How do you find the zeros of #f(x) = 2x^3+2x^2-2x-1# ?
1 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
Explanation:
#f(x) = 2x^3+2x^2-2x-1#
Rational roots theorem
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+-1/2# ,#+-1#
None of these work, so
Discriminant
The discriminant
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
In our example,
#Delta = 16+64+32-108+144 = 148#
Since
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
Trigonometric substitution
Look for a substitution of the form
#4cos^3 theta - 3 cos theta = cos 3 theta#
Let
#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
#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
#t_n = 8 cos (1/3 cos^(-1)(5/32) + (2npi)/3)" "# for#n = 0, 1, 2#
Then
#x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "# for#n = 0, 1, 2#