Answer
(a) 3 or 1 positive real zeros. 1 negative real zero. 0 or 2 complex zeros.
(b) $\frac{p}{q}=\pm1,\pm2,\pm3,\pm6,\pm9,\pm18,\pm\frac{1}{2},\pm\frac{3}{2},\frac{9}{2}$
(c) $x=2,-\frac{1}{2}$
(d) $x=\pm3i$
Work Step by Step
Given $f(x)=2x^4-3x^3+16x^2-27x-18$
(a) For $f(x)$, we can identify 3 sign changes. Based on the Descartes’ rule of signs, we can determine that there are 3 or 1 positive real zeros.
For $f(-x)$, we have $f(-x)=2x^4+3x^3+16x^2+27x-18$, and we can identify 1 sign change, thus there is 1 negative real zero. Since the degree of order is $4$, there are 0 or 2 complex zeros.
(b) Based on the rational zeros theorem, we have $p=\pm1,\pm2,\pm3,\pm6,\pm9,\pm18$ and $q=\pm1,\pm2$, thus we can list the possible rational zeros as $\frac{p}{q}=\pm1,\pm2,\pm3,\pm6,\pm9,\pm18,\pm\frac{1}{2},\pm\frac{3}{2},\frac{9}{2}$
(c) Use synthetic division as shown in the figure, we can find a zero as $x=2,-\frac{1}{2}$ and factor the function as $f(x)=(x-2)(x+\frac{1}{2})(2x^2+18)$. Thus, we can find the rational zeros as $x=2,-\frac{1}{2}$
(d) We can find other complex zeros as $x^2=-9$ thus $x=\pm3i$