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I think this is really weird.

I want to solve the following equation:

$$(2x^2+3x+5)^3+(2x^2+3x+5)^2=0$$

This is a polynomial equation of 6 degrees. It should have 6 roots.

But wolframalpha only returns 4 roots! Anything I miss?

This is the command I use:

solve ((2x^2+3x+5)^3)+((2x^2+3x+5)^2)=0

Edit: Ah, Wolframalpha doesn't return repeated roots. So now case solved.

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Maybe the roots are equal, you have something like (degree 2)^n –  Ronaldo Sep 28 '10 at 12:57
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I think you're missing the fact that $x^2+3x+5=x^2+3x+5,$ as both terms in your brackets are equal. However, are you sure this is what you meant to type as you have $2x^2+3x+5$ in the command line you used in wolframalpha? –  Derek Jennings Sep 28 '10 at 13:04
    
It's actually a decic (tenth-degree) that you have; with two roots of multiplicity 5. –  J. M. Sep 28 '10 at 13:08
    
@J.M., there is a typo, which I have fixed. –  Graviton Sep 28 '10 at 13:09
    
You still have the same expression in each bracketed term, even after the edit, so just write $(2x^2+3x+5)^5=0,$ and you should see what's happening. –  Derek Jennings Sep 28 '10 at 13:11
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1 Answer

up vote 4 down vote accepted

Trying

Solve[(2x^2 + 3x + 5)^3 + (2x^2 + 3x + 5)^2 == 0, x]

on WolframAlpha returns four roots; the truth of the matter is that the equation has two roots of multiplicity two, so that you have six roots in total.

I'm not entirely sure why the multiplicities aren't indicated...

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