Precalculus (6th Edition) Blitzer

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).
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).