## College Algebra (10th Edition)

$f(x)=(x-1)(2x+1)(x+\sqrt{2})(x-\sqrt{2})$ Zeros: $-\displaystyle \sqrt{2},\ \ -\frac{1}{2},\ \ 1,\ \ \sqrt{2},$ all 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 $2,\qquad \pm 1,\pm 2$ $\displaystyle \frac{p}{q}=\pm 1,\pm\frac{1}{2},\pm 2$ Testing with synthetic division, try $x-1$ $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 2 &-1 &-5 &2 &2 \\\hline &2 &1 &-4 & -2 \\\hline 2& 1 &-4 &-2& |\ \ 0 \end{array}$ $f(x)=(x-1)(2x^{3}+x^{2}-4x-2)$ Try synthetic division again, $x+\displaystyle \frac{1}{2}$ $\left.\begin{array}{l} -1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 2 &1 &-4 &-2 \\\hline &-1 &0 & +2 \\\hline 2 & 0 &-4 & |\ \ 0 \end{array}$ $f(x)=(x-1)(x+\displaystyle \frac{1}{2})(2x^{2}-4)=(x-1)(x+\frac{1}{2})\cdot 2(x^{2}-2)$ $=(x-1)(2x+1)(x^{2}-2)$ Solving $x^{2}-2=0,$ $x=\pm\sqrt{2}$, we have the other two zeros. $f(x)=(x-1)(2x+1)(x+\sqrt{2})(x-\sqrt{2})$ Zeros: $-\displaystyle \sqrt{2},\ \ -\frac{1}{2},\ \ 1,\ \ \sqrt{2},$ all with multiplicity 1.