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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Hi,

Can somebody show me a ressource of how to extract the square root of a polynomial that is a square and the same goes for a polynomial that is a cube (to find its cube root)

Thank you!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

Can we see the problem?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

I don't have a problem exactly. Unless you want an example because you don't understand what I wrote ???

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**Maburo****Member**- From: Alberta, Canada
- Registered: 2013-01-08
- Posts: 286

An example would help.

"Pure mathematics is, in its way, the poetry of logical ideas."

-Albert Einstein

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Ok, well if I gave you : (a+b-c)(a+b-c) or (a+b-c)(a+b-c)(a+b-c), each in developped form and I asked you what were the square root of the first and the cube root of the second, you would have : (a+b-c) and (a+b-c).

I want to extract the root in developped form!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

You mean like this:

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Yes. I think that to find the square root we should use the square root algorithm for numbers.... But I'm not sure.The same goes for the cube root, maybe we should use the cube root algorithm.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

Do you have access to some computing power? Do you program or have a CAS?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Nah, I don' even know what is a CAS. But I would want to do it by hand....

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

It will be tedious.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Yes, I know, but I would still like to see how it's done ^^ By curiosity

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

One way would be by a template:

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Ok, byt is there a sort of long division method we could use ???

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

I would guess so but you would have to know the form of the cube root before you start.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Ah, i was trying to find a way to find the root with only the developped form.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

I would use the template provided in post 12. It is a general solver for all trinomials.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

Well, I think I could show you an example of what I was talking about. I read it somewhere :

We have :

a^2+2ab+b^2

Now we know that the first term of the root will be "a"

so...

a^2+2ab+b^2(a

"a" squared cancels the other a^2

2ab+b^2(a

Now, we know that 2ab+b^2=(2a+b)b and so if we can figure out a divisor, we will also know the root. We also know the first term of the divisor because it is always the double of the firm term of the root. (Look at the square root algorithm :http://www.basic-mathematics.com/square-root-algorithm.html)

2ab+b^2(a+b (we find, by luck I guess, that we have "b" for the second term.)

____________

2a+b)2ab+b^2

2ab+b^2

______________

0

Anyway, i was searching for something like what I described.

*Last edited by Al-Allo (2014-07-11 03:41:15)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

Can you extend that to a cubed form?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

bobbym wrote:

Can you extend that to a cubed form?

Well, I don't know. Let's hope that we can....

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

It is easier to use this:

1) Look at the coefficient in front of a^3 in my example. It is 8. Take the cube root of it. So n1 = 2. Follow so far?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

bobbym wrote:

It is easier to use this:

1) Look at the coefficient in front of a^3 in my example. It is 8. Take the cube root of it. So n1 = 2. Follow so far?

yes

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

2) Look at the coefficient in front of c^3 in my example. It is -27. Take the cube root of it. So n3 = -3.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

bobbym wrote:

2) Look at the coefficient in front of c^3 in my example. It is -27. Take the cube root of it. So n3 = -3.

Following

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,394

3) Now take a look at the abc term. What do you see?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 294

bobbym wrote:

3) Now take a look at the abc term. What do you see?

That we have 36 ???

*Last edited by Al-Allo (2014-07-11 03:58:48)*

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