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

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Answer 1 · 2018-05-16T08:15:04

$x = -sqrt(2)$ , multiplicity $1$
$x = sqrt(2)$ , multiplicity $1$
$x = -sqrt(7)$ , multiplicity $1$
$x = sqrt(7)$ , multiplicity $1$

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

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

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

$t = 2 \implies x^2=2 \implies x=\pmsqrt(2)$
$t = 7 \implies x^2=7 \implies x=\pmsqrt(7)$

which are the four solutions.

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