How do you find all the real and complex roots of #3x^3+2x^2+x-4=0#?

1 Answer
May 24, 2016

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.

Explanation:

#f(x) = 3x^3+2x^2+x-4#

This cubic has no rational zeros. It turns out to have one Real zero and a Complex conjugate pair of non-Real zeros.

To simplify it while avoiding some fraction arithmetic, first multiply it by #3^2*3^3 = 243# to find:

#243 f(x) = 729x^3+486x^2+243x-972#

#=(9x)^3+3(9x)^2(2)+x-4#

#=(9x+2)^3+15(9x+2)-1010#

Let #t=9x+2#

We want to solve:

#t^3+15t-1010 = 0#

This substitution to reduce a polynomial to one of simpler form is called a Tschirnhaus transformation.

To solve the simplified cubic, use Cardano's method, letting #t = u+v#

#u^3+v^3+3(uv+5)(u+v)-1010 = 0#

Substitute #v = -5/u# to eliminate the term in #(u+v)# and find:

#u^3-125/u^3-1010 = 0#

Multiply through by #u^3# to get:

#(u^3)^2-1010(u^3)-125=0#

Using the quadratic formula, we find:

#u^3 = (1010+-sqrt(1010^2-(4*1*-125)))/(2*1)#

#=(1010+-sqrt(1020100+500))/2#

#=(1010+-sqrt(1020600))/2#

#=505+-135sqrt(14)#

Since this is Real and the derivation was symmetric in #u# and #v#, we can use one of these values as #u^3# and the other as #v^3# to find the Real root of #t^3+15t-1010 = 0# is:

#t_1 = root(3)(505+135sqrt(14))+root(3)(505-135sqrt(14))#

and 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)/2# is the primitive Complex cube root of #1#.

Then #x = 1/9(t-2)#, so the zeros of the 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)))#