## Precalculus (6th Edition) Blitzer

a. $\frac{p}{q}=\pm1,\pm3,\pm5,\pm15$ b. $x=-1,3$; see figure c. $x=-1,3,-1\pm2i$
a. Given the function $f(x)=x^4-2x^2-16x-15$, we have $p=\pm1,\pm3,\pm5,\pm15$ and $q=\pm1$; thus the possible rational zeros are $\frac{p}{q}=\pm1,\pm3,\pm5,\pm15$ b. Starting from the easiest number, test the possible rational zeros using synthetic divisions; we can find $x=-1,3$ as zeros as shown in the figure (we need to find two zeros to reduce the function to a quadratic form). c. Based on the results from part-b, we have $f(x)=(x+1)(x-3)(x^2+2x+5)$ Thus the zeros are $x=-1,3,-1\pm2i$