Answer
(a) $\pm1,\pm2,\pm5,\pm10$
(b) $\{-2,-1,5\}$
(c) $ƒ(x)=(x+1)(x+2)(x-5)$
Work Step by Step
(a) Based on the given polynomial, we have factors $p=\pm1,\pm2,\pm5,\pm10, q=\pm1$. We can list all possible rational zeros as: $\frac{p}{q}=\pm1,\pm2,\pm5,\pm10$
(b) Use synthetic division and the above possible value, find one zero $x=-1$ as shown in the figure. With quotient $x^2-3-10=0$, we have $(x+2)(x-5)=0$ which gives $x=-2, 5$. Thus the zeros are $\{-2,-1,5\}$
(c) Based on the know zeros and the factor theorem, we can factor $ƒ(x)$ into linear factors as $ƒ(x)=(x+1)(x+2)(x-5)$
