Solve `|x^2+4x+3|=x+1`

Explanation

To solve the equation \( |x^2 + 4x + 3| = x + 1 \), we will first factor the expression inside the absolute value and then consider different cases based on the properties of absolute values.

### Step 1: Factor the quadratic expression
The expression \( x^2 + 4x + 3 \) can be factored as:
\[ x^2 + 4x + 3 = (x + 3)(x + 1) \]
Thus, we rewrite the equation:
\[ |(x + 3)(x + 1)| = x + 1 \]

### Step 2: Consider different cases based on the absolute value
We will consider three cases based on the values of \( x \).

#### Case 1: \( x < -3 \)
In this case, both \( (x + 3) \) and \( (x + 1) \) are negative, so:
\[ |(x + 3)(x + 1)| = -(x + 3)(x + 1) \]
The equation becomes:
\[ -(x + 3)(x + 1) = x + 1 \]
Expanding and simplifying:
\[ -x^2 - 4x - 3 = x + 1 \]
\[ -x^2 - 5x - 4 = 0 \quad \text{(Multiplying through by -1)} \]
\[ x^2 + 5x + 4 = 0 \]
Factoring gives:
\[ (x + 4)(x + 1) = 0 \]
Thus, \( x = -4 \) or \( x = -1 \). Since we are in the case \( x < -3 \), we accept \( x = -4 \) but reject \( x = -1 \).

#### Case 2: \( -3 \leq x < -1 \)
In this case, \( (x + 3) \) is non-negative and \( (x + 1) \) is negative, so:
\[ |(x + 3)(x + 1)| = (x + 3)(- (x + 1)) = -(x + 3)(x + 1) \]
The equation becomes:
\[ -(x + 3)(x + 1) = x + 1 \]
Expanding and simplifying:
\[ -x^2 - 4x - 3 = x + 1 \]
\[ -x^2 - 5x - 4 = 0 \quad \text{(Multiplying through by -1)} \]
\[ x^2 + 5x + 4 = 0 \]
Factoring gives:
\[ (x + 4)(x + 1) = 0 \]
Thus, \( x = -4 \) or \( x = -1 \). Since we are in the case \( -3 \leq x < -1 \), we accept \( x = -4 \) but reject \( x = -1 \).

#### Case 3: \( x \geq -1 \)
In this case, both \( (x + 3) \) and \( (x + 1) \) are non-negative, so:
\[ |(x + 3)(x + 1)| = (x + 3)(x + 1) \]
The equation becomes:
\[ (x + 3)(x + 1) = x + 1 \]
Expanding and simplifying:
\[ x^2 + 4x + 3 = x + 1 \]
\[ x^2 + 3x + 2 = 0 \]
Factoring gives:
\[ (x + 2)(x + 1) = 0 \]
Thus, \( x = -2 \) or \( x = -1 \). Since we are in the case \( x \geq -1 \), we accept \( x = -1 \) but reject \( x = -2 \).

### Final Solutions
From all cases, the solutions are:
- From Case 1: \( x = -4 \)
- From Case 3: \( x = -1 \)

Thus, the final solutions to the equation \( |x^2 + 4x + 3| = x + 1 \) are:
\[ \boxed{-4 \text{ and } -1} \]