What are the zeroes of #f(x)=x^4-6x^2+8# and the multiplicity of each?

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Answer 1 · 2016-11-16T16:56:21

Zeros of #f(x)=x^4-6x^2+8# are #{sqrt2,-sqrt2,2,-2}#

Let us first factorize #f(x)=x^4-6x^2+8#

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

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

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

= #(x^2-(sqrt2)^2)(x^2-2^2)#

= #(x-sqrt2)(x+sqrt2)(x-2)(x+2)#

This means for eac of #x={sqrt2,-sqrt2,2,-2}# we have #f(x)=0#

Hence zeros of #f(x)=x^4-6x^2+8# are #{sqrt2,-sqrt2,2,-2}#