## Precalculus (6th Edition) Blitzer

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).
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).