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

$$(-∞, -3) ∪ (5,+∞)$$
Let $(x + 3)(x - 5) > 0$ be a function $f$. Find the $x$-intercepts by solving $(x + 3)(x - 5) > 0=0$ $x+3=0$ or $x-5=0$ $x = -3$ or $x = 5$ 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: $(-∞, -3)(-3,5)(5, +∞)$ Choose one test value within each interval and evaluate $f$ at that number. $(-∞, -3)$ Test Value = $-4$ $f=(x + 3)(x - 5)$ $f=(-4+3)(-4-5)$ $f=9$ Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -3)$. $(-3,5)$ Test Value = $0$ $f=(x + 3)(x - 5)$ $f=(0+3)(0-5)$ $f=-15$ Conclusion: $f (x) < 0$ for all $x$ in $(-3,5)$. $(5,+∞)$ Test Value = 6 $f=(x + 3)(x - 5)$ $f=(6+3)(6-5)$ $f=9$ Conclusion: $f (x) > 0$ for all $x$ in $(5,+∞)$. Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) > 0$, where $f (x) = (x+3)(x-5)$. Based on the solution above, $f(x)>0$ for all $x$ in $(-∞, -3)$ or $(5,+∞)$. Thus, the solution set of the given inequality is: $$(-∞, -3) ∪ (5,+∞)$$