Find integral 'x's which satisfy the inequality `x^(4) -3x^(3) -x +3 lt 0`

Answer

To solve the inequality \( x^4 - 3x^3 - x + 3 < 0 \), we will follow these steps:

### Step 1: Rewrite the inequality
We start with the inequality:
\[ x^4 - 3x^3 - x + 3 < 0 \]

### Step 2: Factor the polynomial
We can factor the polynomial by grouping. First, we group the terms:
\[ x^4 - 3x^3 - x + 3 = x^3(x - 3) - 1(x - 3) \]
Now, we can factor out \( (x - 3) \):
\[ = (x - 3)(x^3 - 1) \]

### Step 3: Factor \( x^3 - 1 \)
We can factor \( x^3 - 1 \) using the difference of cubes:
\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]
Thus, we can rewrite our expression as:
\[ (x - 3)(x - 1)(x^2 + x + 1) < 0 \]

### Step 4: Analyze the factors
Next, we need to find the roots of the factors:
- The roots of \( (x - 3) \) are \( x = 3 \)
- The roots of \( (x - 1) \) are \( x = 1 \)
- The quadratic \( x^2 + x + 1 \) has no real roots (its discriminant \( b^2 - 4ac = 1 - 4 < 0 \)).

### Step 5: Determine intervals
The critical points from the factors are \( x = 1 \) and \( x = 3 \). We will test the sign of the product in the intervals determined by these points:
1. \( (-\infty, 1) \)
2. \( (1, 3) \)
3. \( (3, \infty) \)

### Step 6: Test the intervals
- For \( x < 1 \) (e.g., \( x = 0 \)):
\[ (0 - 3)(0 - 1)(0^2 + 0 + 1) = (-3)(-1)(1) = 3 > 0 \]
- For \( 1 < x < 3 \) (e.g., \( x = 2 \)):
\[ (2 - 3)(2 - 1)(2^2 + 2 + 1) = (-1)(1)(7) = -7 < 0 \]
- For \( x > 3 \) (e.g., \( x = 4 \)):
\[ (4 - 3)(4 - 1)(4^2 + 4 + 1) = (1)(3)(21) = 63 > 0 \]

### Step 7: Conclusion
The inequality \( (x - 3)(x - 1)(x^2 + x + 1) < 0 \) holds true in the interval \( (1, 3) \). The integer values in this interval are:
\[ x = 2 \]

### Final Answer
The integral \( x \) that satisfies the inequality \( x^4 - 3x^3 - x + 3 < 0 \) is:
\[ \boxed{2} \]