# 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#