How do you find all zeros with multiplicities of $f (x) = x^{4} + 2 x^{2} - 15$ $f(x)=x^4+2x^2-15$ ?

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Answer 1 · 2017-02-21T10:50:41

$x = \pm \sqrt{5} i and \pm \sqrt{3}$ $x=+-sqrt5 i and +-sqrt 3$

$f (x) = x^{4} + 2 x^{2} - 15$ $f(x) = x^4+2x^2-15$

Let say $x^{2} = y$ $x^2=y$
Therefore,

${(x^{2})}^{2} + 2 (x^{2}) - 15 = 0$ $(x^2)^2+2(x^2) - 15 =0$

$y^{2} + 2 y - 15 = 0$ $y^2+2y-15=0$

$(y + 5) (y - 3) = 0$ $(y+5)(y-3)=0$

$y = - 5 and y = 3$ $y = -5 and y =3$

when y =-5, then
$x^{2} = - 5 = 5 i^{2}$ $x^2 = -5 =5i^2$ , where $i^{2} = - 1$ $i^2 = -1$
$x = \pm \sqrt{5} i$ $x=+-sqrt5 i$

when y =3, then
$x^{2} = 3$ $x^2 = 3$ ,
$x = \pm \sqrt{3}$ $x = +-sqrt3$

How do you find all zeros with multiplicities of f(x)=x4+2x2−15f(x)=x^4+2x^2-15?