Answer
(a) 4,2,or 0 positive and 0 negative real zeros.
(b) see graph below.
(c) $0.17,3.93$
(d) $(2.82,-70.31)$
Work Step by Step
(a) It can be seen that $P(x)$ has 4 sign variations, according to Descartes' Rule, there are 4 or 2 or 0 possible positive real zeros. To test negative zeros, we have $P(-x)=2x^4+7x^3+x^2+18x+3$ and there are zero sign variations, which means that there will be 0 negative real zeros. Combining the above, there may be 4,2,or 0 positive and 0 negative real zeros.
(b) Use synthetic division as shown in the figure (b), $4$ is an upper bound because all the numbers in the quotient and remainder line are positive, $-1$ is a lower bound because the signs in the quotient and remainder alternates.
(c) See the graph, two real zeros can be found at $x=0.17,3.93$
(d) The minimum can be found at $(2.82,-70.31)$
