How do you find all zeros with multiplicities of #f(x)=x^6-3x^3-10#?

1 Answer
George C.
Dec 9, 2017

The six zeros are:

#x = root(3)(5)#

#x = -1/2root(3)(5)+-1/2sqrt(3)root(3)(5)i#

#x = -root(3)(2)#

#x = 1/2root(3)(2)+-1/2sqrt(3)root(3)(2)i#

all with multiplicity #1#

Explanation:

#x^6-3x^3-10 = (x^3-5)(x^3+2)#

Then:

#x^3-5 = (x^3-(root(3)(5))^3)#

#color(white)(x^3-5) = (x-root(3)(5))(x^2+root(3)(5)x+root(3)(25))#

#color(white)(x^3-5) = (x-root(3)(5))((x+1/2root(3)(5))^2+(1/2sqrt(3)root(3)(5))^2)#

#color(white)(x^3-5) = (x-root(3)(5))((x+1/2root(3)(5))^2-(1/2sqrt(3)root(3)(5)i)^2)#

#color(white)(x^3-5) = (x-root(3)(5))(x+1/2root(3)(5)-1/2sqrt(3)root(3)(5)i)(x+1/2root(3)(5)+1/2sqrt(3)root(3)(5)i)#

and:

#x^3+2 = (x^3+(root(3)(2))^3)#

#color(white)(x^3+2) = (x+root(3)(2))(x^2-root(3)(2)x+root(3)(4))#

#color(white)(x^3+2) =(x+root(3)(2))((x-1/2root(3)(2))^2+(1/2sqrt(3)root(3)(2))^2)#

#color(white)(x^3+2) =(x+root(3)(2))((x-1/2root(3)(2))^2-(1/2sqrt(3)root(3)(2)i)^2)#

#color(white)(x^3+2) =(x+root(3)(2))(x-1/2root(3)(2)-1/2sqrt(3)root(3)(2)i)(x-1/2root(3)(2)+1/2sqrt(3)root(3)(2)i)#

So the six zeros are:

#x = root(3)(5)#

#x = -1/2root(3)(5)+-1/2sqrt(3)root(3)(5)i#

#x = -root(3)(2)#

#x = 1/2root(3)(2)+-1/2sqrt(3)root(3)(2)i#

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