Question

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

Answer 1

Find the zeros of `F(x)` to find:

`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)`

Explanation:

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)`