## College Algebra (10th Edition)

Zeros: $\displaystyle \ \ \frac{1}{2}$, with multiplicity 1. $f(x)= (2x-1)(x^{2}+1)$
$f(x)$ has at most 3 real zeros, as the degree is 3. Rational zeros: $\displaystyle \frac{p}{q},$ in lowest terms, $p$ must be a factor of $-1,\qquad \pm 1,$ and $q$ must be a factor of $2,\qquad \pm 1,\pm 2$ $\displaystyle \frac{p}{q}=\pm 1,\pm\frac{1}{2}$ Testing with synthetic division, try $x-\displaystyle \frac{1}{2}$ $\left.\begin{array}{l} \displaystyle \frac{1}{2} \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 2 &-1 &2 &-1\\\hline &1 &0 &1 \\\hline 2&0 &2& |\ \ 0 \end{array}$ $f(x)=(x-\displaystyle \frac{1}{2})(2x^{2}+2)=(x-\frac{1}{2})\cdot 2(x^{2}+1)$ $= (2x-1)(x^{2}+1)$ The binomial $x^{2}+1$ has no real zeros, so we can not factor any further. Zeros: $\displaystyle \ \ \frac{1}{2}$, with multiplicity 1. $f(x)= (2x-1)(x^{2}+1)$