Answer
a. $\frac{p}{q}=\pm1,\pm5$
b. $x=1$; see figure.
c. $x=1,\frac{3\pm i\sqrt {11}}{2}$
Work Step by Step
a. Given the function $f(x)=x^3-4x^2+8x-5$, we have $p=\pm1,\pm5$ and $q=\pm1$. Thus the possible rational zeros are $\frac{p}{q}=\pm1,\pm5$
b. Starting from the easiest number, test the possible rational zeros using synthetic division. We can find $x=1$ as a zero (shown in the figure).
c. Based on the results from part-b, we have
$f(x)=x^3-4x^2+8x-5=(x-1)(x^2-3x+5)$
Thus the zeros are $x=1,\frac{3\pm i\sqrt {11}}{2}$
