(3.54. \( \frac{x^{3}-x^{2}+x-1}{x+8} \leqslant 0 \).

To solve the inequality \[ \frac{x^{3}-x^{2}+x-1}{x+8} \leqslant 0, \] we first need to analyze the expression in the numerator, \( x^3 - x^2 + x - 1 \). ### Step 1: Factor the numerator We can try to factor the cubic polynomial \( x^3 - x^2 + x - 1 \). We can use the Rational Root Theorem to test for possible rational roots. Testing \( x = 1 \): \[ 1^3 - 1^2 + 1 - 1 = 1 - 1 + 1 - 1 = 0. \] Since \( x = 1 \) is a root, we can factor \( x - 1 \) out of the polynomial. We can perform polynomial long division or synthetic division to divide \( x^3 - x^2 + x - 1 \) by \( x - 1 \). Using synthetic division: ``` 1 | 1 -1 1 -1 | 1 0 1 ------------------- 1 0 1 0 ``` The result is \( x^2 + 1 \), so we have: \[ x^3 - x^2 + x - 1 = (x - 1)(x^2 + 1). \] ### Step 2: Rewrite the inequality Now we can rewrite the inequality: \[ \frac{(x - 1)(x^2 + 1)}{x + 8} \leqslant 0. \] ### Step 3: Analyze the factors 1. **Numerator**: - \( x - 1 = 0 \) gives \( x = 1 \). - \( x^2 + 1 > 0 \) for all \( x \) (since it is a sum of squares). 2. **Denominator**: - \( x + 8 = 0 \) gives \( x = -8 \). ### Step 4: Determine the sign of the expression We need to find the intervals where the expression is less than or equal to zero. The critical points are \( x = -8 \) and \( x = 1 \). We will test the sign of the expression in the intervals: - \( (-\infty, -8) \) - \( (-8, 1) \) - \( (1, \infty) \) **Interval \( (-\infty, -8) \)**: Choose \( x = -9 \): \[ \frac{(-9 - 1)(-9^2 + 1)}{-9 + 8} = \frac{(-10)(80)}{-1} > 0. \] **Interval \( (-8, 1) \)**: Choose \( x = 0 \): \[ \frac{(0 - 1)(0^2 + 1)}{0 + 8} = \frac{(-1)(1)}{8} < 0. \] **Interval \( (1, \infty) \)**: Choose \( x = 2 \): \[ \frac{(2 - 1)(2^2 + 1)}{2 + 8} = \frac{(1)(5)}{10} > 0. \] ### Step 5: Include critical points - At \( x = -8 \), the expression is undefined. - At \( x = 1 \), the expression equals zero. ### Step 6: Conclusion The solution to the inequality \( \frac{x^{3}-x^{2}+x-1}{x+8} \leqslant 0 \) is: \[ \boxed{(-8, 1]}. \]