This post is used to collate any questions / uncertainties that you may have about Polynomials. Mentors will be looking in on this page, and providing explanations to help you understand. Do not be afraid to ask questions, as there will also be others with similar questions.

If you have a question, start a new comment and ask just that question (you can include related questions). State what you have done, and what you are having trouble with. Do not mix questions across different areas.

As an example,

I was working on question 5 and I got stuck. I can use rational root theorem to show that it has no rational roots. However, I do not know how to show that it has no real roots. Can someone guide me?

## Comments

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TopNewestPlease help with the second solution to E3(a).

I don't understand why we're taking \(n=3k, 3k+1, 3k+2\), or why getting a zero after substituting \(x=\omega\) in \(x^{2n}+x^{n}+1\) means divisibility by \(x^{2}+x+1\). How would this process come to mind? – Siladitya Basu · 2 years, 11 months ago

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Mow let us take \(f(x) = x^{2n}+x^{n}+1\).

So if we have \(f(\omega) = 0\) and \(f(\omega^{2})=0\) ,then we can write by the factor theorem \[(x - \omega)(x - \omega^{2}) | f(x)\]

that is

\[x^{2}+x+1 | f(x)\]

Now the divisibility of \(f(x)\) is totally dependent on the value of \(n\) modulo 3 and hence we have to consider 3 different cases as given by the book. – Eddie The Head · 2 years, 10 months ago

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nwork. Please help. – Siladitya Basu · 2 years, 11 months agoLog in to reply

How many of the problems do you expect us to be able to do and in how much time per problem? I could do all of them except for #37, but problems like #5, #3, #4, #18 took a few days of thought. – Colin Tang · 2 years, 10 months ago

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You should give yourself a pat on the back for solving these problems. They are not easy, and are above the level of what is normally covered in high school. – Calvin Lin Staff · 2 years, 10 months ago

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– Dinesh Chavan · 2 years, 10 months ago

You have a very good start. By the way, what troubles you in ques 37.??Log in to reply

– Colin Tang · 2 years, 10 months ago

It's just that I am not very familiar with functional equations.Log in to reply

Firstly,

a function is not a polynomial! So you can't determine it's coefficient or anything of the sort. So we have a new toolkit:Guess the functions. Normally we guess forlinear and constantfunctions. Guessing these functions provides us with a structure to solve the problem. Gut feel for functional equations is that only linear and constant functions will satisfy, but for some more complex ones, there might be higher powers.Substitute the variables with something else. Remember, you want to cancel stuff out, so if there are \( f \) on both sides of the equation, say \( f(a) + \text{blah} = f(b) \), then set try to manipulate the variables such that \( a=b \).But this problem does not require any of the machinery listed above, it is a \( polynomial \) functional equation, meaning that we can still use our toolkit for polynomials. But step \( 1 \) comes in really handy.

Next, notice that \( x^2 + x + 1 \), the complicated thing in this question, looks extremely like roots of unity. Once we notice

roots of unity, analysing thecomplex zeroessort of transforms this problem from a polynomial functional equation one to one which manipulates complex numbers.Then once we get some roots, (through playing around and trying to force out a contradiction at all steps if possible \( \Rightarrow \) this actually suggests

extremal principle) then we pull out the roots that we already have, (in this case writing \( f(x) = (x^2+1)^m g(x) \) ) at this step what we basically do is tosubstitute everything back into the original equationto solve for \(g(x) \) andremembering our constraintsat the same time. – Anqi Li · 2 years, 10 months agoLog in to reply

Functional Equations Wiki pages?

Can you add these comments to theSelect "Write a summary", and then copy-paste your text into it (with minor edits). Thanks! – Calvin Lin Staff · 2 years, 7 months ago

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– Colin Tang · 2 years, 10 months ago

ThanksLog in to reply

How does one think of a the method given in Example #6 (on page 250) himself? Is there any alternative method? Please help. – Samuel Jones · 2 years, 11 months ago

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Useful Lemma, which is the next installation in the series. – Calvin Lin Staff · 2 years, 11 months ago

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Is there any alternative or easier way to solve 11 in the book other than the factorizing method given in the book? I must say that the questions are brain-wracking and thus this one made me give up factorizing :(. I must also include that few of the questions asked by Calvin to solve had an application of symmetric polynomial functions too, thus he shouldn't have included them! I would like to know methods to simplify eqautions like 11. I quite didnt know how to proceed to prove #4 and #5 though I'd got all of the rest right. I would like to get some exposure to some olympiad proofs as I'd actually learnt form the answers of these questions ( 4& 5)! Thanks :) – Krishna Ar · 2 years, 11 months ago

