## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$f(x)=2(x-2)(x+\sqrt 5)(x-\sqrt 5)$ Real Zeros: $-\sqrt 5, \sqrt 5,2$ and all with multiplicity $1$.
Let us consider that $m$ is a factor of the constant term and $n$ is a factor of the leading coefficient. Then the potential zeros can be expressed by the possible combinations as: $\dfrac{m}{n}$. We see from the given polynomial function that it has at most $1$ real zeros as degree is $3$. The possible factors $m$ of the constant term and $n$ of the leading coefficient are: $m=\pm 1, \pm 2, \pm 5, \pm 10$and $n=\pm 1$ Therefore, the possible rational roots of $f(x)$ are: $\dfrac{m}{n}=\pm 1, \pm 2, \pm 5, \pm 10$ We test with synthetic division; we will try $x-2$. $\left.\begin{array}{l} 2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 1 & -2 &-5&10\\\hline &2&0 &-10\\\hline 1&0 &-5& |\ \ 0\end{array}$ Thus, we have: $f(x)=2(x-2)(x^2-5)\implies f(x)=2(x-2)(x+\sqrt 5)(x-\sqrt 5)$ Real Zeros: $-\sqrt 5, \sqrt 5,2$ and all with multiplicity $1$.