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: Rearrange the inequality
We start with the inequality:
\[ x^4 - 3x^3 - x + 3 < 0 \]

### Step 2: Factor the polynomial
We can try to factor the polynomial. We notice that we can group the terms:
\[ x^4 - 3x^3 - x + 3 = (x^4 - 3x^3) + (-x + 3) \]
This doesn't seem to help, so let's try finding the roots of the polynomial by using the Rational Root Theorem or synthetic division.

### Step 3: Find the roots
Let's test some integer values to find the roots of the polynomial. We can start with \( x = 1 \):
\[ 1^4 - 3(1^3) - 1 + 3 = 1 - 3 - 1 + 3 = 0 \]
So, \( x = 1 \) is a root. Now, we can factor \( x - 1 \) out of the polynomial.

Using synthetic division to divide \( x^4 - 3x^3 - x + 3 \) by \( x - 1 \):
\[ \begin{array}{r|rrrrr} 1 & 1 & -3 & 0 & -1 & 3 \\ & & 1 & -2 & -2 & -3 \\ \hline & 1 & -2 & -2 & -3 & 0 \\ \end{array} \]
This gives us:
\[ x^4 - 3x^3 - x + 3 = (x - 1)(x^3 - 2x^2 - 2x - 3) \]

### Step 4: Factor the cubic polynomial
Next, we need to find the roots of \( x^3 - 2x^2 - 2x - 3 \). We can test \( x = -1 \):
\[ (-1)^3 - 2(-1)^2 - 2(-1) - 3 = -1 - 2 + 2 - 3 = -4 \quad \text{(not a root)} \]
Now let's test \( x = 3 \):
\[ 3^3 - 2(3^2) - 2(3) - 3 = 27 - 18 - 6 - 3 = 0 \]
So, \( x = 3 \) is another root. We can factor \( x - 3 \) out of the cubic polynomial.

Using synthetic division again:
\[ \begin{array}{r|rrrr} 3 & 1 & -2 & -2 & -3 \\ & & 3 & 3 & 3 \\ \hline & 1 & 1 & 1 & 0 \\ \end{array} \]
This gives us:
\[ x^3 - 2x^2 - 2x - 3 = (x - 3)(x^2 + x + 1) \]

### Step 5: Complete factorization
Now we can write the complete factorization of the original polynomial:
\[ x^4 - 3x^3 - x + 3 = (x - 1)(x - 3)(x^2 + x + 1) \]

### Step 6: Analyze the inequality
Now we need to analyze the inequality:
\[ (x - 1)(x - 3)(x^2 + x + 1) < 0 \]
The quadratic \( x^2 + x + 1 \) is always positive (its discriminant is negative).

### Step 7: Find intervals
Now we only need to consider the sign of \( (x - 1)(x - 3) \):
- The roots are \( x = 1 \) and \( x = 3 \).
- The intervals to test are \( (-\infty, 1) \), \( (1, 3) \), and \( (3, \infty) \).

Testing these intervals:
1. For \( x < 1 \) (e.g., \( x = 0 \)): \( (0 - 1)(0 - 3) = (negative)(negative) = positive \)
2. For \( 1 < x < 3 \) (e.g., \( x = 2 \)): \( (2 - 1)(2 - 3) = (positive)(negative) = negative \)
3. For \( x > 3 \) (e.g., \( x = 4 \)): \( (4 - 1)(4 - 3) = (positive)(positive) = positive \)

Thus, the inequality \( (x - 1)(x - 3) < 0 \) is satisfied in the interval \( (1, 3) \).

### Step 8: Find integer solutions
The integral values in the interval \( (1, 3) \) are:
\[ x = 2 \]

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