Answer
a. $\frac{p}{q}=\pm1,\pm2,\pm4,\pm1/2,$
b. positive real zeros: $2,0$, negative real zeros: $2,0$.
c. $x=\pm2$
d. $x=-\frac{1}{2},1$ (and the zeros from c).
Work Step by Step
a. Given the equation
$2x^4+x^3-9x^2-4x+4=0$
we have
$p=\pm1,\pm2,\pm4, q=\pm1,\pm2$
and the possible rational zeros (roots) are
$\frac{p}{q}=\pm1,\pm2,\pm4,\pm1/2$
b. There are 2 sign changes in the equation and based on Descartes’s Rule of Signs, we can determine the possible number of positive real zeros to be: $2,0$.
We examine
$f(-x)=2(-x)^4+(-x)^3-9(-x)^2-4(-x)+4=2x^4-x^3-9x^2+4x+4$
We can find 2 sign changes and based on Descartes’s Rule of Signs, we can determine that the number of negative real zeros is: $2,0$.
c. We can use synthetic division to find two roots as $x=\pm2$, as shown in the figure.
d. Part (c) gives a quotient of $2x^2+x-1=(2x+1)(x-1)$; thus the other two roots are $x=-\frac{1}{2},1$ (and the zeros from c).
