Real solutions of 6th degree polynomial

In this video, I showed how to solve a sixth degree polynomial for real solutions

[ashleyr]

Lovely explanation. Well done and a treat for those both new to and well experienced dealing with these fundamental issues. 😊

[francaisdeuxbaguetteiii]

Very nice video prime newtons ❤, one minor critique is that complex roots only have a complex conjugate if the polynomial has real coefficients, you forgot to specify that

[azraelwai]

It's sad, I didn't get any maths teacher like him during my school/college time. 😢

[ВикторПоплевко-ет]

6:44 if you multiply a positive number by 4 doesn't it always get bigger?

[alexandreb]

Thank you so much for these beautiful mathematics !!!

[mihaipuiu]

Dear NEWTON (?)...I like your writing, your demonstration.I think you are a great teacher, and you look very happy, which is a plus for your students. I have a difficult question...where did you buy your magnificent hat??

[KonBlo]

If the discriminant Δ < 0 <=> no real solutions.

[ragingskull]

"those who stop learning, stop living." wow

[MikooOnYoutube]

One question, wouldn't the other solutions just be the remaining complex cubic roots of 2 and -1, as given by De Moivre's theorem?

[malikahashami]

I used another method by deviding the 2 to 1+1 and bringing one of them to the other side and I got n^6-1=n^3+1
==>(n^3-1)(n^3+1)=n^3+1
==>n^3-1=1
==>n^3=2
This way I didn't get the whole answer I don't know why.

[chelliahRaveedrarajah]

Sir I have a doubt.
Could you please help me to do this question?
a²+b²+c²=1
a³+b³+c³=1
a+b+c=?.

[BukhalovAV]

I don't understand: why do you speak about complex solutions, if the question is in real? 🧐
It solves in less then 1 minute in mind!

[Пузырёк-пе]

The great Russian school mathematicians solve it this way:
n^6 -n^3 = 2 <=> n^6 - n^3 -2 = 0; t = x^3 => t^2 + t - 2 = 0 <=> t1*t2 = -2; t1+t2 = -1 (по теоремушке Виета) <=> t1 = -1; t2 = 2 => x1 = ∛(-1) = -1; x2 = ∛2.
Ответ: -1; ∛2.