If solution set of Inequality `(x^(2)+x-2)(x^(2)+x-16)ge -40` is `x epsilon(-oo, -4]uu[a,b]uu[c,oo)` then `a+b-c` is

Options

A. `-2`
B. `-3`
C. `-4`
D. `-6`

Explanation

To solve the inequality \((x^2 + x - 2)(x^2 + x - 16) \ge -40\), we will follow these steps:

### Step 1: Rearranging the Inequality
We start by rearranging the inequality:
\[ (x^2 + x - 2)(x^2 + x - 16) + 40 \ge 0 \]
This simplifies to:
\[ (x^2 + x - 2)(x^2 + x - 16) + 40 \ge 0 \]

### Step 2: Expanding the Left Side
Next, we expand the left-hand side:
\[ (x^2 + x - 2)(x^2 + x - 16) = x^4 + x^3 - 16x^2 + x^3 + x^2 - 2x - 32 \]
Combining like terms gives us:
\[ x^4 + 2x^3 - 18x^2 - 18x + 32 \]

### Step 3: Setting the Inequality
Now we set up the inequality:
\[ x^4 + 2x^3 - 18x^2 - 18x + 32 \ge 0 \]

### Step 4: Finding Roots
To find the roots of the polynomial, we can use synthetic division or the Rational Root Theorem. We can test \(x = -4\) as a potential root:
\[ (-4)^4 + 2(-4)^3 - 18(-4)^2 - 18(-4) + 32 = 0 \]
Thus, \(x + 4\) is a factor. We can divide the polynomial by \(x + 4\) to find the other factors.

### Step 5: Polynomial Division
Dividing \(x^4 + 2x^3 - 18x^2 - 18x + 32\) by \(x + 4\) gives:
\[ x^4 + 2x^3 - 18x^2 - 18x + 32 = (x + 4)(x^3 - 2x^2 - 10x + 8) \]

### Step 6: Finding Further Roots
Next, we need to factor \(x^3 - 2x^2 - 10x + 8\). We can test for rational roots again and find that \(x = -2\) and \(x = 3\) are roots.

### Step 7: Final Factorization
Thus, we can factor the polynomial completely:
\[ (x + 4)(x + 2)(x - 3)(x + 3) \ge 0 \]

### Step 8: Analyzing Intervals
The critical points are \(x = -4, -3, -2, 3\). We analyze the sign of the product in the intervals:
1. \( (-\infty, -4) \)
2. \( (-4, -3) \)
3. \( (-3, -2) \)
4. \( (-2, 3) \)
5. \( (3, \infty) \)

### Step 9: Testing Intervals
Testing values in each interval:
- For \(x < -4\), the product is positive.
- Between \(-4\) and \(-3\), the product is negative.
- Between \(-3\) and \(-2\), the product is positive.
- Between \(-2\) and \(3\), the product is negative.
- For \(x > 3\), the product is positive.

### Step 10: Solution Set
Thus, the solution set is:
\[ (-\infty, -4] \cup [-3, -2] \cup [3, \infty) \]
This matches the given solution set \(x \in (-\infty, -4] \cup [a, b] \cup [c, \infty)\) where \(a = -3\), \(b = -2\), \(c = 3\).

### Step 11: Finding \(a + b - c\)
Now we calculate:
\[ a + b - c = -3 + (-2) - 3 = -8 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{-8} \]