1. Apr 8, 2009

1. The problem statement, all variables and given/known data

x=2(pi)/7 we will show that this is not constructible and therefore 7-gon is not constructible.

a) show cos4x = cos3x
b) Use the above equation to find a rational quartic polynomial f(y)
where f(cos x) = 0.
c)From f, find a cubic rational polynomial g(y) where g(cos x) = 0

2. Relevant equations

see above

3. The attempt at a solution

im having trouble in part b). i expanded cos4x - cos3x = 0 in terms of cos(x) and I made the substitution y= cos(x) i got the quartic equation 8y^4 + 4y^3 - 8y^(2) -3y + 1 =0.
but when i put y= cos(2(pi)/7)) it dosent come out to 0.

2. Apr 8, 2009

### Billy Bob

You found cos 4x + cos 3x instead of minus. LOL, I made the same error.

3. Apr 8, 2009

oh right it should be -4 and +3 in the above equation, thanks billy do u know how to change this into a cubic equation.

4. Apr 8, 2009

### Billy Bob

Try long division? Divide by y-c, where c is a root of the quartic. Hopefully c is easy to find, by graphing, or guessing.

5. Apr 8, 2009