Find the real zeros (if any) of the polynomial ?

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1 Answer
Jun 23, 2018

#x=0# and #x=-8#

Explanation:

At least #y=0# when #x=0#, since #x=0# gives #ax^5=0# for any a.

Other than that you will have to have #x^6+9=0#, i.e.

#x^6=-9#
This has a complex solution (#x=(-9)^(1/6)# (the 6th root of -9),
which would not be a real zero.

My answer, therefore, would be #x=0# and #x=-8#