## College Algebra (10th Edition)

$f(x)=(x+1)(3x+1)(x^{2}+2)$ Zeros: $-1,\ \ -\displaystyle \frac{1}{3},\ \$with multiplicity $1$
$f(x)$ has at most $4$ real zeros, as the degree is $4$. Rational zeros: $\displaystyle \frac{p}{q},$ in lowest terms, $p$ must be a factor of $2,\qquad \pm 1,\pm 2,$ and $q$ must be a factor of $3,\qquad \pm 1\pm 3$ $\displaystyle \frac{p}{q}=\pm 1,\pm\frac{1}{3},\pm\frac{2}{3},\pm 2$ Testing with synthetic division, try $x+1$ $\left.\begin{array}{l} -1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 3 &4 &7 &8 &2 \\\hline &-3 &-1 &-6 & -2 \\\hline 3& 1 & 6 &2 & |\ \ 0 \end{array}$ $f(x)=(x+1)(3x^{3}+x^{2}+6x+2)$ factor in pairs: $3x^{3}+x^{2}+6x+2=x^{2}(3x+1)+2(3x+1)$ $=(3x+1)(x^{2}+2)$ $x^{2}+2$ has no real zeros and can not be factored any further. $f(x)=(x+1)(3x+1)(x^{2}+2)$ Zeros: $-1,\ \ -\displaystyle \frac{1}{3},\ \$with multiplicity $1$