Answer
a. $\frac{p}{q}=\pm1,\pm2,\pm4$
b. $x=\pm2$; see figure
c. $x=\pm2,1\pm\sqrt 2$
Work Step by Step
a. Given the function $f(x)=x^4-2x^3-5x^2+8x+4$, we have $p=\pm1,\pm2,\pm4$ and $q=\pm1$; thus the possible rational zeros are $\frac{p}{q}=\pm1,\pm2,\pm4$
b. Starting from the easiest number, test the possible rational zeros using synthetic divisions; we can find $x=\pm2$ as a zeros, as shown in the figure (we need to find two zeros to reduce the function to a quadratic form).
c. Based on the results from part-b, we have
$f(x)=(x+2)(x-2)(x^2-2x-1)$
Thus the zeros are $x=\pm2,1\pm\sqrt 2$