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even. That suggests us to apply a substitution \(x=z^2\) to simplify things. Calling the resulting expression \(P(x)\) we have: \[P(x)=x^4+4x^3-10x^2+4x+1=0.\] Now to factorize this, we get help from theRational Root Theorem, which suggests us to try \(\pm 1\). We see that \(P(1)=0\), hence \((x-1)\) is a factor of \(P(x)\) by theRemainder-Factor Theorem. Now rearrange terms and factorize: \[P(x)=x^4-x^3+5x^3-5x^2-5x^2+5x-x+1\] \[=x^3(x-1)+5x^2(x-1)-5x(x-1)-1(x-1)\] \[=(x-1)(x^3+5x^2-5x-1)\] \[=(x-1)\cdot Q(x) ~~~[\text{say}]\] Note that \(Q(x)\) is also monic and the constant term is \(1\). So it's easy to applyRational Root Theoremagain. Since \(Q(1)=0\), \((x-1)\) is a factor of it. We again factorize: \[Q(x)=x^3-x^2+6x^2-6x+x-1=x^2(x-1)+6x(x-1)+1(x-1)=(x-1)(x^2+6x+1)\] Therefore: \[P(x)=(x-1)\cdot Q(x)=(x-1)(x-1)(x^2+6x+1)=(x-1)^2(x^2+6x+1)=0\] Now \(x=1\implies z=\pm\sqrt{x}=\pm 1\) is a root and bash the other factor with Quadratic formula. – Jubayer Nirjhor · 2 years, 11 months agoLog in to reply

– Colin Tang · 2 years, 10 months ago

Huh, you can also do it by noticing that the coefficients are symmetric with respect to the middle term and so you can pull it apart into the sum of two squares which share a common factor.Log in to reply

– Krishna Ar · 2 years, 11 months ago

Ah! Thanks! :)Log in to reply

First, let \[y=z^2\] then we have \[((y+1)^2)^2-(4y)^2=0\] factoring as a difference of squares we have \[(y^2+6y+1)(y^2-2y+1)=0\] then we can find the roots of the two quadratic factors to be \[-3\pm 2\sqrt{2}, 1\] so the solutions are \[z=\pm\sqrt{-3\pm2\sqrt{2}}, \pm{1}\] Sorry if those aren't simplified. – Nathan Ramesh · 2 years, 11 months ago

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– Krishna Ar · 2 years, 11 months ago

Hi! Are you talking about problem no.11? Not example, Problem!Log in to reply

## 11 on page 255, yes.

– Nathan Ramesh · 2 years, 11 months agoLog in to reply

– Krishna Ar · 2 years, 11 months ago

:D. But sorry, I don't get it :( EDIT- Ok....let me try working with the method you use now :)Log in to reply

– Nathan Ramesh · 2 years, 11 months ago

Which step don't you get?Log in to reply

\[ z^8 + 4z^6 - 10z^4 + 4z^2 + 1 = y^4 + 4y^3+ 6y^2 + 4 y + 1 - 16 y^2 = (y+1)^4 - 16y^2 \]

It wasn't obvious how you arrived at the conclusion, and adding the intermediate steps would help.

Very nice observation! – Calvin Lin Staff · 2 years, 11 months ago

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– Krishna Ar · 2 years, 11 months ago

Uhm...I actually didnt get that direct simplification..let me go through again, by this I mean- The expression you have written on the second line-Log in to reply

5 is actually hard. I was surprised it was posted so early in the set. – Calvin Lin Staff · 2 years, 11 months ago

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– Krishna Ar · 2 years, 11 months ago

Yes. Indeed 5 is hard and I actually had to learn from its solutions :(. Could you tell me what would the follow-ups for this course be?Log in to reply

What does #3 ask me to find? – Daniel Lim · 2 years, 11 months ago

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Let \(x_1, x_2 \) be the 2 roots of \( x^2 - 6x + 1 = 0 \).

Let \( A_n = x_1 ^n + x_2 ^ n \) for ever non-negative integer \(n\).

1) Show that \(A_n\) is an integer.

