## Precalculus (6th Edition) Blitzer

a. $\frac{p}{q}=\pm1,\pm2,\pm4$ b. number of positive real zeros is $1$; number of negative real zeros is 2 or 0. c. $x=1$ d. $x=-2$ (multiplicity 2).
a. Given the function $f(x)=x^3+3x^2-4$, we have the possible rational zeros as $\frac{p}{q}=\pm1,\pm2,\pm4$ b. There is 1 sign change and based on Descartes’s Rule of Signs, we can determine the possible number of positive real zeros to be $1$. Examine $f(-x)=(-x)^3+3(-x)^2-4=-x^3+3x^2-4$ We can find 2 sign changes and based on Descartes’s Rule of Signs, we can determine that there are 2 or 0 negative real zeros. c. We can use synthetic division to find a zero as $x=1$, as shown in the figure. d. Part (c) gives a quotient $x^2+4x+4=(x+2)^2$; thus we have $x=-2$ (multiplicity 2).