How do you find all the zeros of $f(x)=x^4-5x^2-36$ ?

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Answer 1 · 2016-08-14T13:27:12

$f(x)$ has zeros $+-3$ and $+-2i$

Note that $9*4 = 36$ and $9-4 = 5$ .

So we find:

$x^4-5x^2-36$

$=(x^2-9)(x^2+4)$

$=(x^2-3^2)(x^2-(2i)^2)$

$=(x-3)(x+3)(x-2i)(x+2i)$

Hence zeros: $+-3$ and $+-2i$

How do you find all the zeros of f(x)=x^4-5x^2-36f(x)=x^4-5x^2-36?