       Re: Factor[1+x^4]

• To: mathgroup at smc.vnet.net
• Subject: [mg26882] Re: Factor[1+x^4]
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Fri, 26 Jan 2001 01:27:27 -0500 (EST)
• References: <94ojm5\$emb@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Tom,

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

(1*(Sqrt - (1 + I)*x)*(Sqrt - (1 - I)*x)*
(Sqrt + (1 - I)*x)*(Sqrt + (1 + I)*x))/4

Factor[1 + x^4, Extension -> {I, Sqrt}]

(1*(Sqrt - (1 + I)*x)*(Sqrt - (1 - I)*x)*
(Sqrt + (1 - I)*x)*(Sqrt + (1 + I)*x))/4

Factor[1 + x^4, Extension -> {I}]

(-I + x^2)*(I + x^2)

Factor[1 + x^4, Extension -> {Sqrt}]

-(-1 + Sqrt*x - x^2)*(1 + Sqrt*x + x^2)

--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Tom Cage" <k5gj at earthlink.net> wrote in message
news:94ojm5\$emb at smc.vnet.net...
> I would like to factor 1+x^4.   Mathematica 3 will only respond with
> In:= Factor[1+x^4]
> Out= 1+x^4
>
>
>     Other systems will give the complex result
>
> 1+x^4 = (x+1/2*Sqrt(2)*I+1/2*Sqrt(2))*
>         (x+1/2*Sqrt(2)*I-1/2*Sqrt(2))*
>         (x-1/2*Sqrt(2)*I+1/2*Sqrt(2))*
>         (x-1/2*Sqrt(2)*I-1/2*Sqrt(2))
>
>
>     How would I factor 1+x^4 with Mathematica
>
>
>

```

• Prev by Date: Re: Overriding Power
• Next by Date: Re: Who can help me?
• Previous by thread: Re: Factor[1+x^4]
• Next by thread: Re: reference