# Simple way to factor sextic

1. Aug 29, 2010

### elfboy

Anyone know of a good way to factor

x^6+x^4(13^.5/2-13/2)+x^2(13-2*13^.5)+3*13^.5/2-13/2 into 2 cubics?

without having to manipulate the messy cubic equation?

2. Aug 29, 2010

### epenguin

These things are intended as exercises in recognising patterns of which you have already dealt with simpler examples, woven into something more elaborate.

The rules are you are supposed to show an effort. Try and deal with different bits of it. For instance can you not do something with the part that has as factor a straight 13? The rest of it looks more difficult. I would ask you check whether you have really transcribed it exactly right. These things (I cannot imagine it is anything but an artificially constructed problem) are meant to work out to something that rhymes with sense.

Last edited: Aug 30, 2010
3. Aug 29, 2010

### elfboy

The equation looks arbitrary but its roots are plus minus sin(pi/13), sin(3pi/13), and sin(4pi/13)

4. Sep 27, 2010

### elfboy

After a month of working on this problem on and off I finally factored it in the way desired

(x^3+bx^2+c*x+(-3*13^.5/2+13/2)^.5)*(x^3-bx^2+c*x-(-3*13^.5/2+13/2)^.5)

13^.5/2-13/2=-b^2+2c
13-2*13^.5=c^2-b(26-6*13^.5)^.5

solve for b and c and plug into the above expression to obtain the two cubics:

(x^3+(13/2+3*13^.5/2)^.5*x^2+13^.5*x+(-3*13^.5/2+13/2)^.5)*
(x^3-(13/2+3*13^.5/2)^.5*x^2+13^.5*x-(-3*13^.5/2+13/2)^.5)=0

i didn't think this was possible to get a nice expression but i've done it