2) Show that \( 5 \) does not divide \(A_n\). – Calvin Lin Staff · 2 years, 11 months ago

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I just don't know what is the problem asking for – Daniel Lim · 2 years, 11 months ago

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How to convert cubic's of the form \(x^3+bx^2+cx+d\) into \(x^3+px+q\) in order to apply the solving method given in the book, The book says it can be done by translation, What does that mean? – Ishan Tarunesh · 2 years, 11 months ago

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Translation means shifting functions. If we translate the function horizontally by \(k\), the roots become \(\alpha + k, \beta + k\) and \(\gamma + k\). Since the coefficient of \(x^2\) in the second cubic is \(0\), the sum of its roots is also \(0\). Try to find a value of \(k\) for which the sum of the new roots become \(0\).

Once that is done, and the new roots are obtained, by Vieta's again, we can get value of \(p\) and \(q\) in terms of \(b, c\) and \(d\). – Pranshu Gaba · 2 years, 11 months ago

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– Samuraiwarm Tsunayoshi · 2 years, 11 months ago

If we shift \(x\) by \(k\), it would change the coefficient of \(x^{3}\), wouldn't it?Log in to reply

In this case, \(x^3 + bx^2 + cx + d\) would become \((x - k)^3 + b(x - k)^2 + c(x-k) +d\). Note that the term of \(x^3\) appears only once, and its coefficient is still \(1\). – Pranshu Gaba · 2 years, 11 months ago

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@Samuraiwarm Tsunayoshi even if the coefficient would have changed, we could have still divided by that coefficient to get a monic cubic – Ishan Tarunesh · 2 years, 11 months ago

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Yes...just as @Pranshu Gaba said you must first translate the polynomial by \(k\) units.and then since we must eliminate the \(x^{2}\) part we must find the value of k for which the co-efficient of \(x^{2}\) is 0.But in this case since the polynomial is monic the co-efficient of \(x^{3}\) will remain unchanged inspite of the transformation...try it out to get the idea :) – Eddie The Head · 2 years, 11 months ago

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I feel that even if a polynomial is not monic, translating it in any direction does not change the leading coefficient. This is because, essentially we are not changing the leading term when we translate the function. What do you think? – Pranshu Gaba · 2 years, 11 months ago

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– Eddie The Head · 2 years, 11 months ago

You are correct it holds for all...nothing special about monic .. :) ......Cheers!Log in to reply

I'm curious to know how others solved #3.

Engel's solution involves taking

modulo 5and I'm not sure how that could come to mind. I used induction instead. – Siladitya Basu · 2 years, 11 months agoLog in to reply

I think that one of the motivations behind using the modulo 5 was to eliminate the \(6S_{n-1}\) part to get \(s_n = s_{n-1}-s_{n-2}\) looks relatively better to infer a solution from.NOw if we can show that this sequence recurs after a value without arriving at 0 then we have already proved that for all n the sequence is not divisible by 5.

If you have solved it in a different way using induction than feel free to post your solution after the due date so that others can learn from it. – Eddie The Head · 2 years, 11 months ago

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I believe seeing other people's solutions after solving, or failing to solve, a problem might be insightful. Also, we might understand shortcomings in our own solutions. However, there seems to be no designated place for discussing the solutions to (and not difficulties with) problems in Engel. Where and when do we post our solutions and discuss them? – Siladitya Basu · 2 years, 11 months ago

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What I learnt page.

Actually Calvin Sir instructed this sprint to be about 2 weeks long and hence if we post our solutions now many will not try the problems on their own.But one thing that you \(\textbf{can}\) do right now is make a generalization of the idea that you used to solve the problem and post it in form of a note!!Also make sure that you post a link to the note in theIf you need help with Latex then I can help. – Eddie The Head · 2 years, 11 months ago

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I didn't play around with the problem after I solved it, so I really don't know in which way to generalize. Should I talk about a monic, integer-coefficient polynomial of

ndegrees; change the divisor from 5 to a broader set of numbers (primes??!); work with other forms that appear in Vieta's formulae\(((x_1x_2)^{n} + (x_2x_3)^{n} + (x_3x_1)^{n})\)? It'd be nice if there was some clarification! – Siladitya Basu · 2 years, 11 months agoLog in to reply

– Eddie The Head · 2 years, 11 months ago

Yeah I think that would be goodLog in to reply

– Jayakumar Krishnan · 2 years, 11 months ago

Where did you learn your math from- Eddie?Log in to reply

– Eddie The Head · 2 years, 11 months ago

I mainly learn from books and online courses -why?Log in to reply

– Jayakumar Krishnan · 2 years, 11 months ago

I asked this because you know a lot and also post lot of nice problems. I wanted to know if you learnt it in school ( age) or were doing it now.Log in to reply

After I'd looked through some of the problems, I still don't know how to start, can anyone here tell me what should I do?

