Okay, so I need to find the sum of the coefficients of the polynomial g(x), which is equal to f(x-1). The original polynomial f(x) is given as 3x⁴ + 5x² - 9x - 2. Hmm, let me think about how to approach this.

First, I remember that when you substitute x-1 into a polynomial, you're essentially shifting the graph of the polynomial to the right by 1 unit. But how does that help me find the sum of the coefficients of g(x)? Wait, the sum of the coefficients of a polynomial is the same as evaluating the polynomial at x=1. Is that right? Let me verify. If I have a polynomial like a_n x^n + ... + a_1 x + a_0, then substituting x=1 gives a_n + ... + a_1 + a_0, which is exactly the sum of the coefficients. So yes, the sum of the coefficients of g(x) is g(1). But since g(x) = f(x-1), substituting x=1 into g gives f(1-1) = f(0). Therefore, the sum of the coefficients of g is equal to f(0). That seems straightforward. Let me check if that makes sense.

Alternatively, if I were to expand f(x-1) completely, then substitute x=1, that should also give the sum of the coefficients. But expanding f(x-1) might be more work. Let me see. If I substitute x=1 into f(x-1), that's f(1-1) = f(0). So f(0) is the constant term of the original polynomial. Let's compute f(0). The original polynomial is 3x⁴ + 5x² -9x -2. Plugging in x=0, all the terms with x in them become zero, so f(0) = -2. Therefore, the sum of the coefficients of g(x) should be -2. Wait, is that correct?

But let me double-check. Let's think about another way. If I expand f(x-1), which is 3(x-1)^4 + 5(x-1)^2 -9(x-1) -2. Then, when I expand each term, the coefficients of the resulting polynomial will be different, but when I sum all the coefficients, which is substituting x=1, it's the same as f(0). So yes, that's consistent. Therefore, the answer is -2.

But maybe I should verify by actually expanding f(x-1) and then summing the coefficients. Let's try that.

First, expand each term:

1. 3(x-1)^4: Let's compute (x-1)^4. Using the binomial theorem, (x-1)^4 = x^4 - 4x^3 + 6x^2 -4x +1. Multiply by 3: 3x^4 -12x^3 +18x^2 -12x +3.

2. 5(x-1)^2: (x-1)^2 = x^2 - 2x +1. Multiply by 5: 5x^2 -10x +5.

3. -9(x-1): Distribute the -9: -9x +9.

4. The constant term is -2.

Now, add all these together:

3x^4 -12x^3 +18x^2 -12x +3

+5x^2 -10x +5

-9x +9

-2

Combine like terms:

3x^4 term: 3x^4

-12x^3 term: -12x^3

18x^2 +5x^2 =23x^2

-12x -10x -9x = -31x

3 +5 +9 -2 =15

So the expanded form of g(x) is 3x^4 -12x^3 +23x^2 -31x +15.

Now, the sum of the coefficients is 3 -12 +23 -31 +15. Let's compute that step by step:

3 -12 = -9

-9 +23 =14

14 -31 = -17

-17 +15 = -2

Yes, that's the same as before. So the sum of the coefficients of g(x) is indeed -2. Therefore, the answer is \boxed{-2}.

But wait, let me check again if there's a mistake in expanding the terms. Let me go through each expansion step again:

First term: 3(x-1)^4. As before, (x-1)^4 is x^4 -4x^3 +6x^2 -4x +1. Multiply by 3: 3x^4 -12x^3 +18x^2 -12x +3. Correct.

Second term: 5(x-1)^2. (x-1)^2 is x^2 -2x +1. Multiply by 5: 5x^2 -10x +5. Correct.

Third term: -9(x-1) = -9x +9. Correct.

Fourth term: -2. Correct.

Adding them together:

3x^4 term: 3x^4.

-12x^3: only term with x^3.

18x^2 +5x^2 =23x^2.

-12x -10x -9x: -31x.

Constants: 3 +5 +9 -2 =15. Correct.

Sum of coefficients: 3 -12 +23 -31 +15. Let's compute again:

3 -12 = -9

-9 +23 =14

14 -31 = -17

-17 +15 = -2. Correct.

So, yes, the answer is definitely -2. That seems solid. Therefore, I can confidently say that the sum of the coefficients of g(x) is -2.

**Final Answer**
\boxed{-2}
Therefore, the sum of the coefficients of \( g \) is \boxed{-2}.
</think>