How do you find all the zeros of #x^3-x^2+2x+1#?

1 Answer
May 18, 2016

Answer:

Use Cardano's method to find Real zero:

#x_1 = 1/3(1+root(3)((-43+9sqrt(29))/2)+root(3)((-43-9sqrt(29))/2))#

and Complex zeros.

Explanation:

Premultiply by #3^3# to cut down on arithmetic involving fractions:

#0 = 3^3 (x^3-x^2+2x+1)#

#=27x^3-27x^2+54x+27#

#=(3x-1)^3+15(3x-1)+43#

Let #t = 3x-1# and solve:

#t^3+15t+43 = 0#

Using Cardano's method, let #t = u + v#

#u^3+v^3+3(uv+5)(u+v)+43 = 0#

Let #v = -5/u# to eliminate the term in #(u+v)#

#u^3-5^3/u^3+43 = 0#

Multiply through by #u^3# to get a quadratic in #u^3#:

#(u^3)^2+43(u^3)-125 = 0#

Use the quadratic formula to find:

#u^3 = (-43+-sqrt(43^2+(4*125)))/2#

#=(-43+-sqrt(1849+500))/2#

#=(-43+-sqrt(2349))/2#

#=(-43+-9sqrt(29))/2#

The derivation was symmetric in #u# and #v#, so we can use one of these roots for #u^3# and the other for #v^3# to find the Real root of our cubic equation in #t# is:

#t_1 = root(3)((-43+9sqrt(29))/2)+root(3)((-43-9sqrt(29))/2)#

and Complex roots:

#t_2 = omega root(3)((-43+9sqrt(29))/2)+omega^2 root(3)((-43-9sqrt(29))/2)#

#t_3 = omega^2 root(3)((-43+9sqrt(29))/2)+omega root(3)((-43-9sqrt(29))/2)#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

Then #x= 1/3(t+1)#, hence zeros of the original cubic in #x#:

#x_1 = 1/3(1+root(3)((-43+9sqrt(29))/2)+root(3)((-43-9sqrt(29))/2))#

#x_2 = 1/3(1+omega root(3)((-43+9sqrt(29))/2)+omega^2 root(3)((-43-9sqrt(29))/2))#

#x_3 = 1/3(1+omega^2 root(3)((-43+9sqrt(29))/2)+omega root(3)((-43-9sqrt(29))/2))#