Solving x^5=1

How do you solve the "simplest" quintic equation x^5=1? We will try to factor x^5-1 and we will encounter how to factor x^4+x^3+x^2+x+1 which is very hard to factor.

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#blackpenredpen #math #calculus #apcalculus

[blackpenredpen]

Try my "extreme quintic equation" next: x^5-5x+3=0

[melvinwarmpf]

2:35
His face when he double thinks about what he said is gold.

[YarinGD]

Everyone says that x⁵ = 0 is easier than this. But i think that x⁵ = x⁵ is much easier

[andracoske]

At 1:03 he said synthetic division so i left the video and spent 3h researching synthetic divisipn and everything else i needed to understand it, and here i am again. Coincidentally now it's 4 am and i have school at 7:45 and im probably gonna kms but idk yet.

[rubyswolf]

I just did one of my end of year exams for year 12 and this video is the reason I was able to correctly answer probably the hardest question on the whole exam! Thank you so much for your videos, please keep making them. The question required finding the root of a cubic and I always forget how to do polynomial long division but I found 1 was a root by just trying it and then solved the cubic by factoring to (x-1)(a^x+bx+c) then expanding again and solving for a, b and c just like you did in this video

[vishalmishra]

Simpler Solution - Use n'th root of unity
x^5 = 1 = e^(i 360° n) so, x = e^(i n 360° / 5) = e ^ (i n 72°) for n = 0, 1, 2, 3, 4 leading to 1 (the only real root) and 2 complex conjugate roots with angles ±72° and ±144°
So, *the 5 roots are* 1, cos 72° ± i sin 72° and cos 144° ± i sin 144°
(Use cos ±72° = sin 18° = (√5-1)/4) and (cos ±144° = -cos 36° = -(√5+1)/4) and similarly the sine counter-parts to express the solution without any un-computed trigonometric functions)

[duggydo]

Does anyone else miss the pokeball mic? Maybe a special return of the pokeball episode before the end of the year?

[sadhanaduttapramanik]

4:26 THAT X² IN THAT STEP STAYED AS "X" FOREVER

Unsatisfactory 😭 once you see it, you can't unsee it

[rayanlima]

8:46 You can use ψ (psi), it's the reverse symmetrical of the golden ratio φ (phi)

[Nikioko]

You can solve this either algebraically or geometrically.
For the algebraic solution, just factor out x - 1 and solve the resulting 4th degree polynomial:
x⁵ - 1 = 0
(x - 1) (x⁴ + x³ + x² + x + 1) = 0
etc.

For the geometrical solution, just use Euler's formula:
xₙ = ⁵√1 ⋅ [cos(n ⋅ 360° / 5) + i ⋅ sin(n ⋅ 360° / 5)]
n = 1→5

So, basically:
x₁ = cos(72°) + i ⋅ sin(72°)
x₂ = cos(144°) + i ⋅ sin(144°)
x₃ = cos(216°) + i ⋅ sin(216°)
x₄ = cos(288°) + i ⋅ sin(288°)
x₅ = cos(360°) + i ⋅ sin(360°)

[Quasarbooster]

I love the connection between phi and this quintic/pentagons

[RealFreshDuke]

"-1/phi" could just be called "1-phi".

[bigwasabi]

I love seeing these kind of videos and knowing how to solve them

[cdkw]

1:10 I am surprised this is my first time seeing this type of method

[xinpingdonohoe]

Hence why I like mutlivalued functions. x=⁵√1, write out the 5 values by rotating by 2π/5 each time from 1, and we're done.

[johndoyle]

One of the best videos I have seen. Excellent mastery of Ceva versus Delanges math when you said that it didn't matter because they were symmetric. Ambiguities and dualities were combined at that point in the calculations.

[The-Last-Rain]

Love the shirt in the thumbnail. For those who do not know, it means algebra in Arabic.

[diskhover]

The golden ratio emerges here, sometimes pi, sometimes e^x, etc.... show up. When I look back, it is those surprising or unexpected little nuggets we find that got me hooked on mathematics early on.

[kkkk-gwl]

we can also divide by x^2 and reduce the equation into quadratic by substituitng x+1/x as t

[Frat]

since b = -1/a shouldn't b be equal to -2/(1 + sqrt(5))?