Solve (x-3)(x+1)leq0

$$( x - 3 ) ( x + 1 ) \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-1,3\end{bmatrix}$$

Solution Steps

For the product to be ≤0, one of the values x+1 and x-3 has to be ≥0 and the other has to be ≤0. Consider the case when x+1\geq 0 and x-3\leq 0.

$$x+1\geq 0 x-3\leq 0$$

The solution satisfying both inequalities is x\in \left[-1,3\right].

$$x\in \begin{bmatrix}-1,3\end{bmatrix}$$

Consider the case when x+1\leq 0 and x-3\geq 0.

$$x-3\geq 0 x+1\leq 0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-1,3\end{bmatrix}$$