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

1 Answer
Jul 23, 2016

Answer:

Real zeros: #1#, #1/3(1+root(3)((83+9sqrt(85))/2)+root(3)((83-9sqrt(85))/2))#

and related Complex zeros.

Explanation:

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

I suspect there may be a typo here. Was #x^4# supposed to be #x^3#? I will proceed assuming that it is correct as given...

First note that since the sum of the coefficients of #f(x)# is zero, #f(1) = 0# and #(x-1)# is a factor:

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

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 case #a=1#, #b=1#, #c=0# and #d=-3#, so we find:

#Delta = 0+0+12-243+0 = -231 < 0#

Since #Delta < 0# this cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

We could use the substitution #t=1/x# to get a simplified cubic, but on this occasion, let's multiply #(x^3+x^2-3)# by #27# and use #t=3x-1#. (That will make it slightly easier to find the roots in #a+bi# form.)

#27(x^3-x^2-3)#

#=27x^3-27x^2-81#

#=(3x)^3-3(3x)^2-81#

#=(3x-1)^3-3(3x-1)-83#

#=t^3-3t-83#

So we want to solve #t^3-3t-83 = 0#

Use Cardano's method:

Let #t=u+v#

#u^3+v^3+3(uv-1)(u+v)-83 = 0#

Eliminate the #(u+v)# term by letting #v = 1/u# to find:

#u^3+1/u^3-83 = 0#

Multiply through by #u^3# to get:

#(u^3)^2-83(u^3)+1 = 0#

By the quadratic formula:

#u^3 = (83+-sqrt(83^2-4(1)(1)))/(2*1)#

#=(83+-sqrt(6889-4))/2#

#=(83+-sqrt(6885))/2#

#=(83+-9sqrt(85))/2#

Since this is Real valued and our derivation was symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to deduce the roots of our cubic in #t#.

Then we can use #x = 1/3(t+1)# to find the Real zero of our original cubic:

#x_1 = 1/3(1+root(3)((83+9sqrt(85))/2)+root(3)((83-9sqrt(85))/2))#

and related Complex zeros:

#x_2 = 1/3(1+omega root(3)((83+9sqrt(85))/2)+omega^2 root(3)((83-9sqrt(85))/2))#

#x_3 = 1/3(1+omega^2 root(3)((83+9sqrt(85))/2)+omega root(3)((83-9sqrt(85))/2))#

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