## College Algebra 7th Edition

$P(x)=(x-1)(x+1)(x+3)$ Zeros: $\ -3,\ -1, \ 1 \$
$P(x)=x^{3}+3x^{2}-x-3$ $P(-x)=-x^{3}+3x^{2}+x-3$ Descart'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:$\quad \pm 1,\pm 3$ candidates for q:$\quad \pm 1,$ Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 3$ Testing with synthetic division, $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr} 1 & 3 & -1 &-3\\\hline &1 & 4 & 3\\\hline 1& 4 & 3&|\ \ 0\end{array}$ $P(x)=(x-1)(x^{2}+4x+3)$ For the trinomial, find two factors of $+3$ with sum $4$ ... ( they are $+1$ and $+3)$ $P(x)=(x-1)(x+1)(x+3)$ Zeros: $\ -3,\ -1, \ 1 \$