Solve the inequality (3x + 11)(x - 1) ext > 0

Set the first factor to zero:

Set the second factor to zero:

The critical points are $x = -\frac{11}{3}$ and $x = 1$.

The critical points are $x = -\frac{11}{3} \approx -3.67$ and $x = 1$. These points divide the number line into three intervals:

1. $x \leq -\frac{11}{3}$
2. $-\frac{11}{3} \leq x \leq 1$
3. $x \geq 1$

Alternatively, we can observe the shape of the quadratic function. When expanded, $(3x + 11)(x - 1) = 3x^2 + 8x - 11$. Since the coefficient of $x^2$ is positive ($3 > 0$), the parabola opens upwards. This means the expression is greater than or equal to zero outside or at its roots.

So, the inequality $(3x + 11)(x - 1) \geq 0$ is satisfied when $x$ is less than or equal to the smaller root, or greater than or equal to the larger root.
This means $x \leq -\frac{11}{3}$ or $x \geq 1$.

The solution set in interval notation is $\left(-\infty, -\frac{11}{3}\right] \cup [1, \infty)$.