Example #13:

Should I expand the equation? – Daniel Lim · 2 years, 11 months ago

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– Nathan Ramesh · 2 years, 11 months ago

What do you expect to find after expanding the equation? What do you notice about the given equation that might let you make progress on the problem? After expanding, what is your plan?Log in to reply

\(x^4-(a^2+b^2)x+a^2b^2=0\)

Edited:

\(x^4-(a^2+b^2)x^2+a^2b^2=0\) – Daniel Lim · 2 years, 11 months ago

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– Nathan Ramesh · 2 years, 11 months ago

Please check your work. Once you're sure you expanded correctly, try to apply things you know about roots of polynomials. (ahem, vietas)Log in to reply

Is this ok?

\(x^4-(a^2+b^2)x^2+a^2b^2=y^2-(a^2+b^2)y+a^2b^2\)

By using Vieta's formula,

Sum of roots = \(a^2+b^2\)

Product of roots = \(a^2b^2\)

So the roots are \(a^2\) and \(b^2\)

Am I right? – Daniel Lim · 2 years, 11 months ago

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– Calvin Lin Staff · 2 years, 11 months ago

I am not too sure what you are doing here. The expansion doesn't seem right to me.Log in to reply

\((x-a)^4\)

\(=[(x-a)^2]^2\)

\(=(x^2-a^2)^2\)

\(=x^4-2a^2x^2+a^4\)

Then I do the same to others

\((x-b)^4=x^4-2b^2x^2+b^4\)

\((a-b)^4=a^4-2a^2b^2+b^4\)

\(x^4-2a^2x^2+a^4+x^4-2b^2x^2+b^4=a^4-2a^2b^2+b^4\)

\(2x^4-2a^2x^2-2b^2x^2=-2a^2b^2\)

\(x^4-(a^2+b^2)x^2+a^2b^2=0\)

Have I done anything wrong here? – Daniel Lim · 2 years, 11 months ago

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\((x-a)^2\not=x^2-a^2\) – Nathan Ramesh · 2 years, 11 months ago

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Thanks – Daniel Lim · 2 years, 11 months ago

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– Nathan Ramesh · 2 years, 11 months ago

I may have epically screwed up, but I have a very different expansion.Log in to reply

Hint:What can we say about \(f(a)\) and \(f(b) \)? – Calvin Lin Staff · 2 years, 11 months agoLog in to reply

\(a\) and \(b\) are the roots of \(f(x)\) – Daniel Lim · 2 years, 11 months ago

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– Daniel Lim · 2 years, 11 months ago

If \(f(a)=0\) and \(f(b)=0\)Log in to reply

– Calvin Lin Staff · 2 years, 11 months ago

Great. And how would you continue, if you are interested in factorizing \(f(x) \)?Log in to reply

– Daniel Lim · 2 years, 11 months ago

\((x-a)(x-b)g(x)\)Log in to reply

Extension on #3

Show why \((\sqrt{2}+1)^n\) is very close to even integers for large integer \(n\). – Nathan Ramesh · 2 years, 11 months ago

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– Krishna Ar · 2 years, 11 months ago

I guess it's something to do with this thing's binomial expansion and then cancelling all the surds to obtain a multiple of 2. I know that I am not fully rightLog in to reply

– Nathan Ramesh · 2 years, 11 months ago

Does anyone wanna do this or should I explain a solution?Log in to reply

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– Nathan Ramesh · 2 years, 11 months ago

That's why I said "for large integer n"Log in to reply

n'. – Sharky Kesa · 2 years, 11 months agoLog in to reply

@Nathan Ramesh ! – Ahaan Rungta · 2 years, 11 months ago

Read: "for large \( n \)." I will post a solution later today if nobody tries it - have some patience,Log in to reply

– Nathan Ramesh · 2 years, 11 months ago

It's time Ahaan :DLog in to reply

When you say "try not to refer to solutions" on the https://brilliant.org/profile/calvin-8u8hog/sets/2-week-sprint-through-polynomials/, does that mean not referring to the solutions to the examples in the text (p. 245-250)? – Colin Tang · 2 years, 11 months ago

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I'm asking you to work on the problems listed on page 254-258. Solutions are given at the end of the chapter and you should not be referring to them.

