think you explained why the average is a constant well enough to satisfy an Oxbridge interviewer, though they would probably point this out and ask you for further clarification.

[@arsenypogosov7206]

What? He just proved it on camera. Why this would not satisfy an Oxbridge interviewer? What else does they need?

[@gerryiles3925]

@@arsenypogosov7206 As I said in my comment, he didn't "explain" it "well enough" in my opinion. These interviews are more about finding out how well you can explain your thinking than about how much you already know or getting the correct answer.

[@mouldyvinegar5665]

@@gerryiles3925 ​ I would be interested in what you think he didn't explain. He showed that 7= -2(x1 + x2 + x3 + x4), and from this it is incredibly obvious what the average is - it is just rearranging the equation, and so obvious your interviewer may just skip that step because there are more important things than testing basic algebra. There is very little to explain, and all his previous steps were clearly laid out and easy to follow.