## Precalculus (6th Edition) Blitzer

a. $\frac{p}{q}=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$ b. $x=1$ (see figure). c. $x=-3,1,4$
a. Given the function $f(x)=x^3-2x^2-11x+12$ we have $p=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$ and $q=\pm1$ Thus the possible rational zeros are $\frac{p}{q}=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$ b. Starting from the easiest number, test the possible rational zeros using synthetic division. We can find $x=1$, as a zero shown in the figure. c. Based on the results from part-b, we have $f(x)=x^3-2x^2-11x+12=(x-1)(x^2-x+12)=(x-1)(x-4)(x+3)$ Thus the zeros are $x=-3,1,4$