What are all the possible rational zeros for #f(x)=x^4+2x^2-8# and how do you find all zeros?

1 Answer
Sep 9, 2016

Answer:

#x=+-sqrt(2)# or #x= +-2i#

Explanation:

#f(x) = x^4+2x^2-8#

The simplest way to solve for the zeros of this function is to use the transformation #z=x^2# the set #f(z) = 0#

Thus: #f(z) = z^2 + 2z -8 =0#

Factorise: #(z-2)(z+4) = 0#

Hence: either #z=2# or #z=-4#

But #z=x^2 -> x=+-sqrt(z)#

#:. # either #x=+-sqrt(2)# or #x=+-sqrt(-4) = +-sqrt(4)i#

#x=+-sqrt(2)# or #x= +-2i#

Therefore we can see that #f(x)# has two irrational zeros and two complex zeros.