To determine the value of \( a \) such that the polynomial \( x^3 + ax + 2a - 2 \) is exactly divisible by \( x + 1 \), we can use the fact that if a polynomial \( f(x) \) is divisible by a binomial \( x - c \), then \( f(c) = 0 \). In this case, we want the polynomial to be divisible by \( x + 1 \), which is the same as \( x - (-1) \). Thus, \( f(-1) = 0 \).
Given the polynomial:
[tex]\[ f(x) = x^3 + ax + 2a - 2 \][/tex]
We substitute \( x = -1 \) into the polynomial:
[tex]\[ f(-1) = (-1)^3 + a(-1) + 2a - 2 \][/tex]
Simplifying the terms, we get:
[tex]\[ f(-1) = -1 - a + 2a - 2 \][/tex]
[tex]\[ f(-1) = -1 + a - 2 \][/tex]
[tex]\[ f(-1) = a - 3 \][/tex]
To ensure the polynomial is divisible by \( x + 1 \), we set \( f(-1) = 0 \):
[tex]\[ a - 3 = 0 \][/tex]
Solving for \( a \):
[tex]\[ a = 3 \][/tex]
Thus, the value of \( a \) is:
[tex]\[ \boxed{3} \][/tex]
