Explanation:
Okay, so I need to solve this equation: Half the value of 3x minus 9 is equal to x plus 37. Let me try to parse that. The problem says "Half the value of 3x - 9 is x + 37." Hmm. So, translating that into an equation, half of (3x - 9) equals x + 37. Let me write that down.

So, half of (3x - 9) would be (1/2)(3x - 9). And that's equal to x + 37. So the equation is:

(1/2)(3x - 9) = x + 37

Alright, now I need to solve for x. Let me think about how to approach this. Maybe I can start by eliminating the fraction to make it easier. If I multiply both sides of the equation by 2, that should get rid of the denominator. Let me try that.

Multiplying both sides by 2:

2 * [ (1/2)(3x - 9) ] = 2 * (x + 37)

Simplifying the left side: The 2 and the 1/2 cancel each other out, so it becomes 3x - 9. On the right side, distributing the 2 gives 2x + 74. So now the equation is:

3x - 9 = 2x + 74

Okay, now I need to solve for x. Let me subtract 2x from both sides to get the x terms on one side. So:

3x - 2x - 9 = 74

Which simplifies to:

x - 9 = 74

Then, adding 9 to both sides:

x = 74 + 9

x = 83

Wait, let me check if that makes sense. Let me plug x = 83 back into the original equation to verify.

Original equation: Half of (3x - 9) equals x + 37.

Left side: (1/2)(3*83 - 9) = (1/2)(249 - 9) = (1/2)(240) = 120

Right side: 83 + 37 = 120

Yes, both sides equal 120. So x = 83 is correct. Hmm, seems straightforward. Let me just go through the steps again to make sure I didn't skip anything or make a mistake in arithmetic.

Starting with (1/2)(3x - 9) = x + 37. Multiply both sides by 2: 3x - 9 = 2x + 74. Subtract 2x: x - 9 = 74. Add 9: x = 83. Yep, that's correct. And plugging back in works. So I think that's the right answer.

But just to make sure, let me try solving it another way. Maybe expanding or something else. Let's see. Alternatively, I could have distributed the 1/2 first. Let me try that approach.

Original equation: (1/2)(3x - 9) = x + 37

Distribute the 1/2: (1/2)*3x - (1/2)*9 = x + 37

Which is (3/2)x - 9/2 = x + 37

Now, maybe subtract x from both sides to get the x terms on the left:

(3/2)x - x - 9/2 = 37

Calculating (3/2)x - x: that's (3/2 - 2/2)x = (1/2)x

So, (1/2)x - 9/2 = 37

Now, add 9/2 to both sides:

(1/2)x = 37 + 9/2

Convert 37 to halves: 37 is 74/2, so 74/2 + 9/2 = 83/2

Therefore, (1/2)x = 83/2

Multiply both sides by 2:

x = 83

Same result. So that confirms it again. So x is indeed 83. I think that's solid. Maybe I can check with another method? Let's see. Maybe using substitution or something else. But I think these two methods are sufficient. Both lead to x = 83, and plugging back in works. So I think that's the correct answer.

Alternatively, if I had made a mistake in arithmetic, maybe I would have gotten a different answer, but since both methods agree and the check works, I think it's safe. Let me just check my arithmetic once more.

Original equation: (1/2)(3x - 9) = x + 37

Multiply both sides by 2: 3x - 9 = 2x + 74

Subtract 2x: x - 9 = 74

Add 9: x = 83. Yep. 3*83 is 249, minus 9 is 240, half of that is 120. 83 + 37 is 120. Perfect. So no issues there.

Another thing to check: Did I interpret the original problem correctly? The problem says "Half the value of 3x - 9 is x + 37." So "half the value of" something is equal to something else. So yes, that translates directly to (1/2)(3x - 9) = x + 37. There's no ambiguity here. If it had said "half of 3x minus 9," that would be different, but it says "half the value of 3x - 9," so the entire expression 3x - 9 is being halved. So that's correct.

Alternatively, if someone misreads it as (3x - 9)/2 = x + 37, which is the same thing. So no confusion there. So yeah, the equation is set up correctly.

Therefore, I think I can confidently say that x = 83 is the correct solution.

