How do you use the rational roots theorem to find all possible zeros of #f(x) = x^3 - 10x^2 + 9x - 24#?

1 Answer
Aug 13, 2016

Answer:

Find #f(x)# has no rational zeros, but we can use Cardano's method to find Real zero:

#x_1 = 1/3(10+root(3)(-919+18sqrt(1406))+root(3)(-919-18sqrt(1406)))#

and related Complex zeros.

Explanation:

#f(x) = x^3-10x^2+9x-24#

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 #-24# and #q# a divisor of the coefficient #1# of the leading term. That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-24#

In addition, using Descartes' rule of signs we find that #f(x)# has #1# or #3# positive Real zeros and no negative Real zeros. So the only possible rational zeros are:

#1, 2, 3, 4, 6, 8, 12, 24#

None of these work, so #f(x)# has no rational zeros.

#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=1#, #b=-10#, #c=9# and #d=-24#, so we find:

#Delta = 8100-2916-96000-15552+38880 = -67488#

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=27f(x)=27x^3-270x^2+243x-648#

#=(3x-10)^3-219(3x-10)-1838#

#=t^3-219t-1838#

where #t=(3x-10)#

#color(white)()#
Cardano's method

We want to solve:

#t^3-219t-1838=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-73)(u+v)-1838=0#

Add the constraint #v=73/u# to eliminate the #(u+v)# term and get:

#u^3+389017/u^3-1838=0#

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

#(u^3)^2-1838(u^3)+389017=0#

Use the quadratic formula to find:

#u^3=(1838+-sqrt((-1838)^2-4(1)(389017)))/(2*1)#

#=(-1838+-sqrt(3378244-1556068))/2#

#=(-1838+-sqrt(1822176))/2#

#=(-1838+-36sqrt(1406))/2#

#=-919+-18sqrt(1406)#

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)(-919+18sqrt(1406))+root(3)(-919-18sqrt(1406))#

and related Complex roots:

#t_2=omega root(3)(-919+18sqrt(1406))+omega^2 root(3)(-919-18sqrt(1406))#

#t_3=omega^2 root(3)(-919+18sqrt(1406))+omega root(3)(-919-18sqrt(1406))#

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

Now #x=1/3(10+t)#. So the zeros of our original cubic are:

#x_1 = 1/3(10+root(3)(-919+18sqrt(1406))+root(3)(-919-18sqrt(1406)))#

#x_2 = 1/3(10+omega root(3)(-919+18sqrt(1406))+omega^2 root(3)(-919-18sqrt(1406)))#

#x_3 = 1/3(10+omega^2 root(3)(-919+18sqrt(1406))+omega root(3)(-919-18sqrt(1406)))#