Answer
(a) $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{6},\pm\frac{1}{8},\pm\frac{3}{8},\pm\frac{1}{12},\pm\frac{1}{24}$
(b) $\{-\frac{3}{2},-\frac{2}{3},\frac{1}{2}\}$
(c) $ƒ(x)=2(2x-1)(2x+3)(3x+2)$
Work Step by Step
(a) Based on the given polynomial, we have factors $p=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12, q=\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24, $. We can list all possible rational zeros as: $\frac{p}{q}=\pm1,\pm2,\pm3,\pm4,\pm6,\pm12,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{6},\pm\frac{1}{8},\pm\frac{3}{8},\pm\frac{1}{12},\pm\frac{1}{24}$
(b) Use synthetic division and the above possible values, find one zero $x=\frac{1}{2}$ as shown in the figure. With quotient $24x^2+52x+24=0$, we have $4(2x+3)(3x+2)=0$ which gives $x=-\frac{3}{2},-\frac{2}{3}$. Thus the zeros are $\{-\frac{3}{2},-\frac{2}{3},\frac{1}{2}\}$
(c) Based on the know zeros and the factor theorem, we can factor $ƒ(x)$ into linear factors as $ƒ(x)=4(2x+3)(3x+2)(x-\frac{1}{2})=2(2x-1)(2x+3)(3x+2)$
