## Precalculus (6th Edition) Blitzer

The given inequality, from Exercise 1, is ${{x}^{2}}+8x+15>0$ >. Solving the equation $f\left( x \right)={{x}^{2}}+8x+15=0$ , the boundary points obtained are $x=-5,-3$. 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 $\left( -\infty ,-5 \right),\left( -5,-3 \right)\text{, and }\left( -3,\infty \right)$. Now, the left-most interval is $\left( -\infty ,-5 \right)$. Thus, any value in this interval can be chosen as the test value for the function and –10 lies in this interval. Thus, $-10$ can be a test value for the leftmost interval on the number line. Hence, the given statement is true.