To find the zeros of the polynomial \( N(x) = x^3 - 3x^2 + 4x + 8 \), we can use the Rational Root Theorem along with synthetic division or polynomial long division to find potential rational roots.
The possible rational roots are the factors of the constant term (8) divided by the factors of the leading coefficient (1). Therefore, the possible rational roots are:
\[ \pm 1, \pm 2, \pm 4, \pm 8 \]
We will test these potential roots in the polynomial:
- Testing \( x = -1 \):
\[ N(-1) = (-1)^3 - 3(-1)^2 + 4(-1) + 8 = -1 - 3 - 4 + 8 = 0 \]
So \( x = -1 \) is a root.
Next, we will perform synthetic division of \( N(x) \) by \( x + 1 \):
\[ \begin{array}{r|rrrr} -1 & 1 & -3 & 4 & 8 \ & & -1 & 4 & -8 \ \hline & 1 & -4 & 0 & 0 \ \end{array} \]
The resulting polynomial is \( x^2 - 4 \).
Now, we need to solve \( x^2 - 4 = 0 \):
\[ x^2 - 4 = (x - 2)(x + 2) = 0 \]
This gives us two additional roots:
\[ x = 2 \quad \text{and} \quad x = -2 \]
Therefore, the zeroes of the polynomial \( N(x) = x^3 - 3x^2 + 4x + 8 \) are:
\[ x = -1, \quad x = 2, \quad x = -2 \]
In exact form, the solutions are:
\[ \boxed{-1, 2, -2} \]