## Precalculus (6th Edition) Blitzer

The points -5 and -3 shown in exercise 1 divide the number line into three intervals: $\left( -\infty ,-5 \right),\left( -5,-3 \right),\left( -3,\infty \right)$.
The given inequality, from Exercise 1, is ${{x}^{2}}+8x+15>0$ >. Solve the equation $f\left( x \right)={{x}^{2}}+8x+15=0$ to obtain the boundary points as $x=-5,-3$. That is, the points, -5 and -3, are the roots of the function and on these values the function changes from positive to negative or negative to positive. Then, the interval is divided into three intervals as $\left( -\infty ,-5 \right),\left( -5,-3 \right)\text{, and }\left( -3,\infty \right)$.