Okay, let's try to figure out the degree of the polynomial f(x) in the equation (x - 1) * f(x) = 3x⁴ + x³ - 25x² + 38x -17. Hmm, so we need to find the degree of f. 

First, I remember that when you multiply two polynomials, the degree of the product is the sum of the degrees of the two polynomials. So, if (x - 1) is a polynomial of degree 1, and f(x) is a polynomial of degree n, then their product should have degree 1 + n. 

Looking at the right-hand side of the equation, which is 3x⁴ + x³ - 25x² + 38x -17. The highest power here is x⁴, so the degree of the right-hand side is 4. 

Since the left-hand side is (x - 1) * f(x), which has degree 1 + degree(f(x)), and the right-hand side has degree 4, that means 1 + degree(f(x)) = 4. Solving for degree(f(x)) gives degree(f(x)) = 4 - 1 = 3. 

Wait, so f(x) is a cubic polynomial? Let me check if there's any catch here. For example, could there be a case where multiplying by (x - 1) reduces the degree? But no, multiplying by a first-degree polynomial can't reduce the degree. If (x - 1) is a factor of the right-hand side, then f(x) would be the quotient when dividing the right-hand side by (x - 1). But regardless of whether (x - 1) is a factor or not, the degree of the product is still 1 + degree(f(x)). 

So even if (x - 1) is a factor of the right-hand side, the degree of the quotient f(x) would still be 4 - 1 = 3. Because when you divide a degree 4 polynomial by a degree 1 polynomial, you get a degree 3 polynomial. 

Let me verify this with an example. Suppose I have (x - 1) * (ax³ + bx² + cx + d) = 3x⁴ + x³ - 25x² + 38x -17. Expanding the left side: x*(ax³ + bx² + cx + d) - 1*(ax³ + bx² + cx + d) = ax⁴ + bx³ + cx² + dx - ax³ - bx² - cx - d. Combining like terms: ax⁴ + (b - a)x³ + (c - b)x² + (d - c)x - d. 

Set this equal to the right-hand side: 3x⁴ + x³ - 25x² + 38x -17. So equating coefficients:

ax⁴: a = 3

x³: b - a = 1 => b = a + 1 = 3 + 1 = 4

x²: c - b = -25 => c = b -25 = 4 -25 = -21

x: d - c = 38 => d = c +38 = -21 +38 = 17

constant term: -d = -17 => d =17, which matches. 

So f(x) would be 3x³ +4x² -21x +17, which is indeed a cubic polynomial, degree 3. 

Therefore, the degree of f(x) is 3. 

Just to make sure there's no trick here, maybe (x - 1) is a factor of the right-hand side. Let's check if x=1 is a root of the right-hand side polynomial. Substitute x=1 into 3(1)^4 + (1)^3 -25(1)^2 +38(1) -17 = 3 +1 -25 +38 -17 = (3+1) + (-25+38) -17 = 4 +13 -17 = 0. 

Oh, so x=1 is a root. That means (x -1) is a factor, so when we divide the right-hand side by (x -1), we get a cubic polynomial, which is f(x). So even though (x -1) is a factor, the degree is still 3. 

Therefore, the answer is 3.
Therefore, the degree of $f$ is $\boxed{3}$.
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