How do you find the zeros of #x^5 - 5x^4 - x^3 + x^2 + 4 = 0#?

1 Answer
Aug 7, 2016

Answer:

Zeros:

#x_1 = 1#

#x_2 = -1#

#x_3 = 1/3(5+root(3)(179+12sqrt(114))+root(3)(179-12sqrt(114)))#

and two related Complex zeros.

Explanation:

#x^5-5x^4-x^3+x^2+4 = 0#

First note that the sum of the coefficients is zero.

That is:

#1-5-1+1+4 = 0#

So #x_1=1# is a zero and #(x-1)# a factor:

#x^5-5x^4-x^3+x^2+4 = (x-1)(x^4-4x^3-5x^2-4x-4)#

The sum of the coefficients of the remaining quartic with signs reversed on terms of odd degree is zero.

That is:

#1+4-5+4-4 = 0#

So #x_2=-1# is a zero and #(x+1)# a factor:

#x^4-4x^3-5x^2-4x-4=(x+1)(x^3-5x^2-4)#

By the rational root theorem, any rational zeros of the remaining cubic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-4# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational zeros are:

#+-1, +-2, +-4#

None of these work, so the cubic only has irrational zeros.

Let: #f(x) = x^3-5x^2-4#

#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=-5#, #c=0# and #d=-4#, so we find:

#Delta = 0+0-2000-432+0 = -2432#

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^3-135x^2-108#

#=(3x-5)^3-75(3x-5)-358#

#=t^3-75t-358#

where #t=(3x-5)#

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

We want to solve:

#t^3-75t-358=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-25)(u+v)-358=0#

Add the constraint #v=25/u# to eliminate the #(u+v)# term and get:

#u^3+15625/u^3-358=0#

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

#(u^3)^2-358(u^3)+15625=0#

Use the quadratic formula to find:

#u^3=(358+-sqrt((-358)^2-4(1)(15625)))/(2*1)#

#=(-358+-sqrt(128164-62500))/2#

#=(-358+-sqrt(65664))/2#

#=(-358+-24sqrt(114))/2#

#=-119+-12sqrt(114)#

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_3=root(3)(-119+12sqrt(114))+root(3)(-119-12sqrt(114))#

and related Complex roots:

#t_4=omega root(3)(-119+12sqrt(114))+omega^2 root(3)(-119-12sqrt(114))#

#t_5=omega^2 root(3)(-119+12sqrt(114))+omega root(3)(-119-12sqrt(114))#

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

Now #x=1/3(5+t)#. So the zeros of #f(x)# are:

#x_3 = 1/3(5+root(3)(-119+12sqrt(114))+root(3)(-119-12sqrt(114)))#

#x_4 = 1/3(5+omega root(3)(-119+12sqrt(114))+omega^2 root(3)(-119-12sqrt(114)))#

#x_5 = 1/3(5+omega^2 root(3)(-119+12sqrt(114))+omega root(3)(-119-12sqrt(114)))#