Question

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

Answer 1

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`