With regards to

You should be reading through the solutions to the examples. – Calvin Lin Staff · 2 years, 11 months ago

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@Calvin Lin Generally ,when solving Olympiad problems, after how many hours of trying should one look at the solutions? – Rishabh Raj · 2 years, 11 months ago

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When you read solutions, don't simply see how it's done. An important aspect (which is often ignored), is to dissect and analyze them. Ask yourself:

1. Do I know enough to come up with this solution? If yes, why didn't I? If no, what do I need to know?

2. Is the solution motivated by some principle? What aspects of the question gave it away?

3. If I saw a similar (or even the exact same) problem in 6 months, can I solve it within an hour? – Calvin Lin Staff · 2 years, 11 months ago

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doing stuff(not just sitting idle)! If you are doing USAMO, I'd say do stuff for at least 1-1.5 hours. Same for IMO. – Ahaan Rungta · 2 years, 11 months agoLog in to reply

– Finn Hulse · 2 years, 11 months ago

Oh dude not to brag but I feel like the USAMO problems are really transparent. Like you can just look at it and know exactly what to do. AIME problems for me are more challenging because you have to actually problem solve.Log in to reply

– Nathan Ramesh · 2 years, 11 months ago

Then why don't you make USAMO and win? They aren't transparent at all.Log in to reply

– Finn Hulse · 2 years, 11 months ago

Well first I need to get through AIME. :DLog in to reply

– David Lee · 2 years, 11 months ago

No first u need 2 get through AMC 10 ggLog in to reply

– Finn Hulse · 2 years, 11 months ago

Yeah. Dude I don't think that will be a problem this year. I've been practicing. :DLog in to reply

– Nathan Ramesh · 2 years, 11 months ago

Generally you shouldn't read the solutions until after solving it. The beauty of olympiad problems is that there is a lot to learn from solving them. If you cannot solve one after a few days put it away and come back to it. After solving other problems you will get ideas.Log in to reply

I haven't encountered roots of unity before. I think the material in the book is too insufficient to understand for first-timers like me. Can somebody guide me to a good resource for roots of unity? Thanks. – Ananay Agarwal · 2 years, 11 months ago

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Roots of Unity is also a pretty good resource. – Anqi Li · 2 years, 11 months ago

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– Daniel Lim · 2 years, 11 months ago

I read through the notes but I don't understand how to change something into exponential form, can anyone tell me how?Log in to reply

If you are comfortable with basic trigonometry and complex numbers, you should be able to graph them on the complex plane given a unit circle. Given this, you should look at the AoPSWiki - this link is probably the best short fundamental explanation of roots of unity I have seen on the internet. Also, here is a nice blog post on AoPS expanding on roots of unity.

Lastly, if you are looking for a more visual approach, read this. Hope that helps! – Ahaan Rungta · 2 years, 11 months ago

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– Ahaan Rungta · 2 years, 11 months ago

The topic doesn't exist, Nathan.Log in to reply

– Finn Hulse · 2 years, 11 months ago

Wait what topic did he say existed? :DLog in to reply

– Nathan Ramesh · 2 years, 11 months ago

I guess I'll just try again laterLog in to reply

– Ahaan Rungta · 2 years, 11 months ago

Not sure what this comment means.Log in to reply

– Nathan Ramesh · 2 years, 11 months ago

It's not posting correctly I think... I'm trying to link to an aops post explaining the roots of unity filterLog in to reply

Click here – Eddie The Head · 2 years, 11 months ago

Be sure to check out the khan academy videos on the roots of unity!!!There Sal makes an excellent job to explain them for beginners...... Also check out this video by Patrick JMTLog in to reply

– Finn Hulse · 2 years, 11 months ago

I'll admit they are a little confusing. If you're not one to memorize formulas, you can easily derive or find any root of unity by just drawing the unit circle and placing \(n\) equidistant points along the circumference (for the \(n\)th roots of unity). You can use basic trig from there. :DLog in to reply

@Ananay Agarwal is looking for something geometrical to get an intuition about it. – Ahaan Rungta · 2 years, 11 months ago

The issue is that some students who are trying to learn roots of unity without De Moivre's Theorem will have trouble doing this. I thinkLog in to reply

