Fundamental theorem of algebra | Polynomial and rational functions | Algebra II | Khan Academy

preview_player
Показать описание


Missed the previous lesson?

Algebra II on Khan Academy: Your studies in algebra 1 have built a solid foundation from which you can explore linear equations, inequalities, and functions. In algebra 2 we build upon that foundation and not only extend our knowledge of algebra 1, but slowly become capable of tackling the BIG questions of the universe. We'll again touch on systems of equations, inequalities, and functions...but we'll also address exponential and logarithmic functions, logarithms, imaginary and complex numbers, conic sections, and matrices. Don't let these big words intimidate you. We're on this journey with you!

About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.

For free. For everyone. Forever. #YouCanLearnAnything

Subscribe to Khan Academy’s Algebra II channel:
Рекомендации по теме
Комментарии
Автор

We would love to see if you make a proof of Fundamental theorem of algebra video

Kashif_Javaid
Автор

Please do a video on solving polynomials, especially on how a polynomial above the 4th degree cannot be solved with a finite number of operations (add, subtract, multiply divide, power, root).

UltimateBargains
Автор

4:54 what about the polynomial
(x + 3i) * (x + 4i) * (x - 2i)? This is a third degree polynomial with three imaginary roots, and none of its roots is a conjugate of any of its other roots.

nafrost
Автор

The really cool thing about polynomials is that if the polynomial is of degree 2k+1, then there ALWAYS exists a REAL root. (a result in real analysis)

alvinlepik
Автор

What would this function look like if we would plot it on the complex plane ? Could we also visualise it's roots like we do on a classic x, y plane ?

MaskedJackson
Автор

Okay. But I'm still confused on how to solve the problem

alexarmstrong
Автор

To address some of the comments about the single points, the point would have to be x^2=0, and when a zero is squared on a graph, it doesn't actually just become zero. It "bounces" off the x-axis, making "two" points

elizabetheid
Автор

Can you give a link to a proof of this theorem?

modolief
Автор

This is a good explanation, thank you.

bananapudding
Автор

I love math, and you, Khan academy ❤️

dorfinhorg
Автор

This 5 minutes video seems like half an hour

mohammedarsalaan
Автор

could do a video on the proof of this?

brianpark
Автор

Is this proof sine, cosine and tangent are infinite series?

Questiala
Автор

Good Explanation however it would help if you actually did a problem for more helpand not just an easy one but a difficult one 

Diedeux
Автор

What happens then if the curve just touches the axis and there is only one distinct solution?

OmegaCraftable
Автор

"Imaginary roots always come in pairs." What if you have an equation like y=x^2, which has one x-intercept? Wouldn't that be one real solution (x=0) and one imaginary solution?

AndrewTyberg
Автор

The first comment has now been reserved.

SupBrosNBozos
Автор

what about f(x) = x^2. doesn't this have root? i.e x = 0

estratz
Автор

Why is it always the same guy, same voice?

LoversGrief
Автор

the way you move the cursor is way disturbing.. it makes me feel anxious lol 😆

theapatriciaantonio