**Final Answer**
The value of $ x $ is \boxed{83}.

To solve the equation:

> "Half the value of $ 3x - 9 $ is $ x + 37 $,"

we begin by translating the verbal statement into a mathematical equation. The phrase "half the value of $ 3x - 9 $" means we take one-half of the expression $ 3x - 9 $, which is:

$$
\frac{1}{2}(3x - 9)
$$

This is equal to $ x + 37 $, so the equation becomes:

$$
\frac{1}{2}(3x - 9) = x + 37
$$

---

### Step 1: Eliminate the Fraction

To simplify the equation, we can eliminate the fraction by multiplying both sides by 2:

$$
2 \cdot \left( \frac{1}{2}(3x - 9) \right) = 2 \cdot (x + 37)
$$

Simplifying both sides:

$$
3x - 9 = 2x + 74
$$

---

### Step 2: Solve for $ x $

Now, subtract $ 2x $ from both sides to isolate the variable:

$$
3x - 2x - 9 = 74
$$

$$
x - 9 = 74
$$

Add 9 to both sides:

$$
x = 74 + 9 = 83
$$

---

### Step 3: Verify the Solution

To ensure the solution is correct, substitute $ x = 83 $ back into the original equation:

**Left side:**

$$
\frac{1}{2}(3x - 9) = \frac{1}{2}(3 \cdot 83 - 9) = \frac{1}{2}(249 - 9) = \frac{1}{2}(240) = 120
$$

**Right side:**

$$
x + 37 = 83 + 37 = 120
$$

Since both sides are equal, the solution is verified.

---

### Final Answer

$$
\boxed{83}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: Half the value of 3x minus 9 is equal to x plus 37. Let me try to parse that. The problem says "Half the value of 3x - 9 is x + 37." Hmm. So, translating that into an equation, half of (3x - 9) equals x + 37. Let me write that down.

So, half of (3x - 9) would be (1/2)(3x - 9). And that's equal to x + 37. So the equation is:

(1/2)(3x - 9) = x + 37

Alright, now I need to solve for x. Let me think about how to approach this. Maybe I can start by eliminating the fraction to make it easier. If I multiply both sides of the equation by 2, that should get rid of the denominator. Let me try that.

Multiplying both sides by 2:

2 * [ (1/2)(3x - 9) ] = 2 * (x + 37)

Simplifying the left side: The 2 and the 1/2 cancel each other out, so it becomes 3x - 9. On the right side, distributing the 2 gives 2x + 74. So now the equation is:

3x - 9 = 2x + 74

Okay, now I need to solve for x. Let me subtract 2x from both sides to get the x terms on one side. So:

3x - 2x - 9 = 74

Which simplifies to:

x - 9 = 74

Then, adding 9 to both sides:

x = 74 + 9

x = 83

Wait, let me check if that makes sense. Let me plug x = 83 back into the original equation to verify.

Original equation: Half of (3x - 9) equals x + 37.

Left side: (1/2)(3*83 - 9) = (1/2)(249 - 9) = (1/2)(240) = 120

Right side: 83 + 37 = 120

Yes, both sides equal 120. So x = 83 is correct. Hmm, seems straightforward. Let me just go through the steps again to make sure I didn't skip anything or make a mistake in arithmetic.

Starting with (1/2)(3x - 9) = x + 37. Multiply both sides by 2: 3x - 9 = 2x + 74. Subtract 2x: x - 9 = 74. Add 9: x = 83. Yep, that's correct. And plugging back in works. So I think that's the right answer.

But just to make sure, let me try solving it another way. Maybe expanding or something else. Let's see. Alternatively, I could have distributed the 1/2 first. Let me try that approach.

