(2x+3)•|x+1|=

To simplify the expression \((2x + 3) \cdot |x + 1|\), we need to consider the properties of absolute values, as they impact the expression depending on the value of \(x\).

Case 1: \(x + 1 \geq 0\) (which means \(x \geq -1\))

In this case, \(|x + 1| = x + 1\). Therefore, the expression becomes:

\[ (2x + 3) \cdot |x + 1| = (2x + 3) \cdot (x + 1) \]

Now we can distribute:

\[ = 2x^2 + 2x + 3x + 3 = 2x^2 + 5x + 3 \]

Case 2: \(x + 1 < 0\) (which means \(x < -1\))

Here, \(|x + 1| = -(x + 1)\). Thus, the expression becomes:

\[ (2x + 3) \cdot |x + 1| = (2x + 3) \cdot (-(x + 1)) \]

Distributing this, we have:

\[ = - (2x + 3)(x + 1) = - (2x^2 + 2x + 3x + 3) = - (2x^2 + 5x + 3) = -2x^2 - 5x - 3 \]

Summary of the Results

The expression \( (2x + 3) \cdot |x + 1| \) can be summarized as follows:

\[ (2x + 3) \cdot |x + 1| = \begin{cases} 2x^2 + 5x + 3 & \text{if } x \geq -1 \\ -2x^2 - 5x - 3 & \text{if } x < -1 \end{cases} \]