## College Algebra 7th Edition

$P(x)=(x+1)^{2}(x-2)$ Zeros: $-1,\ 2\ \ \$
$P(x)=x^{3}-3x-2$ $P(-x)=-x^{3}+3x-2$ Decscart's rule of signs: P(x) has $1$ sign changes $\Rightarrow 1$ positive zeros. P(-x) has $2$ sign changes $\Rightarrow$ 2 or 0 negative zeros. Rational Zeros Theorem: candidates for p:$\quad \pm 1,\pm 2$ candidates for q:$\quad \pm 1,$ Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2$ Testing with synthetic division, $\left.\begin{array}{l} -1\ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr} 1 &0& -3&-2\\\hline &-1& 1 &2\\\hline 1& -1& -2&|\ \ 0\end{array}$ Remainder th: $x=-1$ is a zero $\Rightarrow (x+1)$ is a factor (Factor th.). $P(x)=(x+1)(x^{2}-x-2)$ For the trinomial, find two factors of $-2$ with sum $-1$ ... ( they are $-2$ and $+1)$ $P(x)=(x+1)(x+1)(x-2)=(x+1)^{2}(x-2)$ Zeros: $-1,\ 2\ \ \$