Original equation: (1/2)(3x - 9) = x + 37

Distribute the 1/2: (1/2)*3x - (1/2)*9 = x + 37

Which is (3/2)x - 9/2 = x + 37

Now, maybe subtract x from both sides to get the x terms on the left:

(3/2)x - x - 9/2 = 37

Calculating (3/2)x - x: that's (3/2 - 2/2)x = (1/2)x

So, (1/2)x - 9/2 = 37

Now, add 9/2 to both sides:

(1/2)x = 37 + 9/2

Convert 37 to halves: 37 is 74/2, so 74/2 + 9/2 = 83/2

Therefore, (1/2)x = 83/2

Multiply both sides by 2:

x = 83

Same result. So that confirms it again. So x is indeed 83. I think that's solid. Maybe I can check with another method? Let's see. Maybe using substitution or something else. But I think these two methods are sufficient. Both lead to x = 83, and plugging back in works. So I think that's the correct answer.

Alternatively, if I had made a mistake in arithmetic, maybe I would have gotten a different answer, but since both methods agree and the check works, I think it's safe. Let me just check my arithmetic once more.

Original equation: (1/2)(3x - 9) = x + 37

Multiply both sides by 2: 3x - 9 = 2x + 74

Subtract 2x: x - 9 = 74

Add 9: x = 83. Yep. 3*83 is 249, minus 9 is 240, half of that is 120. 83 + 37 is 120. Perfect. So no issues there.

Another thing to check: Did I interpret the original problem correctly? The problem says "Half the value of 3x - 9 is x + 37." So "half the value of" something is equal to something else. So yes, that translates directly to (1/2)(3x - 9) = x + 37. There's no ambiguity here. If it had said "half of 3x minus 9," that would be different, but it says "half the value of 3x - 9," so the entire expression 3x - 9 is being halved. So that's correct.

Alternatively, if someone misreads it as (3x - 9)/2 = x + 37, which is the same thing. So no confusion there. So yeah, the equation is set up correctly.

Therefore, I think I can confidently say that x = 83 is the correct solution.

**Final Answer**
The value of $ x $ is \boxed{83}.

To solve the equation:

> "Half the value of $ 3x - 9 $ is $ x + 37 $,"

we begin by translating the verbal statement into a mathematical equation. The phrase "half the value of $ 3x - 9 $" means we take one-half of the expression $ 3x - 9 $, which is:

$$
\frac{1}{2}(3x - 9)
$$

This is equal to $ x + 37 $, so the equation becomes:

$$
\frac{1}{2}(3x - 9) = x + 37
$$

---

### Step 1: Eliminate the Fraction

To simplify the equation, we can eliminate the fraction by multiplying both sides by 2:

$$
2 \cdot \left( \frac{1}{2}(3x - 9) \right) = 2 \cdot (x + 37)
$$

Simplifying both sides:

$$
3x - 9 = 2x + 74
$$

---

### Step 2: Solve for $ x $

Now, subtract $ 2x $ from both sides to isolate the variable:

$$
3x - 2x - 9 = 74
$$

$$
x - 9 = 74
$$

Add 9 to both sides:

$$
x = 74 + 9 = 83
$$

---

### Step 3: Verify the Solution

To ensure the solution is correct, substitute $ x = 83 $ back into the original equation:

**Left side:**

$$
\frac{1}{2}(3x - 9) = \frac{1}{2}(3 \cdot 83 - 9) = \frac{1}{2}(249 - 9) = \frac{1}{2}(240) = 120
$$

**Right side:**

$$
x + 37 = 83 + 37 = 120
$$

Since both sides are equal, the solution is verified.

---

### Final Answer

$$
\boxed{83}
$$

Solution: We translate the problem to the equation $\frac{1}{2}(3x-9) = x+37$. Multiplying both sides by 2 gives $3x-9 = 2x+74$. Subtracting $2x$ from both sides yield $x-9 = 74$. Adding $9$ to both sides yields $ x = \boxed{83}$.


Explanation:
Okay, so I need to figure out the degree of the polynomial f. The problem says that (x - 1) multiplied by f(x) equals this polynomial: 3x⁴ + x³ - 25x² + 38x - 17. Hmm, let me think about how to approach this.

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 I have (x - 1) times f(x), then the degree of the product should be equal to the degree of (x - 1) plus the degree of f(x).

Let me write that down. Let’s denote the degree of f(x) as n. Then, the degree of (x - 1) is 1, since it's a linear polynomial. Therefore, the degree of the product (x - 1) * f(x) should be 1 + n.

