How do you use find the zeroes of $f(x)=x^4-4x^2-45$ ?

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Answer 1 · 2016-08-13T12:04:42

$f(x)$ has zeros: $3, -3, sqrt(5)i, -sqrt(5)i$

Treat as a quadratic in $x^2$ then use the difference of squares identity:

$a^2-b^2 = (a-b)(a+b)$

Note that $9*5=45$ and $9-5=4$

Hence:

$x^4-4x^2-45$

$=(x^2-9)(x^2+5)$

$=(x^2-3^2)(x^2-(sqrt(5)i)^2)$

$=(x-3)(x+3)(x-sqrt(5)i)(x+sqrt(5)i)$

Hence zeros:

$3, -3, sqrt(5)i, -sqrt(5)i$

How do you use find the zeroes of f(x)=x^4-4x^2-45f(x)=x^4-4x^2-45?