Solve (x+2)(x+3)+3(x+2)leq0

$$(x+2)(x+3)+3(x+2) \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-6,-2\end{bmatrix}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+2 by x+3 and combine like terms.

$$x^{2}+5x+6+3\left(x+2\right)\leq 0$$

Use the distributive property to multiply 3 by x+2.

$$x^{2}+5x+6+3x+6\leq 0$$

Combine 5x and 3x to get 8x.

$$x^{2}+8x+6+6\leq 0$$

Add 6 and 6 to get 12.

$$x^{2}+8x+12\leq 0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}+8x+12=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 8 for b, and 12 for c in the quadratic formula.

$$x=\frac{-8±\sqrt{8^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$x=\frac{-8±4}{2}$$

Solve the equation x=\frac{-8±4}{2} when ± is plus and when ± is minus.

$$x=-2 x=-6$$

Rewrite the inequality by using the obtained solutions.

$$\left(x+2\right)\left(x+6\right)\leq 0$$

For the product to be ≤0, one of the values x+2 and x+6 has to be ≥0 and the other has to be ≤0. Consider the case when x+2\geq 0 and x+6\leq 0.

$$x+2\geq 0 x+6\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x+2\leq 0 and x+6\geq 0.

$$x+6\geq 0 x+2\leq 0$$

The solution satisfying both inequalities is x\in \left[-6,-2\right].

$$x\in \begin{bmatrix}-6,-2\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-6,-2\end{bmatrix}$$