But the problem gives me the product as 3x⁴ + x³ - 25x² + 38x - 17. Let me check the degree of that polynomial. The highest power of x here is 4, so the degree is 4.

So, according to the earlier reasoning, 1 + n = 4. Solving for n, that would be n = 4 - 1 = 3. Therefore, the degree of f(x) should be 3? Wait, is that all? Let me make sure there isn't a trick here.

But hold on, maybe there's something else. For example, if (x - 1) is a factor of the polynomial on the right-hand side, then f(x) would be that polynomial divided by (x - 1). But if (x - 1) is a factor, then when we divide, there shouldn't be any remainder. However, the problem doesn't state that (x - 1) is a factor, but actually, it's given that (x - 1) multiplied by f(x) equals that polynomial. Therefore, (x - 1) must be a factor, so the division should result in a polynomial f(x) with no remainder.

But maybe there's a case where even if (x - 1) is not a factor, but the product is given? Wait, no. The equation is (x - 1) * f(x) equals that polynomial. Therefore, (x - 1) must divide that polynomial. Therefore, the division should be exact.

But maybe I need to check if (x - 1) is actually a factor of the given polynomial? Let me try plugging in x = 1 into the polynomial to see if it equals zero. If it does, then (x - 1) is a factor.

Let me compute the value of the polynomial at x = 1:

3(1)^4 + (1)^3 - 25(1)^2 + 38(1) - 17.

Calculating each term:

3(1) = 3

1^3 = 1

-25(1)^2 = -25

38(1) = 38

-17 remains.

Adding them up: 3 + 1 = 4; 4 - 25 = -21; -21 + 38 = 17; 17 - 17 = 0.

Oh, so the polynomial evaluates to zero when x = 1. Therefore, (x - 1) is indeed a factor. Therefore, f(x) is the quotient when we divide the given polynomial by (x - 1).

So, if I divide 3x⁴ + x³ - 25x² + 38x - 17 by (x - 1), the quotient will be f(x). Since the degree of the original polynomial is 4 and we are dividing by a degree 1 polynomial, the quotient should have degree 4 - 1 = 3. Therefore, f(x) is a cubic polynomial, so its degree is 3.

But let me confirm this by actually performing the division, just to make sure there isn't a mistake in my reasoning. Maybe I can do polynomial long division or synthetic division. Let me try synthetic division since it's faster for linear factors.

For synthetic division with root x = 1, the coefficients of the polynomial are 3, 1, -25, 38, -17.

Set up the coefficients:

1 | 3   1   -25   38   -17

Bring down the 3.

Multiply 3 by 1 = 3. Add to next coefficient: 1 + 3 = 4.

Multiply 4 by 1 = 4. Add to next coefficient: -25 + 4 = -21.

Multiply -21 by 1 = -21. Add to next coefficient: 38 + (-21) = 17.

Multiply 17 by 1 = 17. Add to next coefficient: -17 + 17 = 0.

So the remainder is 0, as expected, and the coefficients of the quotient are 3, 4, -21, 17. Therefore, the quotient polynomial is 3x³ + 4x² - 21x + 17.

Therefore, f(x) is 3x³ + 4x² - 21x + 17, which is indeed a cubic polynomial. Therefore, the degree of f(x) is 3.

Wait, but let me check if there's any possibility that f(x) could have a higher degree? For example, if there was some cancellation? But no, because when you multiply (x - 1) by f(x), the leading term of the product is the leading term of (x - 1) times the leading term of f(x).

Let me think. Suppose f(x) is degree n. Then, the leading term of (x - 1) * f(x) is x * (leading term of f(x)). So, if the original polynomial has leading term 3x⁴, then the leading term of f(x) must be 3x³, since x * 3x³ = 3x⁴. Therefore, the leading term of f(x) is 3x³, so the degree of f(x) is 3.

Therefore, regardless of the other terms, the degree is 3. So even if I didn't do the division, I could have reasoned that the degree of f(x) is 3.

But just to make sure, let me check the division result. If f(x) is 3x³ + 4x² - 21x + 17, then multiplying by (x - 1):

First, multiply 3x³ by (x - 1): 3x⁴ - 3x³.

