How do you find all zeros with multiplicities of #f(x)=x^5-2x^4-4x+8#?

1 Answer
George C.
May 29, 2017

The zeros are #2#, #+-sqrt(2)# and #+-sqrt(2)i#

Explanation:

Given:

#f(x) = x^5-2x^4-4x+8#

Note that the ratio between the first and second terms is the same as that between the third and fourth terms, so we can factor this quadrinomial by grouping:

#x^5-2x^4-4x+8#

#= (x^5-2x^4)-(4x-8)#

#= x^4(x-2)-4(x-2)#

#= (x^4-4)(x-2)#

#= ((x^2)^2-2^2)(x-2)#

#= (x^2-2)(x^2+2)(x-2)#

#= (x^2-(sqrt(2))^2)(x^2-(sqrt(2)i)^2)(x-2)#

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

Hence zeros:

#+-sqrt(2)#, #+-sqrt(2)i#, #2#

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