Answer
a. $\frac{p}{q}=\pm1,\pm2,\pm1/2,\pm1/4$
b. $1$, $1$; see explanations.
c. $x=\pm\frac{1}{2}$
d. $x=\pm\sqrt 2i$ (and the zeros from c).
Work Step by Step
a. Given the equation
$4x^4+7x^2-2=0$
we have
$p=\pm1,\pm2, q=\pm1,\pm2,\pm4$
and the possible rational zeros (roots) are
$\frac{p}{q}=\pm1,\pm2,\pm1/2,\pm1/4$
b. There is 1 sign change in the equation and based on Descartes’s Rule of Signs, we can determine the possible number of positive real zeros to be $1$
We examine
$f(-x)=4(-x)^4+7(-x)^2-2=4x^4+7x^2-2$
We can find 1 sign change and based on Descartes’s Rule of Signs, we can determine that the number of negative real zeros is $1$.
c. We can use synthetic division to find two roots as $x=\pm\frac{1}{2}$, as shown in the figure.
d. Part (c) gives a quotient of $4x^2+8=4(x^2+2)$; thus the other two roots are $x=\pm\sqrt 2i$ (and the zeros from c).