– Finn Hulse · 2 years, 11 months ago

Yeah. That's what I was trying to say, is that you can easily construct the roots on a unit circle in the complex plane. I still have to do that sometimes. :DLog in to reply

– Ahaan Rungta · 2 years, 11 months ago

Yes, okay, I see what you're saying. I was just pointing out that writing stuff in \( \text{cis} \) form is not the most obvious thing for some students. For me, it's very automatic at this point because I have practiced a lot, but I am sure that, the first time I saw roots of unity and \( \text{cis} \) form, I was a bit confused for at least one moment.Log in to reply

– Finn Hulse · 2 years, 11 months ago

Same. I'm actually really terrible at trig and complex numbers, I have to rely fully on my intuition to solve them. Like converting rectangular to polar, I have to derive everything on the spot. I can't really memorize stuff like that. :PLog in to reply

So Is this note still active? May I ask a doubt? – Swapnil Das · 1 year, 10 months ago

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Can someone please explain the example problem solution for this article - I'm having trouble following it.

Thanks – Soham Karwa · 2 years, 11 months ago

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Sure! I commit to helping anyone in these topics to the maximum possible. – Krishna Ar · 2 years, 11 months ago

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@Mursalin Habib @Sreejato Bhattacharya @Shaan Vaidya @anup navin @Sudeep Salgia @Anqi Li @Dinesh Chavan @Trevor B. @Eddie The Head @Sharky Kesa @Pranjal Prashant @Pranshu Gaba @Finn Hulse @Mardokay Mosazghi @Zong Zhang @Samuraiwarm Tsunayoshi @David Lee @Ahaan Rungta @Krishna Ar @Jubayer Nirjhor @Vagish Jha @Nathan Ramesh @José Marín Guzmán @Aayush Mani

Thanks for volunteering to be mentors. When the members have any questions about the material, they will post their question here. Please check in when you can, and help to clear out doubts. If this post become too long / unwieldly, I will come up with an alternative.

If any of you would like to post relevant notes and problems, you can state the links in a comment and I'd review them. (This includes posts made in the past) – Calvin Lin Staff · 2 years, 11 months ago

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– Jayakumar Krishnan · 2 years, 11 months ago

This could be a bit off-topic, but since this topic would be covered later here, can anyone explain or suggest any good e-reosurce to learn cyclic and symmetric factorization? I believe these along with roots of unity are used to simplify lots of complex equations! thnx!Log in to reply

You should start a new comment for every question you have! To answer your question, here are some resources I can think of:

Also, I recommend taking small symmetric and cyclic polynomials and multiplying them and seeing how the expanded polynomials relate to the factors, etc. and just getting some experience. Hope that helps! – Ahaan Rungta · 2 years, 11 months ago

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@Ahaan Rungta i am not really good at symmetric and cyclic factorization but this really helped. – Mardokay Mosazghi · 2 years, 11 months ago

Thanks for the adviceLog in to reply

– Jayakumar Krishnan · 2 years, 11 months ago

Thanks a lot Ahaan! :)Log in to reply

Hopefully, I'll come around and help at combinatorics and number theory. ^__^

Edit: I am completely confused what this topic is, so I just called it "this kind of algebra" . XD – Samuraiwarm Tsunayoshi · 2 years, 11 months ago

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Second, this is not abstract algebra.

Abstract algebrais a totally different field of math (pun intended)! This is basic polynomial algebra, and considering you are Level 5, there is no reason that you should not be able to help. This thread is for Levels 2-3 so you are free to spectate but you should probably let the level 2-3s be the students.Third, even I have my weak subjects. For example, I am better in algebra and number theory than I am in combinatorics. So don't feel bad if you are slightly less active in some of these sprints. Good to have you on board! :) – Ahaan Rungta · 2 years, 11 months ago

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– Samuraiwarm Tsunayoshi · 2 years, 11 months ago

I got lvl 5 algebra because of the completely different topics like algebraic identities, solving system of equations, little bit inequality, or stuffs like that. I can't even know what is the \(i\) thing all about other than \(\sqrt{-1}\). T__TLog in to reply

– Ahaan Rungta · 2 years, 11 months ago

That's fine; then you can be a spectator!Log in to reply

@Suyeon Khim @Silas Hundt @Peter Taylor @David Mattingly @Calvin Lin @Arron Kau and all other staff members – David Lee · 2 years, 11 months ago

When a topic is deleted it actually should disappear, this long list makes my head hurtLog in to reply