Re: Factor question

• To: mathgroup at smc.vnet.net
• Subject: [mg71303] Re: Factor question
• From: "dimitris" <dimmechan at yahoo.com>
• Date: Tue, 14 Nov 2006 05:06:20 -0500 (EST)
• References: <ej7320\$iur\$1@smc.vnet.net>

```More on this subject...
See first the following commands.
Any comments on the outputs below?

sols = x /. Solve[x^2 - 1 == 0]
Factor[x^2 - 1, Extension -> sols]

sols = x /. Solve[x^3 - 1 == 0]
Factor[x^3 - 1, Extension -> sols]

sols = x /. Solve[x^4 - 1 == 0]
Factor[x^4 - 1, Extension -> sols]

sols = x /. Solve[x^5 - 1 == 0]
Factor[x^5 - 1, Extension -> sols]

sols = x /. Solve[x^6 - 1 == 0]
Factor[x^6 - 1, Extension -> sols]

sols = x /. Solve[x^7 - 1 == 0]
Factor[x^7 - 1, Extension -> sols]

sols = x /. Solve[x^8 - 1 == 0]
Factor[x^8 - 1, Extension -> sols]

sols = x /. Solve[x^9 - 1 == 0]
Factor[x^9 - 1, Extension -> sols]

sols = x /. Solve[x^10 - 1 == 0]
Factor[x^10 - 1, Extension -> sols]

sols = x /. Solve[x^11 - 1 == 0]
Timing[Factor[x^11 - 1, Extension -> sols]]

sols = x /. Solve[x^12 - 1 == 0]
Timing[Factor[x^12 - 1, Extension -> sols]]

sols = x /. Solve[x^13 - 1 == 0]
Timing[Factor[x^13 - 1, Extension -> sols]]

Dimitris

dimitris wrote:
> Why Factor gives the follwing output below?
>
> sols = x /. Solve[x^6 - 1 == 0]
> {-1, 1, -(-1)^(1/3), (-1)^(1/3), -(-1)^(2/3), (-1)^(2/3)}
>
> Factor[x^6 - 1, Extension -> %]
> (-1 + (-1)^(1/3) - x)*((-1)^(1/3) - x)*(-1 + x)*(1 + x)*(-1 +
> (-1)^(1/3) + x)*((-1)^(1/3) + x)
>
> that is why it writes e.g. (-1 + (-1)^(1/3) instead of (-1)^(2/3)
>
> Block[{Message}, FullSimplify[(-1)^(2/3) == -1 + (-1)^(1/3)]]
> True
>
> How is possible to get with Factor the output as follows?
>
> Times @@ Apply[Plus, Solve[x^6 - 1 == 0], 2]
> (-1 + x)*(1 + x)*(-(-1)^(1/3) + x)*((-1)^(1/3) + x)*(-(-1)^(2/3) +
> x)*((-1)^(2/3) + x)
>
> or
>
> Times @@ Apply[Plus, Flatten[Solve[x^6 - 1 == 0], 1], {1}]
> (-1 + x)*(1 + x)*(-(-1)^(1/3) + x)*((-1)^(1/3) + x)*(-(-1)^(2/3) +
> x)*((-1)^(2/3) + x)]
>
> Regards
> Dimitris

```

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