## College Algebra 7th Edition

$P(x)=(x-2)(x-5)(x+3)^{2}$ Zeros: $\ -3,\ 2, \ 5. \$
$P(x)=x^{4}-x^{3}-23x^{2}-3x+90$ $P(-x)=x^{4}+x^{3}-23x^{2}+3x+90$ Descart's rule of signs: P(x) has 2 sign changes $\Rightarrow$ 2 or 0 positive zeros. P(-x) has 2 sign changes $\Rightarrow$ 2 or 0 negative zeros. Rational Zeros Theorem: candidates for p:$\quad \pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 9,\pm 10,\pm 15,\pm 30,\pm 45,\pm 90$ candidates for q:$\quad \pm 1,$ Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 9,\pm 10,\pm 15,\pm 30,\pm 45,\pm 90$ Testing with synthetic division, $\left.\begin{array}{l} 2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr} 1 & -1 & -23 & -3 & 90 \\\hline & 2 & 2 & -42 & -90\\\hline 1& 1 & -21 & -45&|\ \ 0\end{array}$ $P(x)=(x-2)(x^{3}+x^{2}-21x-45)$ Again, testing with synthetic division, $\left.\begin{array}{l} 5 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr} 1& 1 & -21 & -45\\\hline & 5 & 30& 45\\\hline 1& 6 & 9 &|\ \ 0\end{array}$ $P(x)=(x-2)(x-5)(x^{2}+6x+9)$ ... recognize a perfect square $P(x)=(x-2)(x-5)(x+3)^{2}$ Zeros: $\ -3,\ 2, \ 5. \$