## Precalculus (6th Edition) Blitzer

To find possibilities for negative real zeros, count the number of sign changes in the equation for f (-x). We obtain this equation by replacing x with -x in the given function. For example, consider the polynomial function$f\left( x \right)={{x}^{3}}+2{{x}^{2}}+5x+4$…… (1) Replace x with –x in equation (1). \begin{align} & f\left( -x \right)={{\left( -x \right)}^{3}}+2{{\left( -x \right)}^{2}}+5\left( -x \right)+4 \\ & f\left( -x \right)=-{{x}^{3}}+2{{x}^{2}}-5x+4 \\ \end{align} The equation with $f\left( -x \right)$is $f\left( -x \right)=-{{x}^{3}}+2{{x}^{2}}-5x+4$ ….. (2) Now, we see that there are three variations in sign. The number of negative real zeros of f (x) is either equal to the number of sign changes, 3, or is less than this number by an even integer. This means that either there are 3 negative real zeros or there is 3 - 2 = 1 negative real zero.