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

1 Answer
George C.
Jun 8, 2016

Use a trigonometric solution to find:

#x_k = 1/3(-2+2sqrt(13) cos(1/3(arccos((11sqrt(13))/338) + 2kpi)))#

#k = 0, 1, 2#

Explanation:

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

Using the rational root theorem we find that the only possible rational zeros are:

#+-1#, #+-3#

None of these work, so none of the zeros are rational.

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=2#, #c=-3# and #d=-3#, so we find:

#Delta = 36+108+96-243+324 = 321#

Since #Delta > 0# this cubic has #3# Real zeros.

As a result, pure algebraic methods like Cardano's method will yield results containing irreducible cube roots of Complex numbers. So choose to use a trigonometric solution instead.

First, use a Tschirnhaus transformation to eliminate the square term...

#27f(x) = 27x^3+54x^2-81x-81 = (3x+2)^3-39(3x+2)-11#

Let #t = (3x+2)#

We want to solve:

#t^3-39t-11 = 0#

Next, use a substitution of the form:

#t = 2sqrt(13) cos theta#

The multiplier #2sqrt(13)# is chosen so that our cubic contains a multiple of #4cos^3theta - 3cos theta = cos 3 theta# as follows:

#0 = (2sqrt(13) cos theta)^3 - 39 (2sqrt(13) cos theta) - 11#

#=26sqrt(13)(4 cos^3 theta - 3 cos theta) - 11#

#=26sqrt(13)cos 3theta - 11#

So:

#cos 3 theta = 11/(26sqrt(13)) = (11sqrt(13))/338#

#theta = +-1/3(arccos((11sqrt(13))/338) + 2kpi)#

#cos theta = cos(1/3(arccos((11sqrt(13))/338) + 2kpi))#

#t = 2sqrt(13) cos(1/3(arccos((11sqrt(13))/338) + 2kpi))#

#x_k = 1/3(-2+2sqrt(13) cos(1/3(arccos((11sqrt(13))/338) + 2kpi)))#

which has distinct values for #k = 0, 1, 2#

#x_0 ~~ 1.4605048700187635#

#x_1 ~~ -2.699628148275318#

#x_2 ~~ -0.7608767217434455#

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