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#