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Let \(f(x)\) be a polynomial of degree \(4\) with rational coefficients which has \(1+2\sqrt{3}\) and \(3-\sqrt{2}\) as roots, and such that \(f(0) = -154.\) Find \(f(1).\)

 

Thank you for all your help <3

 Feb 16, 2020

Best Answer 

 #7
avatar+29252 
+5

Like so:

 

 Feb 18, 2020
 #1
avatar+108727 
+1

You just finished telling us that you know all about polynomials now.

 Feb 16, 2020
 #2
avatar+278 
+1

That was for one question, but ok. Sorry

 Feb 17, 2020
 #3
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+2

You did not write it on one question. You wrote a general comment.

You did thank your answerers, which was nice, but even so, the comment sounded arrogant.

 

The comment should have been on the thread that you were talking about.

It was not appropriate to place it as a new 'question'.

 

There was a reply left saying you should have shown 'us' how to do it.

This reply was whited out because the person was using 'language' to display his/her annoyance.

But he was right.

If you ask a question here, then find out how to do it elsewhere, you should share your new knowledge here.

 

https://web2.0calc.com/questions/scratch-the-polynomial-help-i-figured-it-out

Melody  Feb 17, 2020
edited by Melody  Feb 17, 2020
edited by Melody  Feb 17, 2020
edited by Melody  Feb 17, 2020
 #4
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+1

I would help you here but I do not know how to do it.

 Feb 17, 2020
 #5
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+1

Sorry for my annoyance.

 Feb 17, 2020
 #6
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0

Thats ok, we all annoy others at times.

Melody  Feb 18, 2020
 #7
avatar+29252 
+5
Best Answer

Like so:

 

Alan Feb 18, 2020
 #8
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+1

Thanks Alan   cool

Melody  Feb 18, 2020
 #10
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+1

Thanks Alan,

I understand that roots come in pairs for parabolas but I did not realize that was still true for degrees higher than 2.

 

I am still confused by it.  Why is it so?      What happens if the degree is odd?

Melody  Feb 19, 2020
 #11
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+4

The question specified that the polynomial had rational coefficients.  The only way that could occur here is if the irrational terms in sqrt(3) and sqrt(2) were eliminated by complementary terms.  In general, an n'th order polynomial will have n roots (counting possible repeated ones).  Therefore with an odd degree polynomial with rational coefficients you couldn't have an odd number of roots that all contained irrational terms; the irrational ones would have to appear in complementary pairs. 

Alan  Feb 19, 2020
 #12
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0

Ok, got it.  Thanks Alan  cool

Melody  Feb 19, 2020
 #13
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+2

Alan i don't think thats right. There are 3rd degree polynomials with rational coefficients that have only irrational roots (for example x3-3x+1)

Guest Feb 19, 2020
 #15
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+4

True!  Perhaps I just got lucky!

 

Alan  Feb 19, 2020
 #16
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Hi guest,

 

I looked here....... but how do you know for sure that they are all irrational ?

 

Melody  Feb 19, 2020
 #17
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0

ok thanks Alan.

Melody  Feb 19, 2020
 #18
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+5

I guess I shouldn't have put the emphasis on the word "irrational" in my earlier reply.  However, I think the terms with the surds should come in  complementary pairs.  For example, the cubic can be rewritten as follows:

 

All solutions are (I suspect) irrational, but two are in the form a + b√3 and a - b√3.

Alan  Feb 19, 2020
edited by Alan  Feb 19, 2020
 #19
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+1

a and b here are not rational

Guest Feb 19, 2020
 #20
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+3

Correct. The solutions are irrational, the coefficients of the cubic polynomial are rational. 

Alan  Feb 19, 2020
 #21
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+1

im talking about what you said about a + b√3 and a - b√3.

Guest Feb 19, 2020
 #22
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ok thanks Alan.

Melody  Feb 19, 2020
 #9
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+1

Thank you very much, Alan!

 Feb 18, 2020
 #14
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+1

I just had fun drawing this curve

 

\(y=a(x^4-8x^3+8x^2+52x-77)\)

 

In the case of this picture 'a' is negative

 

 

 

 

 Feb 19, 2020
 #23
avatar+278 
+1

cool

 Feb 20, 2020

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