## Intermediate Algebra for College Students (7th Edition)

$$(-∞, -2) ∪ (4,+∞)$$
Let $(x-4)(x+2)>0$ be a function $f$. Find the $x$-intercepts by solving $(x-4)(x+2)=0$ $x-4=0$ or $x+2=0$ $x = 4$ or $x = -2$ These $x$-intercepts serve as boundary points that separate the number line into intervals. The boundary points of this equation therefore divide the number line into three intervals: $(-∞, -2)(-2,4)(4, +∞)$ Choose one test value within each interval and evaluate $f$ at that number. $(-∞, -2)$ Test Value = $-3$ $f=(x-4)(x+2)$ $f=(-3-4)(-3+2)$ $f=7$ Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -2)$. $(-2,4)$ Test Value = $0$ $f=(x-4)(x+2)$ $f=(0-4)(0+2)$ $f=-8$ Conclusion: $f (x) < 0$ for all $x$ in $(-2,4)$. $(4,+∞)$ Test Value = $5$ $f=(x-4)(x+2)$ $f=(5-4)(5+2)$ $f=7$ Conclusion: $f (x) > 0$ for all $x$ in $(4,+∞)$. Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) > 0$, where $f (x) = (x-4)(x+2)$. Based on the solution above, $f(x)>0$ for all $x$ in $(-∞, -2)$ or $(4,+∞)$. Thus, the solution set of the given inequality is: $$(-∞, -2) ∪ (4,+∞)$$