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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)^n2010-09-28
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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?2010-09-28
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    It's actually a decic (tenth-degree) that you have; with two roots of multiplicity 5.2010-09-28
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    @J.M., there is a typo, which I have fixed.2010-09-28
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    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.2010-09-28
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    @Derek, Yup. I think it's clear now. See my updated question and the thing I miss.2010-09-28
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    @Ngu I'm pleased you solved it :-)2010-09-28

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4

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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    "Professor: What is a double root of a degree 3 polynomial? Student: This is a number such that you substitute it once and get 0, substitute it the second time and get 0 again, but when you do it the third time you won't get 0."2015-10-06