Please use the same nickname throughout your different posts.
You are making things way too complicated.
You are looking at a difference of squares
x^6 - 64
= (x^3 -8)(x^3 + 8)
now I see a sum and a difference of cubes
= (x-2)(x^2 + 2x + 4)(x+2)(x^2 -2x + 4)
the two quadratics don't factor over the rationals so they will not yield rational zeros
x = 2 or x = -2 are the rational zeros
Did I solve this problem correctly?
Directions: State the possible rational zeros for each function.
Question:f(x) = x6 - 64
Answer:
Constant term:64 Factors:1,2,4,8,16,32,64
Leading coefficient:1 Factors: 1
±1,2,4,8,16,32,64/1
= ±1, ±2, ±4, ±8, ±16, ±32, ±64
3 answers
NO, NO , NO.
f(x)=0=x^6-64
x^6=64
take sixth root of each side,
x=+-2
Now there are other roots, but not real, I suspect you have not had DeMoirves theorem, but in polar comples plane
x^6=64@360deg
x=2@n60 where n=0,1,2,3,4,5, and the n=0 or 3 gives the +-2 real roosts, with no imaginary part.
f(x)=0=x^6-64
x^6=64
take sixth root of each side,
x=+-2
Now there are other roots, but not real, I suspect you have not had DeMoirves theorem, but in polar comples plane
x^6=64@360deg
x=2@n60 where n=0,1,2,3,4,5, and the n=0 or 3 gives the +-2 real roosts, with no imaginary part.
As I read it, the directions were to list the possible rational zeros, not to actually find them.
Looks to me like Truda did it just right.
Looks to me like Truda did it just right.
2 answers