## College Algebra (10th Edition)

$f(x)=(x-1)^{2}(x+1)(x+2)$ Zeros: $-1,\ \ -2,\ \$with multiplicity $1$, and $1\ \$ with multiplicity $2$.
$f(x)$ 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 $1,\qquad \pm 1$ $\displaystyle \frac{p}{q}=\pm 1,\pm 2$ Testing with synthetic division, try $x-1$ $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 1 &1 &-3 &-1 &2 \\\hline &1 &2 &-1 & -2 \\\hline 1& 2 &-1 &-2& |\ \ 0 \end{array}$ $f(x)=(x-1)(x^{3}+2x^{2}-x-2)$ Try synthetic division again, $x-1$ $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 1 &2 &-1 &-2 \\\hline &1 &3 & +2 \\\hline 1 & 3 &2 & |\ \ 0 \end{array}$ $f(x)=(x-1)^{2}(x+3x+2)=$ ... factors of 2 whose sum is 3 ... are 1 and 2: $=(x-1)^{2}(x+1)(x+2)$ $f(x)=(x-1)^{2}(x+1)(x+2)$ Zeros: $-1,\ \ -2,\ \$with multiplicity $1$, and $1\ \$ with multiplicity $2$.