## College Algebra (10th Edition)

$f(x)=(x-2)(x-3+2i)(x-3-2i)$ Zeros: $2,\ \ 3+2i,\ \ 3-2i$
Degree 3: there are 3 complex zeros. First, rational zero candidates: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 13,\pm 26}{\pm 1}$ Trying synthetic division, ... $x-2$ $\left.\begin{array}{l} 2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 1 & -8 &25 & 26 \\\hline & 2 & -12 & -26 \\\hline 1 & -6 & -13 & |\ \ 0 \end{array}$ $f(x)=(x-2)(x^{2}-6x-13)$ ... factoring the trinomial, we can't find two factors of -13 that add up to -6, so we use the quadratic formula. $x=\displaystyle \frac{6\pm\sqrt{36-4(13)}}{2}=\frac{6\pm\sqrt{-16}}{2}=\frac{6\pm 4i}{2}=3\pm 2i$ $f(x)=(x-2)(x-3+2i)(x-3-2i)$ Zeros: $2,\ \ 3+2i,\ \ 3-2i$