How do you find the zeros of #f(x) = 2x^3+2x^2-2x-1# ?

1 Answer
Oct 27, 2016

Answer:

Use a trigonometric substitution method to find zeros:

#x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "#

for #n = 0, 1, 2#.

Explanation:

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

#color(white)()#
Rational roots theorem

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-1# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2#, #+-1#

None of these work, so #f(x)# has no rational zeros.

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

#Delta = 16+64+32-108+144 = 148#

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

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

#=(6x+2)^3-48(6x+2)-20#

#=t^3-48t-20#

where #t=(6x+2)#

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Trigonometric substitution

Look for a substitution of the form #t = k cos theta# with #k# chosen so that the resulting cubic contains:

#4cos^3 theta - 3 cos theta = cos 3 theta#

Let #t = k cos theta#

#0 = t^3-48t-20#

#color(white)(0) = k^3 cos^3 theta - 48k cos theta - 20#

#color(white)(0) = k (k^2 cos^3 theta - 48 cos theta) - 20#

Let #k=8#

#0 = 8 (64 cos^3 theta - 48 cos theta) - 20#

#color(white)(0) = 128 (4 cos^3 theta - 3 cos theta) - 20#

#color(white)(0) = 128 cos 3 theta - 20#

So:

#cos 3 theta = 20/128 = 5/32#

Hence:

#3 theta = +-cos^(-1)(5/32) + 2npi#

So:

#theta = +-1/3 cos^(-1)(5/32) + (2npi)/3#

Note that #cos (-theta) = cos (theta)#, so we can discard the #+-# to get distinct solutions:

#t_n = 8 cos (1/3 cos^(-1)(5/32) + (2npi)/3)" "# for #n = 0, 1, 2#

Then #x = 1/6(t-2)#

#x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "# for #n = 0, 1, 2#