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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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: 104,766

Can we see the problem?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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: 287

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: 324

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: 104,766

You mean like this:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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
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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.****If it ain't broke, fix it until it is.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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
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It will be tedious.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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
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One way would be by a template:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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: 104,766

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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
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Can you extend that to a cubed form?

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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
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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?

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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
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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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

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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