To solve the polynomial inequality [tex]\( x^2 + 8x + 12 > 0 \)[/tex], we need to follow a series of steps to find its solution set. Here's a detailed, step-by-step solution:
1. Factor the quadratic expression: First, we need to factor the quadratic expression on the left-hand side of the inequality.
[tex]\[
x^2 + 8x + 12 = (x + 6)(x + 2)
\][/tex]
2. Set up the inequality: After factoring, we rewrite the inequality as:
[tex]\[
(x + 6)(x + 2) > 0
\][/tex]
3. Determine the critical points: The critical points are where the expression is equal to zero, which occur at the roots of the equation:
[tex]\[
x + 6 = 0 \implies x = -6
\][/tex]
[tex]\[
x + 2 = 0 \implies x = -2
\][/tex]
Thus, our critical points are [tex]\( x = -6 \)[/tex] and [tex]\( x = -2 \)[/tex].
4. Test the intervals: These critical points divide the real number line into three intervals:
[tex]\[
(-\infty, -6), \quad (-6, -2), \quad \text{and} \quad (-2, \infty)
\][/tex]
We need to test a point in each interval to determine where the product [tex]\((x + 6)(x + 2)\)[/tex] is positive.
- For the interval [tex]\((-\infty, -6)\)[/tex], we can choose [tex]\( x = -7 \)[/tex]:
[tex]\[
(-7 + 6)(-7 + 2) = (-1)(-5) = 5 > 0
\][/tex]
- For the interval [tex]\((-6, -2)\)[/tex], we can choose [tex]\( x = -4 \)[/tex]:
[tex]\[
(-4 + 6)(-4 + 2) = 2 \cdot (-2) = -4 < 0
\][/tex]
- For the interval [tex]\((-2, \infty)\)[/tex], we can choose [tex]\( x = 0 \)[/tex]:
[tex]\[
(0 + 6)(0 + 2) = 6 \cdot 2 = 12 > 0
\][/tex]
5. Combine the results: The inequality [tex]\( (x + 6)(x + 2) > 0 \)[/tex] is satisfied for the intervals [tex]\((-\infty, -6)\)[/tex] and [tex]\((-2, \infty)\)[/tex].
6. Express the solution set in interval notation: Combining the intervals where the inequality holds true, the solution set is:
[tex]\[
(-\infty, -6) \cup (-2, \infty)
\][/tex]
Thus, the solution set for the inequality [tex]\( x^2 + 8x + 12 > 0 \)[/tex] is:
[tex]\[ \boxed{(-\infty, -6) \cup (-2, \infty)} \][/tex]
