How do you find all the zeros of #f(x)=x^3+3x^2-5x+8#?

1 Answer
Jul 16, 2016

Answer:

Use Cardano's method to find Real zero:

#x_1 = 1/3(-3+root(3)((-405+3sqrt(12081))/2) + root(3)((-405-3sqrt(12081))/2))#

and related Complex zeros.

Explanation:

#f(x) = x^3+3x^2-5x+8#

From the rational root theorem we can deduce that the only possible rational zeros are:

#+-1, +-2, +-4, +-8#

None of these work, so there are no rational zeros.

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=3#, #c=-5# and #d=8#, so we find:

#Delta = 225+500-864-1728-2160=-4027#

Since #Delta < 0#, this cubic has one Real zero and two non-Real Complex zeros. As a result, Cardano's method will work well.

We can simplify the problem by making the substitution #t=x+1#:

#x^3+3x^2-5x+8#

#= (x^3+3x^2+3x+1)-8(x+1)+15#

#=(x+1)^3-8(x+1)+15#

#= t^3-8t+15#

Using Cardano's method, let #t=u+v#, then we want to solve:

#u^3+v^3+(3uv-8)(u+v)+15 = 0#

Add the constraint #v = 8/(3u)# to eliminate the term in #(u+v)# and get:

#u^3+512/(27u^3)+15 = 0#

Multiply through by #27u^3# and rearrange slightly to get a quadratic in #u^3#:

#27(u^3)^2+405(u^3)+512 = 0#

Use the quadratic formula to find:

#u^3=(-405+-sqrt(405^2-4(27)(512)))/(2*27)#

#=(-405+-sqrt(164025-55296))/54#

#=(-405+-sqrt(108729))/54#

#=(-405+-3sqrt(12081))/54#

Hence Real zero:

#t_1 = root(3)((-405+3sqrt(12081))/54) + root(3)((-405-3sqrt(12081))/54)#

#= 1/3(root(3)((-405+3sqrt(12081))/2) + root(3)((-405-3sqrt(12081))/2))#

and related Complex zeros:

#t_2 = 1/3(omega root(3)((-405+3sqrt(12081))/2) + omega^2 root(3)((-405-3sqrt(12081))/2))#

#t_3 = 1/3(omega^2 root(3)((-405+3sqrt(12081))/2) + omega root(3)((-405-3sqrt(12081))/2))#

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

Then #x = t-1#, hence zeros of #f(x)#:

#x_1 = 1/3(-3+root(3)((-405+3sqrt(12081))/2) + root(3)((-405-3sqrt(12081))/2))#

#x_2 = 1/3(-3+omega root(3)((-405+3sqrt(12081))/2) + omega^2 root(3)((-405-3sqrt(12081))/2))#

#x_3 = 1/3(-3+omega^2 root(3)((-405+3sqrt(12081))/2) + omega root(3)((-405-3sqrt(12081))/2))#