How do you factor completely #F(x) = 2x^3 - 7x + 1#?

This cubic function has three irrational real zeros, for which we can find expressions using a trigonometric substitution...

Let #x = k cos theta#

Then:

#F(x) = F(k cos theta) = 2k^3 cos^3 theta - 7k cos theta + 1#

I would like this to contain something like:

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

So I would like:

#(2k^3)/(7k) = 4/3#

Choose #k=sqrt(42)/3#

Then:

#(2k^3)/4 = (7sqrt(42))/9#

So:

#2k^3 cos^3 theta - 7k cos theta + 1 = (7sqrt(42))/9(4 cos^3 theta - 3 cos theta) + 1#

#color(white)(2k^3 cos^3 theta - 7k cos theta + 1) = (7sqrt(42))/9cos 3 theta + 1#

This has zeros when:

#cos 3 theta = -9/(7sqrt(42)) = -(3sqrt(42))/98#

That is when:

#3 theta = +-(cos^(-1)(-(3sqrt(42))/98) + 2npi)" "# for integer values of #n#

So:

#cos theta = cos(1/3(+-cos^(-1)(-(3sqrt(42))/98)+2npi))#

Hence using #x = k cos theta# we find distinct zeros of #F(x)#:

#x_n = sqrt(42)/3 cos(1/3cos^(-1)(-(3sqrt(42))/98)+(2npi)/3)#

for #n = 0, 1, 2#

Hence:

#F(x) = 2(x-x_0)(x-x_1)(x-x_2)#

where:

#x_n = sqrt(42)/3 cos(1/3cos^(-1)(-(3sqrt(42))/98)+(2npi)/3)#