Then, multiply 4x² by (x - 1): 4x³ - 4x².

Then, multiply -21x by (x - 1): -21x² + 21x.

Then, multiply 17 by (x - 1): 17x - 17.

Now, add all these together:

3x⁴ - 3x³ + 4x³ - 4x² - 21x² + 21x + 17x - 17.

Combine like terms:

For x⁴: 3x⁴.

For x³: (-3x³ + 4x³) = 1x³.

For x²: (-4x² -21x²) = -25x².

For x terms: (21x + 17x) = 38x.

Constants: -17.

So altogether, 3x⁴ + x³ -25x² + 38x -17, which matches the given polynomial. Therefore, the division is correct, and f(x) is indeed a cubic polynomial.

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

But just to recap, the key steps were:

1. Recognize that multiplying two polynomials adds their degrees.

2. Since the product is degree 4, and (x - 1) is degree 1, then f(x) must be degree 3.

3. Verified by synthetic division that dividing the given polynomial by (x - 1) gives a cubic polynomial with no remainder.

4. Checked by multiplying back to confirm the result.

Therefore, the answer is 3.

I don't think there's any other possible way this could be approached. Maybe someone might think that if there was a remainder, but the problem states that (x - 1) * f(x) equals that polynomial, so there can't be a remainder. Therefore, the division is exact, and the degree is indeed 3.

Another way to think about it: if f(x) had a degree higher than 3, say 4, then multiplying by (x - 1) would result in a degree 5 polynomial, which contradicts the given degree 4. Similarly, if f(x) had degree less than 3, say 2, then multiplying by (x - 1) would result in degree 3, which is less than 4. Therefore, the only possibility is degree 3.

So, all methods point to the degree of f(x) being 3.

**Final Answer**
The degree of $ f $ is \boxed{3}.

To determine the degree of the polynomial $ f(x) $, we are given the equation:

$$
(x - 1) \cdot f(x) = 3x^4 + x^3 - 25x^2 + 38x - 17
$$

---

### Step 1: Understand the Degree Relationship

When two polynomials are multiplied, the degree of the product is the **sum** of the degrees of the two polynomials. That is:

$$
\deg((x - 1) \cdot f(x)) = \deg(x - 1) + \deg(f(x))
$$

We are told that the product is a polynomial of degree 4. The polynomial $ x - 1 $ is linear, so its degree is 1. Let the degree of $ f(x) $ be $ n $. Then:

$$
1 + n = 4 \Rightarrow n = 3
$$

So, the degree of $ f(x) $ is **3**.

---

### Step 2: Confirm by Polynomial Division

To ensure this is correct, we can perform **synthetic division** of the given polynomial by $ x - 1 $, since the equation implies that $ x - 1 $ is a factor of the polynomial.

The polynomial is:

$$
3x^4 + x^3 - 25x^2 + 38x - 17
$$

Using synthetic division with root $ x = 1 $, we proceed as follows:

```
1 |  3   1   -25   38   -17
    |      3    4   -21    17
    -------------------------
      3   4   -21   17     0
```

The remainder is 0, confirming that $ x - 1 $ is indeed a factor. The quotient is:

$$
3x^3 + 4x^2 - 21x + 17
$$

This is a cubic polynomial, so the degree of $ f(x) $ is 3.

---

### Step 3: Final Verification

To double-check, we can multiply the quotient $ f(x) = 3x^3 + 4x^2 - 21x + 17 $ by $ x - 1 $:

$$
(x - 1)(3x^3 + 4x^2 - 21x + 17) = 3x^4 + x^3 - 25x^2 + 38x - 17
$$

This matches the original polynomial, confirming that the division is correct and that $ f(x) $ is indeed a cubic polynomial.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the degree of the polynomial f. The problem says that (x - 1) multiplied by f(x) equals this polynomial: 3x⁴ + x³ - 25x² + 38x - 17. Hmm, let me think about how to approach this.

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 I have (x - 1) times f(x), then the degree of the product should be equal to the degree of (x - 1) plus the degree of f(x).

