Answer
(a) $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12, \pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{1}{6}$
(b) $\{-4,-\frac{1}{3},\frac{3}{2}\}$
(c) $ƒ(x)=(x+4)(3x+1)(2x-3)$
Work Step by Step
(a) Based on the given polynomial, we have factors $p=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12, q=\pm1,\pm2,\pm3,\pm6$. We can list all possible rational zeros as: $\frac{p}{q}=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12, \pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{1}{6}$
(b) Use synthetic division and the above possible values, find one zero $x=-4$ as shown in the figure. With quotient $6x^2-7x-3=0$, we have $(3x+1)(2x-3)=0$ which gives $x=-\frac{1}{3},\frac{3}{2}$. Thus the zeros are $\{-4,-\frac{1}{3},\frac{3}{2}\}$
(c) Based on the know zeros and the factor theorem, we can factor $ƒ(x)$ into linear factors as $ƒ(x)=(x+4)(3x+1)(2x-3)$
