How do you find the zeros of #f(x) = 7x^3 + 4x^2 - 3x + 3#?

1 Answer
Feb 19, 2016

Answer:

Use a substitution to get into the form #t^3+pt+q = 0# then Cardano's method...

Explanation:

When a cubic has one irrational Real zero and two non-Real Complex zeros, I like to solve using Cardano's method.

First, I would like to have a cubic with integer coefficients and no square term. To get there from our #f(x)#, multiply by #3^3*7^2 = 1323# and substitute #t = 21x+4# as follows:

#1323(7x^3+4x^2-3x+3)#

#=9261x^3+5292x^2-3969x+3969#

#=(9261x^3+5292x^2+1008x+64)-(4977x+948)+4853#

#=(21^3x^3+(3*21^2*4)x^2+(3*21*4^2)x+4^3)-((237*21)x+(237*4))+4853#

#=(21x+4)^3-237(21x+4)+4853#

#=t^3-237t+4853#

So we want to solve #t^3-237t+4853 = 0#

Next substitute #t = u + v# to get:

#0 = (u+v)^3-237(u+v)+4853#

#=u^3+v^3+(3uv-237)(u+v)+4853#

#=u^3+v^3+3(uv-79)(u+v)+4853#

Add the constraint #v = 79/u# and multiply through by #u^3# to get:

#(u^3)^2+4853(u^3)+493039 = 0#

Use the quadratic formula to find:

#u^3 = (-4853+-sqrt(4853^2-4*1*493039))/2#

#=(-4853+-sqrt(21579453))/2#

#=(-4853+-63 sqrt(5437))/2#

Since the derivation was symmetric in #u# and #v#, one of these roots can be taken as #u^3# and the other #v^3# to give us the Real root:

#t_1 = root(3)((-4853+63 sqrt(5437))/2)+root(3)((-4853-63 sqrt(5437))/2)#

and the Complex roots:

#t_2 = omega root(3)((-4853+63 sqrt(5437))/2)+ omega^2 root(3)((-4853-63 sqrt(5437))/2)#

#t_3 = omega^2 root(3)((-4853+63 sqrt(5437))/2)+ omega root(3)((-4853-63 sqrt(5437))/2)#

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

Then #x = (t-4)/21#, so the zeros of #f(x)# are:

#x_1 = 1/21(-4+ root(3)((-4853+63 sqrt(5437))/2)+root(3)((-4853-63 sqrt(5437))/2))#

#x_2 = 1/21(-4+ omega root(3)((-4853+63 sqrt(5437))/2)+ omega^2 root(3)((-4853-63 sqrt(5437))/2))#

#x_3 = 1/21(-4+ omega^2 root(3)((-4853+63 sqrt(5437))/2)+ omega root(3)((-4853-63 sqrt(5437))/2))#