How do you find all the zeros of #F(X)= 3x^2 - 2x^2 + x +4#?

Question

Serenity Jones · Accepted Answer

Use Cardano's method to find Real zero:

#x_1 = 1/9(2+root(3)(505+135sqrt(14))+root(3)(505-135sqrt(14)))#

and related Complex zeros...

Assuming that there's a typo in the question... #f(x) = 3x^3-2x^2+x+4# #color(white)()# Descriminant The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula: #Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd# In our example, #a=3#, #b=-2#, #c=1# and #d=4#, so we find: #Delta = 4-12+128-3888-432 = -4200# Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another. #color(white)()# Tschirnhaus transformation To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation. #0=243f(x)=729x^3-486x^2+243x+972# #=(9x-2)^3+15(9x-2)+1010# #=t^3+15t+1010# where #t=(9x-2)# #color(white)()# Cardano's method We want to solve: #t^3+15t+1010=0# Let #t=u+v#. Then: #u^3+v^3+3(uv+5)(u+v)+1010=0# Add the constraint #v=-5/u# to eliminate the #(u+v)# term and get: #u^3-125/u^3+1010=0# Multiply through by #u^3# and rearrange slightly to get: #(u^3)^2+1010(u^3)-125=0# Use the quadratic formula to find: #u^3=(-1010+-sqrt((1010)^2-4(1)(-125)))/(2*1)# #=(1010+-sqrt(1020100+500))/2# #=(1010+-sqrt(1020600))/2# #=(1010+-270sqrt(14))/2# #=505+-135sqrt(14)# Since this is Real and the derivation is symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find Real root: #t_1=root(3)(505+135sqrt(14))+root(3)(505-135sqrt(14))# and related Complex roots: #t_2=omega root(3)(505+135sqrt(14))+omega^2 root(3)(505-135sqrt(14))# #t_3=omega^2 root(3)(505+135sqrt(14))+omega root(3)(505-135sqrt(14))# where #omega=-1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#. Now #x=1/9(2+t)#. So the zeros of our original cubic are: #x_1 = 1/9(2+root(3)(505+135sqrt(14))+root(3)(505-135sqrt(14)))# #x_2 = 1/9(2+omega root(3)(505+135sqrt(14))+omega^2 root(3)(505-135sqrt(14)))# #x_3 = 1/9(2+omega^2 root(3)(505+135sqrt(14))+omega root(3)(505-135sqrt(14)))#