How do you find all the zeros of #4x^3-2x^2-x+6#?

1 Answer
Jul 6, 2016

Answer:

Use Cardano's method to find Real zero:

#x_1 = 1/6(1+root(3)((-313+3sqrt(10857))/2) + root(3)((-313-3sqrt(10857))/2))#

and related Complex zeros.

Explanation:

#f(x) = 4x^3-2x^2-x+6#

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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=4#, #b=-2#, #c=-1# and #d=6#, so we find:

#Delta = 4+16+192-15552+864=-14476#

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

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Tschirnhaus transformation

To simplify the problem, eliminate any square term using a linear substitution. This is a simple form of what is called a Tschirnhaus transformation. To reduce the amount of arithmetic involving fractions, first multiply by #2*3^3 = 54# ...

#54 f(x) = 216x^3-108x^2-54x+324#

#=(6x-1)^3-12(6x-1)+313#

#=t^3-12t+313#

where #t = 6x-1#

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Cardano's method

To solve #t^3-12t+313 = 0#, use the substitution #t=u+v# to get:

#u^3+v^3+3(uv-4)(u+v)+313 = 0#

To eliminate the term in #(u+v)# add the constraint #v = 4/u# to get:

#u^3+64/u^3+313 = 0#

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

#(u^3)^2+313(u^3)+64 = 0#

Using the quadratic formula, we find:

#u^3 = (-313+-sqrt(313^2-(4*1*64)))/(2*1)#

#=(-313+-sqrt(97969-256))/2#

#=(-313+-sqrt(97713))/2#

#=(-313+-3sqrt(10857))/2#

Since these roots are Real and the derivation was 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)((-313+3sqrt(10857))/2) + root(3)((-313-3sqrt(10857))/2)#

and related Complex roots:

#t_2 = omega root(3)((-313+3sqrt(10857))/2) + omega^2 root(3)((-313-3sqrt(10857))/2)#

#t_3 = omega^2 root(3)((-313+3sqrt(10857))/2) + omega root(3)((-313-3sqrt(10857))/2)#

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

Then #x = 1/6(1+t)#, hence zeros of our original cubic:

#x_1 = 1/6(1+root(3)((-313+3sqrt(10857))/2) + root(3)((-313-3sqrt(10857))/2))#

#x_2 = 1/6(1+omega root(3)((-313+3sqrt(10857))/2) + omega^2 root(3)((-313-3sqrt(10857))/2))#

#x_3 = 1/6(1+omega^2 root(3)((-313+3sqrt(10857))/2) + omega root(3)((-313-3sqrt(10857))/2))#