Answer :

To find the value of [tex]\( a \)[/tex] such that 2 is a zero of the polynomial [tex]\( f(x) = a x^2 - 3(a-1) x - 1 \)[/tex], follow these steps:

1. Substitute [tex]\( x = 2 \)[/tex] into the polynomial: Since 2 is a zero of the polynomial, it should satisfy the equation [tex]\( f(2) = 0 \)[/tex].

[tex]\[ f(2) = a (2)^2 - 3(a-1) (2) - 1 \][/tex]

2. Simplify the expression: Substitute [tex]\( x = 2 \)[/tex] into the polynomial and simplify.

[tex]\[ f(2) = a (2^2) - 3(a-1) (2) - 1 = a (4) - 3(a-1) (2) - 1 \][/tex]

[tex]\[ f(2) = 4a - 3 [2(a - 1)] - 1 = 4a - 3 (2a - 2) - 1 \][/tex]

[tex]\[ f(2) = 4a - 6a + 6 - 1 = -2a + 5 \][/tex]

3. Set the simplified expression equal to 0: Since 2 is a zero of the polynomial, we have:

[tex]\[ -2a + 5 = 0 \][/tex]

4. Solve for [tex]\( a \)[/tex]:

[tex]\[ -2a + 5 = 0 \][/tex]

[tex]\[ -2a = -5 \][/tex]

[tex]\[ a = \frac{5}{2} \][/tex]

Hence, the value of [tex]\( a \)[/tex] is [tex]\(\boxed{\frac{5}{2}}\)[/tex].