Let me write that down. Let’s denote the degree of f(x) as n. Then, the degree of (x - 1) is 1, since it's a linear polynomial. Therefore, the degree of the product (x - 1) * f(x) should be 1 + n.

But the problem gives me the product as 3x⁴ + x³ - 25x² + 38x - 17. Let me check the degree of that polynomial. The highest power of x here is 4, so the degree is 4.

So, according to the earlier reasoning, 1 + n = 4. Solving for n, that would be n = 4 - 1 = 3. Therefore, the degree of f(x) should be 3? Wait, is that all? Let me make sure there isn't a trick here.

But hold on, maybe there's something else. For example, if (x - 1) is a factor of the polynomial on the right-hand side, then f(x) would be that polynomial divided by (x - 1). But if (x - 1) is a factor, then when we divide, there shouldn't be any remainder. However, the problem doesn't state that (x - 1) is a factor, but actually, it's given that (x - 1) multiplied by f(x) equals that polynomial. Therefore, (x - 1) must be a factor, so the division should result in a polynomial f(x) with no remainder.

But maybe there's a case where even if (x - 1) is not a factor, but the product is given? Wait, no. The equation is (x - 1) * f(x) equals that polynomial. Therefore, (x - 1) must divide that polynomial. Therefore, the division should be exact.

But maybe I need to check if (x - 1) is actually a factor of the given polynomial? Let me try plugging in x = 1 into the polynomial to see if it equals zero. If it does, then (x - 1) is a factor.

Let me compute the value of the polynomial at x = 1:

3(1)^4 + (1)^3 - 25(1)^2 + 38(1) - 17.

Calculating each term:

3(1) = 3

1^3 = 1

-25(1)^2 = -25

38(1) = 38

-17 remains.

Adding them up: 3 + 1 = 4; 4 - 25 = -21; -21 + 38 = 17; 17 - 17 = 0.

Oh, so the polynomial evaluates to zero when x = 1. Therefore, (x - 1) is indeed a factor. Therefore, f(x) is the quotient when we divide the given polynomial by (x - 1).

So, if I divide 3x⁴ + x³ - 25x² + 38x - 17 by (x - 1), the quotient will be f(x). Since the degree of the original polynomial is 4 and we are dividing by a degree 1 polynomial, the quotient should have degree 4 - 1 = 3. Therefore, f(x) is a cubic polynomial, so its degree is 3.

But let me confirm this by actually performing the division, just to make sure there isn't a mistake in my reasoning. Maybe I can do polynomial long division or synthetic division. Let me try synthetic division since it's faster for linear factors.

For synthetic division with root x = 1, the coefficients of the polynomial are 3, 1, -25, 38, -17.

Set up the coefficients:

1 | 3   1   -25   38   -17

Bring down the 3.

Multiply 3 by 1 = 3. Add to next coefficient: 1 + 3 = 4.

Multiply 4 by 1 = 4. Add to next coefficient: -25 + 4 = -21.

Multiply -21 by 1 = -21. Add to next coefficient: 38 + (-21) = 17.

Multiply 17 by 1 = 17. Add to next coefficient: -17 + 17 = 0.

So the remainder is 0, as expected, and the coefficients of the quotient are 3, 4, -21, 17. Therefore, the quotient polynomial is 3x³ + 4x² - 21x + 17.

Therefore, f(x) is 3x³ + 4x² - 21x + 17, which is indeed a cubic polynomial. Therefore, the degree of f(x) is 3.

Wait, but let me check if there's any possibility that f(x) could have a higher degree? For example, if there was some cancellation? But no, because when you multiply (x - 1) by f(x), the leading term of the product is the leading term of (x - 1) times the leading term of f(x).

Let me think. Suppose f(x) is degree n. Then, the leading term of (x - 1) * f(x) is x * (leading term of f(x)). So, if the original polynomial has leading term 3x⁴, then the leading term of f(x) must be 3x³, since x * 3x³ = 3x⁴. Therefore, the leading term of f(x) is 3x³, so the degree of f(x) is 3.

Therefore, regardless of the other terms, the degree is 3. So even if I didn't do the division, I could have reasoned that the degree of f(x) is 3.

