How do you use the rational roots theorem to find all possible zeros of #f(x)=2x^3+7x^2+8x-8#?

Question

Elias Ackerman · Accepted Answer

Find #f(x)# has no rational zeros.

Use Cardano's method to find Real zero:

#x_1 = 1/6(-7+root(3)(-593+12sqrt(2442))+root(3)(-593-12sqrt(2442)))#

and related Complex zeros.

#f(x) = 2x^3+7x^2+8x-8# By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-8# and #q# a divisor of the coefficient #2# of the leading term. So the only possible rational zeros are: #+-1/2, +-1, +-2, +-4, +-8# None of these works, so #f(x)# has no rational zeros. We can find the irrational zeros using Cardano's method. #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=2#, #b=7#, #c=8# and #d=-8#, so we find: #Delta = 3136-4096+10976-6912-16128 = -13024# 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=108f(x)=216x^3+756x^2+864x-864# #=(6x+7)^3-3(6x+7)-1186# #=t^3-3t-1186# where #t=(6x+7)# #color(white)()# Cardano's method We want to solve: #t^3-3t-1186=0# Let #t=u+v#. Then: #u^3+v^3+3(uv-1)(u+v)-1186=0# Add the constraint #v=1/u# to eliminate the #(u+v)# term and get: #u^3+1/u^3-1186=0# Multiply through by #u^3# and rearrange slightly to get: #(u^3)^2-1186(u^3)+1=0# Use the quadratic formula to find: #u^3=(1186+-sqrt((-1186)^2-4(1)(1)))/(2*1)# #=(-1186+-sqrt(1406596-4))/2# #=(-1186+-sqrt(1406592))/2# #=(-1186+-24sqrt(2442))/2# #=-593+-12sqrt(2442)# 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)(-593+12sqrt(2442))+root(3)((-593-12sqrt(2442))# and related Complex roots: #t_2=omega root(3)(-593+12sqrt(2442))+omega^2 root(3)(-593-12sqrt(2442))# #t_3=omega^2 root(3)(-593+12sqrt(2442))+omega root(3)(-593-12sqrt(2442))# where #omega=-1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#. Now #x=1/6(-7+t)#. So the zeros of our original cubic are: #x_1 = 1/6(-7+root(3)(-593+12sqrt(2442))+root(3)(-593-12sqrt(2442)))# #x_2 = 1/6(-7+omega root(3)(-593+12sqrt(2442))+omega^2 root(3)(-593-12sqrt(2442)))# #x_3 = 1/6(-7+omega^2 root(3)(-593+12sqrt(2442))+omega root(3)(-593-12sqrt(2442)))#