Answer
a. $\frac{p}{q}=\pm1,\pm2,\pm3,\pm6,$
b. positive real zeros: $2,0$, negative real zeros: $2,0$.
c. $x=\pm1$
d. $x=-2,3$ (and the zeros from c).
Work Step by Step
a. Given the equation
$x^4-x^3-7x^2+x+6=0$
we have
$p=\pm1,\pm2,\pm3,\pm6, q=\pm1$
and the possible rational zeros (roots) are
$\frac{p}{q}=\pm1,\pm2,\pm3,\pm6,$
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
$f(-x)=(-x)^4-(-x)^3-7(-x)^2+(-x)+6=x^4+x^3-7x^2-x+6$
We can find 2 sign changes and based on Descartes’s Rule of Signs, we can determine that the number of negative real zeros are $2,0$.
c. We can use synthetic division to find two roots as $x=\pm1$, as shown in the figure.
d. Part (c) gives a quotient $x^2-x-6=(x-3)(x+2)$; thus the other two roots are $x=-2,3$ (and the zeros from c).
