How do you find all the zeros of #f(x)=x^3-2x^2+x+1#?

1 Answer
George C.
Aug 8, 2016

Use Cardano's method to find Real zero:

#x_1 = 1/3(2+root(3)((29+3sqrt(93))/2)+root(3)((29-3sqrt(93))/2))#

and Complex conjugate pair of related Complex root.

Explanation:

#f(x) = x^3-2x^2+x+1#

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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=-2#, #c=1# and #d=1#, so we find:

#Delta = 4-4+32-27-36 = -31#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

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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^3-54x^2+27x+27#

#=(3x-2)^3-3(3x-2)+29#

#=t^3-3t+29#

where #t=(3x-2)#

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Cardano's method

We want to solve:

#t^3-3t+29=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-1)(u+v)+29=0#

Add the constraint #v=1/u# to eliminate the #(u+v)# term and get:

#u^3+1/u^3+29=0#

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

#(u^3)^2+29(u^3)+1=0#

Use the quadratic formula to find:

#u^3=(-29+-sqrt((29)^2-4(1)(1)))/(2*1)#

#=(29+-sqrt(841-4))/2#

#=(29+-sqrt(837))/2#

#=(29+-3sqrt(93))/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)((29+3sqrt(93))/2)+root(3)((29-3sqrt(93))/2)#

and related Complex roots:

#t_2=omega root(3)((29+3sqrt(93))/2)+omega^2 root(3)((29-3sqrt(93))/2)#

#t_3=omega^2 root(3)((29+3sqrt(93))/2)+omega root(3)((29-3sqrt(93))/2)#

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

Now #x=1/3(2+t)#. So the roots of our original cubic are:

#x_1 = 1/3(2+root(3)((29+3sqrt(93))/2)+root(3)((29-3sqrt(93))/2))#

#x_2 = 1/3(2+omega root(3)((29+3sqrt(93))/2)+omega^2 root(3)((29-3sqrt(93))/2))#

#x_3 = 1/3(2+omega^2 root(3)((29+3sqrt(93))/2)+omega root(3)((29-3sqrt(93))/2))#

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