If `(x + 1)^(2) - 2x > 2(x + 1) + 2`, then x cannot be which of the following?

Options

A. `-5`
B. `-3`
C. `0`
D. `3`

Answer

To solve the inequality \((x + 1)^{2} - 2x > 2(x + 1) + 2\), we will follow these steps:

### Step 1: Expand both sides of the inequality
Start by expanding the left-hand side and right-hand side of the inequality.

\[ (x + 1)^{2} - 2x > 2(x + 1) + 2 \]

Expanding the left side:
\[ x^{2} + 2x + 1 - 2x = x^{2} + 1 \]

Expanding the right side:
\[ 2x + 2 + 2 = 2x + 4 \]

### Step 2: Set up the simplified inequality
Now, substitute the expanded forms back into the inequality:

\[ x^{2} + 1 > 2x + 4 \]

### Step 3: Rearrange the inequality
Rearranging the inequality gives us:

\[ x^{2} - 2x + 1 - 4 > 0 \]

This simplifies to:

\[ x^{2} - 2x - 3 > 0 \]

### Step 4: Factor the quadratic expression
Next, we factor the quadratic expression:

\[ (x - 3)(x + 1) > 0 \]

### Step 5: Find the critical points
The critical points from the factors are \(x = 3\) and \(x = -1\). These points will help us determine the intervals to test.

### Step 6: Test intervals
We will test the intervals created by the critical points:
1. \( (-\infty, -1) \)
2. \( (-1, 3) \)
3. \( (3, \infty) \)

**Interval 1: \( (-\infty, -1) \)**
Choose \(x = -2\):
\[ (-2 - 3)(-2 + 1) = (-5)(-1) = 5 > 0 \quad \text{(True)} \]

**Interval 2: \( (-1, 3) \)**
Choose \(x = 0\):
\[ (0 - 3)(0 + 1) = (-3)(1) = -3 < 0 \quad \text{(False)} \]

**Interval 3: \( (3, \infty) \)**
Choose \(x = 4\):
\[ (4 - 3)(4 + 1) = (1)(5) = 5 > 0 \quad \text{(True)} \]

### Step 7: Conclusion of intervals
The inequality holds true for the intervals:
\[ (-\infty, -1) \cup (3, \infty) \]

### Step 8: Determine which values x cannot take
Now we check the options given:
- Option 1: \(-5\) (in \((-∞, -1)\)) - can take
- Option 2: \(-3\) (in \((-∞, -1)\)) - can take
- Option 3: \(0\) (in \((-1, 3)\)) - cannot take
- Option 4: \(3\) (boundary point, not included) - cannot take

### Final Answer
Thus, \(x\) cannot take the values \(0\) and \(3\).

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