Answer
(a) 2 or 0 positive real zeros. one negative real zero. 0 or 2 complex zeros.
(b) $\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2}$
(c) $-1, \frac{1}{2}, 3$
(d) no other complex zeros.
Work Step by Step
Given $f(x)=2x^3-5x^2-4x+3$
(a) For $f(x)$, we can identify two sign changes. Based on the Descartes’ rule of signs, we can determine that there are 2 or 0 positive real zeros.
For $f(-x)$, we have $f(-x)=2(-x)^3-5(-x)^2-4(-x)+3=-2x^3-5x^2+4x+3$, and we can identify one sign change, thus there is one negative real zero. Since the degree of order is $3$, there are 0 or 2 complex zeros.
(b) Based on the rational zeros theorem, we have $p=\pm1,\pm3$ and $q=\pm1,\pm2$, thus we can list the possible rational zeros as $\frac{p}{q}=\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2}$
(c) Use synthetic division as shown in the figure, we can find a zero as $x=3$ and factor the function as $f(x)=(x-3)(2x^2+x-1)=(x-3)(2x-1)(x+1)$. Thus, we can find all the rational zeros as $x=-1, \frac{1}{2}, 3$
(d) We can not find other complex zeros.