Find the rational zeros and then the other zeros f(x)=x^3+27

The sum of cubes is a special form that has the factorization

... x³ + a³ = (x +a)(x² -ax +a²)

For a=3, the factorization is

... x³ + 3³ = (x +3)(x² -3x +9)

The first of these factors is zero for x=-3. The second of these factors is zero for

... x = (3±√-27)/2 = (3/2)(1±i√3) . . . . . from the quadratic formula

The zeros are {-3, (3/2)(1-i√3), (3/2)(1+i√3)}.

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation:

step-by-step explanation:

if you meant, "on the way from (3, 2) to (3, 14):

x does not change; y increases by 12.

ensure that you have copied down this problem correctly.   .