How do you find all the zeros of #f(x) = 4x^3 – 5x^2 + 9x – 6#?

1 Answer
Mar 8, 2016

Answer:

Multiply by #432#, substitute #t = 12x-5#, then use Cardano's method,

Explanation:

#f(x) = 4x^3-5x^2+9x-6#

First multiply through by #3^3*4^2 = 432# to cut down on the number of fractions we need to work with.

#0 = 432 f(x)#

#= 1728x^3-2160x^2+3888x-2592#

#=(12x-5)^3+249(12x-5)-1222#

Substitute #t = 12x-5#

#t^3+249t-1222 = 0#

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

#u^3+v^3+(3uv+249)(u+v)-1222 = 0#

Add the constraint #v = -249/(3u)# to eliminate the term in #(u+v)#

#u^3-249^3/(27u^3)-1222 = 0#

Multiply through by #27u^3# and rearrange to get:

#27(u^3)^2-32994(u^3)-15438249 = 0#

Use the quadratic formula to find:

#u^3 = (32994+-sqrt(2755934928))/54#

#=(32994+-972 sqrt(2917))/54#

#=611+-18sqrt(2917)#

Due to the symmetry of the derivation in #u# and #v#, we can take one of these roots as the value of #u^3# and the other as #v^3# to get the Real root:

#t_1 = root(3)(611+18sqrt(2917)) + root(3)(611-18sqrt(2917))#

and Complex roots:

#t_2 = omega root(3)(611+18sqrt(2917)) + omega^2 root(3)(611-18sqrt(2917))#

#t_3 = omega^2 root(3)(611+18sqrt(2917)) + omega root(3)(611-18sqrt(2917))#

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

Then you can derive the roots of the original equation:

#x_1 = (t_1+5)/12=1/12(5+root(3)(611+18sqrt(2917)) + root(3)(611-18sqrt(2917)))#

#x_2 = 1/12(5+omega root(3)(611+18sqrt(2917)) + omega^2 root(3)(611-18sqrt(2917)))#

#x_3 = 1/12(5+omega^2 root(3)(611+18sqrt(2917)) + omega root(3)(611-18sqrt(2917)))#