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
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
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.
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.
To determine the possible rational zeros of the function f(x) = x^6 - 64, you correctly used a combination of the constant term and the leading coefficient.
First, you found the factors of the constant term, which in this case is 64. The factors of 64 are 1, 2, 4, 8, 16, 32, and 64.
Then, you found the factors of the leading coefficient, which is 1. The only factor of 1 is 1 itself.
To find the possible rational zeros, you took all the factors of the constant term (±1, ±2, ±4, ±8, ±16, ±32, ±64) and divided them by the factors of the leading coefficient (±1), resulting in the possible rational zeros: ±1, ±2, ±4, ±8, ±16, ±32, ±64.
Therefore, based on your explanation, it appears that you have correctly determined the possible rational zeros for the given function. Well done!