## Precalculus (6th Edition) Blitzer

a. $\frac{p}{q}=\pm1,\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{1}{6}$ b. positive real zeros: $2,0$. negative real zeros: 1. c. $x=-1$ see figure. d. $x=\frac{1}{2}, \frac{1}{3}$ (and $x=-1$ from c).
a. Given the function $f(x)=6x^3+x^2-4x+1$ we have the possible rational zeros as $\frac{p}{q}=\pm1,\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{1}{6}$ b. There are 2 sign changes and based on Descartes’s Rule of Signs, we can determine the possible number of positive real zeros to be $2,0$. We examine $f(-x)=6(-x)^3+(-x)^2-4(-x)+1=-6x^3+x^2+4x+1$ We can find 1 sign change and based on Descartes’s Rule of Signs, we can determine that there is exactly one negative real zero. c. We can use synthetic division to find a zero as $x=-1$, as shown in the figure. d. Part (c) gives a quotient $6x^2-5x+1=(2x-1)(3x-1)$; thus the other two zeros are $x=\frac{1}{2}, \frac{1}{3}$ (along with $x=-1$ from c).