## College Algebra (10th Edition)

$f(x)=2(x+2)(x+\sqrt{5})(x-\sqrt{5})$ Zeros: $-\sqrt{5},\ \ -2,\ \ \sqrt{5},$ all with multiplicity 1.
$f(x)=3x^{3}+6x^{2}-15x-30=3(x^{3}+2x^{2}-5x-10)$ $f(x)$ 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 $-10,\qquad \pm 1,\pm 2,\pm 5,\pm 10$ and $q$ must be a factor of $2,\qquad \pm 1$ $\displaystyle \frac{p}{q}=\pm 1,\pm 2,\pm 5,\pm 10$ Testing with synthetic division, try $x+2$ $\left.\begin{array}{l} -2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 1 &2 &-5 &-10 \\\hline &-2 &0 &+10 \\\hline 1&0 &-5& |\ \ 0 \end{array}$ $f(x)=2(x+2)(x^{2}-5)$ Solving $x^{2}-5=0,$ $x=\pm\sqrt{5}$, we have the other two zeros. $f(x)=2(x+2)(x+\sqrt{5})(x-\sqrt{5})$ Zeros: $-\sqrt{5},\ \ -2,\ \ \sqrt{5},$ all with multiplicity 1.