Question

what is the solution set to the inequality $5(x - 2)(x + 4) > 0$?
○ ${x| x > -4 \text{ and } x < 2}$
○ ${x| x < -4 \text{ or } x > 2}$
○ ${x| x < -2 \text{ or } x > 4}$
○ ${x| x > -2 \text{ or } x < 4}$

Explanation:

Step1: Analyze the coefficient and roots

The inequality is \(5(x - 2)(x + 4)>0\). The coefficient \(5\) is positive, so the sign of the left - hand side is determined by \((x - 2)(x + 4)\). The roots of the equation \((x - 2)(x + 4)=0\) are \(x = 2\) and \(x=-4\). These roots divide the number line into three intervals: \((-\infty,-4)\), \((-4,2)\), and \((2,\infty)\).

Step2: Test intervals

• For the interval \((-\infty,-4)\), let's take a test point, say \(x=-5\). Then \((-5 - 2)(-5 + 4)=(-7)\times(-1) = 7>0\). Since \(5>0\) and \((x - 2)(x + 4)>0\) in this interval, \(5(x - 2)(x + 4)>0\) holds.
• For the interval \((-4,2)\), let's take a test point, say \(x = 0\). Then \((0 - 2)(0 + 4)=(-2)\times4=-8<0\). Since \(5>0\) and \((x - 2)(x + 4)<0\) in this interval, \(5(x - 2)(x + 4)<0\) does not satisfy the inequality.
• For the interval \((2,\infty)\), let's take a test point, say \(x = 3\). Then \((3 - 2)(3 + 4)=(1)\times(7) = 7>0\). Since \(5>0\) and \((x - 2)(x + 4)>0\) in this interval, \(5(x - 2)(x + 4)>0\) holds.

Answer:

\(\{x|x < - 4\text{ or }x>2\}\) (the second option)