## College Algebra 7th Edition

$P(x)=(x-1)^{2}(x+2)(x-2)$ Zeros: $\ -2,\ 1, \ 2 \$
$P(x)=x^{4}-2x^{3}-3x^{2}+8x-4$ $P(-x)=x^{4}+2x^{3}-3x^{2}-8x-4$ Descart's rule of signs: P(x) has 3 sign changes $\Rightarrow$ 3 or 1 positive zeros. P(-x) has 1 sign changes $\Rightarrow$ 1 negative zeros. Rational Zeros Theorem: candidates for p:$\quad \pm 1,\pm 2,\pm 4$ candidates for q:$\quad \pm 1,$ Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 4$ Testing with synthetic division, $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr} 1 & -2 & -3 & 8 & -4\\\hline & 1 & -1 & -4 & 4\\\hline 1& -1 & -4 & 4&|\ \ 0\end{array}$ $P(x)=(x-1)(x^{3}-x^{2}-4x+4)$ Again, testing with synthetic division, $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr} 1& -1 & -4 & 4\\\hline & 1 & 0 & -4\\\hline 1& 0 & -4 &|\ \ 0\end{array}$ $P(x)=(x-1)^{2}(x^{2}-4)$ ... recognize a difference of squares $P(x)=(x-1)^{2}(x+2)(x-2)$ Zeros: $\ -2,\ 1, \ 2 \$