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