That's smart 👌

Much more direct than the case-by-case analysis. Cool!

Thank you for sharing. I never knew it existed.

The answer would be 0, not 14^2 - 2, even if glossing over the word 'real,' via Vieta's formulae.

Any polynomial is divisible by x-r. Write p(x)=a(n)*x^n+a(n-1)*x^(n-1)+...+a2*x²+a1*x+a0=an(x-r1)(x-r2)...(x-rn). Distribute the parentheses on the right side and Bob's your mother's brother, it works for any power.

is that he should do in order the case1 and 2

Thank you!

Sure, but the derivative P'(x) = 4x³ − 8x − 1 is a cubic which has three real roots and which can therefore only be solved trigonometrically (or numerically). Not really worth it. But, I agree that the explanation in the video about the number of real zeros of the polynomial P(x) = x⁴ − 4x² − x + 1 may not have been entirely clear if you haven't seen something like this before. The purpose of rewriting the polynomial as P(x) = x⁴(1 − 4/x² − 1/x³ + 1/x⁴) is to note that | −4/x² − 1/x³ + 1/x⁴ | < 1 for any sufficiently large |x| so −1 < −4/x² − 1/x³ + 1/x⁴ < 1 and therefore 0 < 1 − 4/x² − 1/x³ + 1/x⁴ < 2 for any sufficiently large |x| so P(x) will be _positive_ for any sufficiently large |x|. And since P(−1) = −1, P(0) = 1, P(1) = −3 this means there will be at least one zero on each of the four intervals (−∞, −1), (−1, 0), (0, 1), (1, ∞) And since there can be no more than four zeros, the conclusion follows that P(x) has _exactly one zero_ on each of these four intervals.

Good thought, but that wouldn't really address the question of the roots. For example, for a quartic polynomial ( with leading coefficient positive ) to have 4 real roots ( as does the quartic that Mr. Prime Newtons considers here ) it is certainly NECESSARY that it also has 1 maximum and 2 minimums. But, conversely, that is NOT a sufficient condition for having 4 real roots. i.e. having 1 max and 2 mins does not guarantee 4 real roots, and in fact it does not guarantee any real roots ! To see this, imagine that the final numerical term in this quartic was the number 10 instead of 1. That would simply shift the graph UP by nine units. It would still have 1 max and 2 mins, but would not have any real roots.

A root of an equation is a number that makes the expression equal 0.

@@xinpingdonohoe3978 Thank you very much! ❤️🙏 I need to take more time with this one to comprehend what the professor is doing.

Now I get it. We call it "Nullstellen" (the spots, where the function is 0). I still don't understand, why it is called "roots" in English. 🤔

@@kragiharproots of tree are seen at the ground, ground is x-axis?

@@lcex1649 Ahhhh. Now I get it. ❤️🙏

We don't have what to plug in

When a polynomial curve "just barely touches the x-axis", i.e. when it is TANGENT to the x-axis, then that implies a double real root at that location. So, to answer your "N-shaped" curve question : That cubic polynomial would have 3 real roots ( all positive by the way you described it ) and the two roots farthest to the right would be exactly equal to each other ( hence the term "double root" ).

To add : Tangency always requires a double-root, even when it is, say, a circle that is tangent to a parabola.

@@GWaters-xr1fv That is the answer that I was expecting that academia would endorse, however, such an answer should really undergo further review. Out of the 3 possible roots, if two of those roots are guaranteed to always be equal to each other (whenever the given assumptions happen to be true), then, in truth, such a degree-3 polynomial only has 2 roots (both of which are Real numbers.) When a degree-3 polynomial only has 2 roots, and both of the roots are Real numbers, I would prefer if academia would refer to such an instance as a depressed or degenerative case. In any case, thank you for your informed reply.

No. The function you give can be written as f(x) = (x − 2)(x − 3) so we have f(2) = 0 and f(3) = 0. The graph of your function crosses the x-axis at x = 2 and at x = 3. The value of your function is negative for any real x between 2 and 3. You can graph your function online at the desmos website.

For power sums there are also Newton - Girard formulas so we can apply Newton - Girard formula and then Vieta formula

Too much work... ESPECIALLY in a tournament where u only have such limited time

Genuinely, what are you talking about? Some strange patronisation to the уоungеr students is all you're providing.

@@xinpingdonohoe3978 It is not usual to relate solutions of a polynomial to concepts of calculus - it may not connect with the younger audience.

@@Arkapravo sure, if you're 6 it's not normal to do that, but if you're watching videos of Harvard-MIT tournaments you should be expecting *something* clever to occur.

@@xinpingdonohoe3978 Well put.

@@xinpingdonohoe3978 yeah, I am a fool