How do you find all the real and complex roots of #f(x)= x^3-x+3#?

#f(x) = x^3-x+3#

Using the rational root theorem, we find that the only possible rational zeros of #f(x)# are: #+-1#, #+-3#. None of these are zeros, so #f(x)# has no rational zeros.

In fact it turns out to have one Real zero and two non-Real Complex zeros.

We can use Cardano's method to find all three as follows:

Let #x = u+v#. Then #f(x) = 0# becomes:

#u^3+v^3+(3uv-1)(u+v)+3 = 0#

Let #v=1/(3u)# to eliminate the term in #(u+v)# and get:

#u^3+1/(27u^3)+3 = 0#

Multiply through by #27u^3# to get:

#27(u^3)^2+81(u^3)+1 = 0#

Using the quadratic formula, we get:

#u^3=(-81+-sqrt(81^2-4*27*1))/(2*27)#

#=(-81+-sqrt(6561-108))/54#

#=(-81+-sqrt(6453))/54#

#=(-81+-3sqrt(717))/54#

#=(-324+-12sqrt(717))/216#

Since these roots are Real and the derivation was symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find the Real zero:

#x_1 = 1/6(root(3)(-324+12sqrt(717))+root(3)(-324-12sqrt(717)))#

and Complex zeros:

#x_2 = 1/6(omega root(3)(-324+12sqrt(717))+omega^2 root(3)(-324-12sqrt(717)))#

#x_3 = 1/6(omega^2 root(3)(-324+12sqrt(717))+omega root(3)(-324-12sqrt(717)))#