Solve (x-4)(x+2) > 0: Finding Positive Values of a Quadratic Function

Look at the following function:

$y=\left(x-4\right)\left(x+2\right)$

Determine for which values of $x$ the is true:

$f(x) > 0$

Solution: We begin by finding the roots of the function $y = (x-4)(x+2)$.

Step 1: Find the roots by solving $(x-4)(x+2) = 0$.

• Root 1: $x - 4 = 0$ implies $x = 4$.
• Root 2: $x + 2 = 0$ implies $x = -2$.

Step 2: The function changes sign at the roots, so we analyze the intervals determined by these roots: $(-\infty, -2)$, $(-2, 4)$, and $(4, \infty)$.

Step 3: Determine where $y > 0$ within these intervals.

• Select a test point from the interval $(-\infty, -2)$, e.g., $x = -3$: $y = (-3-4)(-3+2) = (-7)(-1) = 7$ which is positive.
• Select a test point from the interval $(-2, 4)$, e.g., $x = 0$: $y = (0-4)(0+2) = (-4)(2) = -8$ which is negative.
• Select a test point from the interval $(4, \infty)$, e.g., $x = 5$: $y = (5-4)(5+2) = (1)(7) = 7$ which is positive.

Step 4: Conclude that the function is positive in the intervals $(-\infty, -2)$ and $(4, \infty)$.

Therefore, the solution to the problem is that $f(x) > 0$ when $x < -2$ or $x > 4$.

Upon reviewing the problem's given correct answer, identify any typographical error in it.

Consequently, the function is positive for $x > 4$ or $x < -2$.