How do you solve #x^3+x^2+x-2 = 0# ?

2 Answers
Sean
Nov 24, 2017

Are you sure you entered the problem correctly?

Explanation:

#x^3+x^2+x-2=0#

#x^3+x^2-2x+3x+1-3=0#

#x^3+(x^2-2x+1)+3x-3=0#

#x^3+(x-1)^2+3(x-1)=0#

#x^3+(x-1)(x-1+3)=0#

#x^3+(x-1)(x+2)=0#

It does not factor beyond this. Are you sure you entered the problem correctly?

George C.
Nov 24, 2017

This cubic has real root:

#x = 1/3(-1+root(3)((61+3sqrt(417))/2)+root(3)((61-3sqrt(417))/2))#

and related complex roots...

Explanation:

Given:

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

Discriminant

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

#Delta = 1-4+8-108-36 = -139#

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

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+27x^2+27x-54#

#=(3x+1)^3+6(3x+1)-61#

#=t^3+6t-61#

where #t=(3x+1)#

Cardano's method

We want to solve:

#t^3+6t-61=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv+2)(u+v)-61=0#

Add the constraint #v=-2/u# to eliminate the #(u+v)# term and get:

#u^3-8/u^3-61=0#

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

#(u^3)^2-61(u^3)-8=0#

Use the quadratic formula to find:

#u^3=(61+-sqrt((-61)^2-4(1)(-8)))/(2*1)#

#=(61+-sqrt(3721+32))/2#

#=(61+-sqrt(3753))/2#

#=(61+-3sqrt(417))/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)((61+3sqrt(417))/2)+root(3)((61-3sqrt(417))/2)#

and related Complex roots:

#t_2=omega root(3)((61+3sqrt(417))/2)+omega^2 root(3)((61-3sqrt(417))/2)#

#t_3=omega^2 root(3)((61+3sqrt(417))/2)+omega root(3)((61-3sqrt(417))/2)#

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

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

#x_1 = 1/3(-1+root(3)((61+3sqrt(417))/2)+root(3)((61-3sqrt(417))/2))#

#x_2 = 1/3(-1+omega root(3)((61+3sqrt(417))/2)+omega^2 root(3)((61-3sqrt(417))/2))#

#x_3 = 1/3(-1+omega^2 root(3)((61+3sqrt(417))/2)+omega root(3)((61-3sqrt(417))/2))#

If you like, you can use these values to give a factorisation:

#0 = x^3+x^2+x-2 = (x-x_1)(x-x_2)(x-x_3)#

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