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

1 Answer
Aug 16, 2016

Answer:

Use Cardano's method to find Real zero:

#x_1 = 1/27(1+root(3)((6235+27sqrt(46221))/2)+root(3)((6235-27sqrt(46221))/2))#

and related Complex zeros.

Explanation:

#color(white)()#

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

#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=-9#, #b=1#, #c=4# and #d=-3#, so we find:

#Delta = 16+2304+12-19683+1944 = -15407#

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=-2187f(x)=19683x^3-2187x^2-8748x+6561#

#=(27x-1)^3-327(27x-1)+6235#

#=t^3-327t+6235#

where #t=(27x-1)#

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

We want to solve:

#t^3-327t+6235=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-109)(u+v)+6235=0#

Add the constraint #v=109/u# to eliminate the #(u+v)# term and get:

#u^3+1295029/u^3+6235=0#

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

#(u^3)^2+6235(u^3)+1295029=0#

Use the quadratic formula to find:

#u^3=(-6235+-sqrt((6235)^2-4(1)(1295029)))/(2*1)#

#=(6235+-sqrt(38875225-5180116))/2#

#=(6235+-sqrt(33695109))/2#

#=(6235+-27sqrt(46221))/2#

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)((6235+27sqrt(46221))/2)+root(3)((6235-27sqrt(46221))/2)#

and related Complex roots:

#t_2=omega root(3)((6235+27sqrt(46221))/2)+omega^2 root(3)((6235-27sqrt(46221))/2)#

#t_3=omega^2 root(3)((6235+27sqrt(46221))/2)+omega root(3)((6235-27sqrt(46221))/2)#

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

Now #x=1/27(1+t)#. So the zeros of our original cubic are:

#x_1 = 1/27(1+root(3)((6235+27sqrt(46221))/2)+root(3)((6235-27sqrt(46221))/2))#

#x_2 = 1/27(1+omega root(3)((6235+27sqrt(46221))/2)+omega^2 root(3)((6235-27sqrt(46221))/2))#

#x_3 = 1/27(1+omega^2 root(3)((6235+27sqrt(46221))/2)+omega root(3)((6235-27sqrt(46221))/2))#