`-x^2+2x+8lt0`

Options

A. `[-oo,-2)uu(4,oo]`
B. `(-oo,-2)uu(4,oo)`
C. `(-2,4)`
D. `(-oo,-4)uu(2,oo)`

Answer

To solve the inequality \(-x^2 + 2x + 8 < 0\), we will follow these steps:

### Step 1: Rewrite the inequality
First, we can rewrite the given inequality by multiplying through by -1 (remember to reverse the inequality sign):

\[ x^2 - 2x - 8 > 0 \]

### Step 2: Factor the quadratic expression
Next, we need to factor the quadratic expression \(x^2 - 2x - 8\). We look for two numbers that multiply to \(-8\) (the constant term) and add to \(-2\) (the coefficient of \(x\)). These numbers are \(-4\) and \(2\).

Thus, we can factor the expression as:

\[ (x - 4)(x + 2) > 0 \]

### Step 3: Find the critical points
To find the critical points, we set each factor equal to zero:

1. \(x - 4 = 0 \Rightarrow x = 4\)
2. \(x + 2 = 0 \Rightarrow x = -2\)

So, the critical points are \(x = -2\) and \(x = 4\).

### Step 4: Test intervals
Now, we will test the intervals determined by the critical points to see where the product \((x - 4)(x + 2)\) is greater than zero. The intervals to test are:

1. \( (-\infty, -2) \)
2. \( (-2, 4) \)
3. \( (4, \infty) \)

**Testing the intervals:**

- For the interval \( (-\infty, -2) \), choose \(x = -3\):
\[ (-3 - 4)(-3 + 2) = (-7)(-1) = 7 > 0 \quad \text{(True)} \]

- For the interval \( (-2, 4) \), choose \(x = 0\):
\[ (0 - 4)(0 + 2) = (-4)(2) = -8 < 0 \quad \text{(False)} \]

- For the interval \( (4, \infty) \), choose \(x = 5\):
\[ (5 - 4)(5 + 2) = (1)(7) = 7 > 0 \quad \text{(True)} \]

### Step 5: Write the solution
From our tests, we find that the inequality holds true in the intervals \( (-\infty, -2) \) and \( (4, \infty) \). Therefore, the solution to the inequality \(-x^2 + 2x + 8 < 0\) is:

\[ (-\infty, -2) \cup (4, \infty) \]

### Final Answer
The range of \(x\) is \( (-\infty, -2) \cup (4, \infty) \).

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