1. **State the problem:** Solve the inequality $ (x-2)(x+1) \leq 3(x+1) $. 2. **Rewrite the inequality:** Expand the left side and keep the right side as is: $$ (x-2)(x+1) \leq 3(x+1) $$ $$ x^2 + x - 2x - 2 \leq 3x + 3 $$ $$ x^2 - x - 2 \leq 3x + 3 $$ 3. **Bring all terms to one side:** $$ x^2 - x - 2 - 3x - 3 \leq 0 $$ $$ x^2 - 4x - 5 \leq 0 $$ 4. **Factor the quadratic:** $$ x^2 - 4x - 5 = (x - 5)(x + 1) $$ 5. **Rewrite the inequality:** $$ (x - 5)(x + 1) \leq 0 $$ 6. **Analyze the inequality:** The product of two factors is less than or equal to zero when one factor is non-positive and the other is non-negative. The critical points are $x = 5$ and $x = -1$. 7. **Test intervals:** - For $x < -1$, both $(x-5)$ and $(x+1)$ are negative, so product is positive. - For $-1 \leq x \leq 5$, $(x-5)$ is negative or zero, $(x+1)$ is positive or zero, so product is less than or equal to zero. - For $x > 5$, both factors are positive, product positive. 8. **Solution:** $$ \boxed{-1 \leq x \leq 5} $$ This is the set of $x$ values satisfying the inequality.