Answer
(a) $\pm1,\pm2,\pm4,\pm8,\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{8}{3},\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm\frac{8}{5},\pm\frac{1}{15},\pm\frac{2}{15},\pm\frac{4}{15},\pm\frac{8}{15}$
(b) $\{-4,-\frac{2}{5},\frac{1}{3}\}$
(c) $ƒ(x)=(x+4)(3x-1)(5x+2)$
Work Step by Step
(a) Based on the given polynomial, we have factors $p=\pm1,\pm2,\pm4,\pm8, q=\pm1,\pm3,\pm5,\pm15$. We can list all possible rational zeros as: $\frac{p}{q}=\pm1,\pm2,\pm4,\pm8,\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{8}{3},\pm\frac{1}{5},\pm\frac{2}{5},\pm\frac{4}{5},\pm\frac{8}{5},\pm\frac{1}{15},\pm\frac{2}{15},\pm\frac{4}{15},\pm\frac{8}{15}$
(b) Use synthetic division and the above possible values, find one zero $x=-4$ as shown in the figure. With quotient $15x^2+1x-2=0$, we have $(3x-1)(5x+2)=0$ which gives $x=-\frac{2}{5},\frac{1}{3}$. Thus the zeros are $\{-4,-\frac{2}{5},\frac{1}{3}\}$
(c) Based on the know zeros and the factor theorem, we can factor $ƒ(x)$ into linear factors as $ƒ(x)=(x+4)(3x-1)(5x+2)$
