How do you list all possible roots and find all factors of #x^4+2x^3-8x^2+16x-23#?

1 Answer
Dec 21, 2017

Answer:

Here is a solution for #x^4-2x^3+8x^2+16x-23=0#

(#x^4+2x^3-8x^2+16x-23=0# is much more complicated)

Explanation:

Assuming typographical errors in the problem, let us solve:

#x^4-2x^3+8x^2+16x-23=0#

By the rational roots theorem, any rational zeros of this quartic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-23# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-23#

Note that #x=1# is a solution, since the sum of the coefficients is #0#, i.e. #1-2+8+16-23 = 0#

Then:

#x^4-2x^3+8x^2+16x-23= (x-1)(x^3-x^2+7x+23)#

To find the zeros for the remaining cubic, proceed as follows:

#f(x) = x^3-x^2+7x+23#

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=7# and #d=23#, so we find:

#Delta = 49-1372+92-14283-2898 = -18412#

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+189x+621#

#=(3x-1)^3+60(3x-1)+682#

#=t^3+60t+682#

where #t=(3x-1)#

Cardano's method

We want to solve:

#t^3+60t+682=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv+20)(u+v)+682=0#

Add the constraint #v=-20/u# to eliminate the #(u+v)# term and get:

#u^3-8000/u^3+682=0#

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

#(u^3)^2+682(u^3)-8000=0#

Use the quadratic formula to find:

#u^3=(-682+-sqrt((682)^2-4(1)(-8000)))/(2*1)#

#=(682+-sqrt(465124+32000))/2#

#=(682+-sqrt(497124))/2#

#=341+-3sqrt(13809)#

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)(341+3sqrt(13809))+root(3)(341-3sqrt(13809))#

and related Complex roots:

#t_2=omega root(3)(341+3sqrt(13809))+omega^2 root(3)(341-3sqrt(13809))#

#t_3=omega^2 root(3)(341+3sqrt(13809))+omega root(3)(341-3sqrt(13809))#

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)(341+3sqrt(13809))+root(3)(341-3sqrt(13809)))#

#x_2 = 1/3(1+omega root(3)(341+3sqrt(13809))+omega^2 root(3)(341-3sqrt(13809)))#

#x_3 = 1/3(1+omega^2 root(3)(341+3sqrt(13809))+omega root(3)(341-3sqrt(13809)))#

Then the complete factorisation of our original quartic takes the form:

#x^4-2x^3+8x^2+16x-23 = (x-1)(x-x_1)(x-x_2)(x-x_3)#