Answer
(a) 2 or 0 positive real zeros, one negative real zero, 0 or 2 complex zeros.
(b) $\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{8}{3},\pm\frac{1}{6}$
(c) $-\frac{1}{2}, \frac{4}{3}, 6$
(d) no other complex zeros.
Work Step by Step
Given $f(x)=6x^3-41x^2+26x+24$
(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 $f(-x)=6(-x)^3-41(-x)^2+26(-x)+24=-6x^3-41x^2-26x+24$, and we can identify one sign change, thus there is one negative real zeros. Since the degree of order is $3$, there are 0 or 2 complex zeros.
(b) Based on the rational zeros theorem, we have $p=\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24$ and $q=\pm1,\pm2,\pm3,\pm6$, thus we can list the possible rational zeros as $\frac{p}{q}=\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{4}{3},\pm\frac{8}{3},\pm\frac{1}{6}$
(c) Use synthetic division as shown in the figure, we can find a zero as $x=6$ and factor the function as $f(x)=(x-6)(6x^2-5x-4)=(x-6)(2x+1)(3x-4)$. Thus, we can find all the rational zeros as $x=-\frac{1}{2}, \frac{4}{3}, 6$
(d) We can not find other complex zeros.