But just to make sure, let me check the division result. If f(x) is 3x³ + 4x² - 21x + 17, then multiplying by (x - 1):

First, multiply 3x³ by (x - 1): 3x⁴ - 3x³.

Then, multiply 4x² by (x - 1): 4x³ - 4x².

Then, multiply -21x by (x - 1): -21x² + 21x.

Then, multiply 17 by (x - 1): 17x - 17.

Now, add all these together:

3x⁴ - 3x³ + 4x³ - 4x² - 21x² + 21x + 17x - 17.

Combine like terms:

For x⁴: 3x⁴.

For x³: (-3x³ + 4x³) = 1x³.

For x²: (-4x² -21x²) = -25x².

For x terms: (21x + 17x) = 38x.

Constants: -17.

So altogether, 3x⁴ + x³ -25x² + 38x -17, which matches the given polynomial. Therefore, the division is correct, and f(x) is indeed a cubic polynomial.

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

But just to recap, the key steps were:

1. Recognize that multiplying two polynomials adds their degrees.

2. Since the product is degree 4, and (x - 1) is degree 1, then f(x) must be degree 3.

3. Verified by synthetic division that dividing the given polynomial by (x - 1) gives a cubic polynomial with no remainder.

4. Checked by multiplying back to confirm the result.

Therefore, the answer is 3.

I don't think there's any other possible way this could be approached. Maybe someone might think that if there was a remainder, but the problem states that (x - 1) * f(x) equals that polynomial, so there can't be a remainder. Therefore, the division is exact, and the degree is indeed 3.

Another way to think about it: if f(x) had a degree higher than 3, say 4, then multiplying by (x - 1) would result in a degree 5 polynomial, which contradicts the given degree 4. Similarly, if f(x) had degree less than 3, say 2, then multiplying by (x - 1) would result in degree 3, which is less than 4. Therefore, the only possibility is degree 3.

So, all methods point to the degree of f(x) being 3.

**Final Answer**
The degree of $ f $ is \boxed{3}.

To determine the degree of the polynomial $ f(x) $, we are given the equation:

$$
(x - 1) \cdot f(x) = 3x^4 + x^3 - 25x^2 + 38x - 17
$$

---

### Step 1: Understand the Degree Relationship

When two polynomials are multiplied, the degree of the product is the **sum** of the degrees of the two polynomials. That is:

$$
\deg((x - 1) \cdot f(x)) = \deg(x - 1) + \deg(f(x))
$$

We are told that the product is a polynomial of degree 4. The polynomial $ x - 1 $ is linear, so its degree is 1. Let the degree of $ f(x) $ be $ n $. Then:

$$
1 + n = 4 \Rightarrow n = 3
$$

So, the degree of $ f(x) $ is **3**.

---

### Step 2: Confirm by Polynomial Division

To ensure this is correct, we can perform **synthetic division** of the given polynomial by $ x - 1 $, since the equation implies that $ x - 1 $ is a factor of the polynomial.

The polynomial is:

$$
3x^4 + x^3 - 25x^2 + 38x - 17
$$

Using synthetic division with root $ x = 1 $, we proceed as follows:

```
1 |  3   1   -25   38   -17
    |      3    4   -21    17
    -------------------------
      3   4   -21   17     0
```

The remainder is 0, confirming that $ x - 1 $ is indeed a factor. The quotient is:

$$
3x^3 + 4x^2 - 21x + 17
$$

This is a cubic polynomial, so the degree of $ f(x) $ is 3.

---

### Step 3: Final Verification

To double-check, we can multiply the quotient $ f(x) = 3x^3 + 4x^2 - 21x + 17 $ by $ x - 1 $:

$$
(x - 1)(3x^3 + 4x^2 - 21x + 17) = 3x^4 + x^3 - 25x^2 + 38x - 17
$$

This matches the original polynomial, confirming that the division is correct and that $ f(x) $ is indeed a cubic polynomial.

---

### Final Answer

$$
\boxed{3}
$$

Solution: Since the product of $f$ and a polynomial with degree 1 equals a polynomial with degree 4, we know that $f$ is a polynomial with degree $4-1=\boxed{3}$.