Answer
(a) $\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30$
(b) $\{-5,-3,2\}$
(c) $ƒ(x)=(x-2)(x+3)(x+5)$
Work Step by Step
(a) Based on the given polynomial, we have factors $p=\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30 q=\pm1$. We can list all possible rational zeros as: $\frac{p}{q}=\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30$
(b) Use synthetic division and the above possible values, find one zero $x=2$ as shown in the figure. With quotient $x^2+8x+15=0$, we have $(x+3)(x+5)=0$ which gives $x=-3, -5$. Thus the zeros are $\{-5,-3,2\}$
(c) Based on the know zeros and the factor theorem, we can factor $ƒ(x)$ into linear factors as $ƒ(x)=(x-2)(x+3)(x+5)$
