Answer
a. $\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{8},\pm\frac{3}{8},\pm\frac{1}{16},\pm\frac{3}{16},\pm\frac{1}{32},\pm\frac{3}{32} $
b. $-\frac{1}{8}, \frac{3}{4}, 1$
Work Step by Step
a. Based on the given equation, we have
$p=\pm1,\pm3$
and
$q=\pm1,\pm2,\pm4,\pm8,\pm16,\pm32$
Thus the possible rational zeros are
$\frac{p}{q}=\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{8},\pm\frac{3}{8},\pm\frac{1}{16},\pm\frac{3}{16},\pm\frac{1}{32},\pm\frac{3}{32} $
b. Using the given information, we can identify $x=1$ as a zero. Now, we use synthetic division as shown. We can factor the equation as
$(x-1)(32x^2-20x-3)=0$
or
$(x-1)(8x+1)(4x-3)=0$
Thus the solutions are
$x=-\frac{1}{8}, \frac{3}{4}, 1$
