# How do you find all zeros with multiplicities of f(x)=x^4-9x^2+14?

May 16, 2018

$x = - \sqrt{2}$, multiplicity $1$
$x = \sqrt{2}$, multiplicity $1$
$x = - \sqrt{7}$, multiplicity $1$
$x = \sqrt{7}$, multiplicity $1$

#### Explanation:

Substitute $t = {x}^{2}$ (which implies ${t}^{2} = {x}^{4}$) to get

${t}^{2} - 9 t + 14 = 0$

The solutions of this equation are $t = 2$ and $t = 7$. Remember that $t = {x}^{2}$, so you have

$t = 2 \setminus \implies {x}^{2} = 2 \setminus \implies x = \setminus \pm \sqrt{2}$
$t = 7 \setminus \implies {x}^{2} = 7 \setminus \implies x = \setminus \pm \sqrt{7}$

which are the four solutions.