Question

How do you factor `-2x^3+6x^2-3`?

Answer 1

`-2x^3+6x^2-3 = -2(x-x_0)(x-x_1)(x-x_2)`

where:

`x_n = 1+2cos(1/3 cos^(-1)(1/4)+(2npi)/3)`

Explanation:

Given:

`f(x) = -2x^3+6x^2-3`

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 `-3` and `q` a divisor of the coefficient `-2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2, +-1, +-3/2, +-3`

Trying each in ascending order, we find:

`f(-3) = 105`

`f(-3/2) = 69/4`

`f(-1) = 5`

`f(-1/2) = -5/4`

`f(1/2) = -7/4`

`f(1) = 1`

`f(3/2) = 15/4`

`f(3) = -3`

So `f(x)` has no rational zeros, but it has irrational zeros in each of the intervals: `(-1, -1/2)`, `(1/2, 1)`, `(3/2, 3)`

`color(white)()`
Tschirnhaus transformation

To make the cubic easier to solve, use a linear substitution called a Tschirnhaus transformation as follows:

`(-4)f(x) = 8x^3-24x^2+12`

`color(white)((-4)f(x)) = (2x-2)^3-12(2x-2)-4`

`color(white)((-4)f(x)) = t^3-12t-4`

where `t = 2x-2`

`color(white)()`
Trigonometric substitution

Next we make a substitution:

`t = k cos theta`

where `k` is chosen so that the resulting expansion contains:

`4 cos^3 theta - 3 cos theta = cos 3 theta`

Let `k=4`

Then:

`0 = t^3-12t-4`

`color(white)(0) = (k cos theta)^3-12(k cos theta)-4`

`color(white)(0) = k(k^2 cos^3 theta-12 cos theta)-4`

`color(white)(0) = 4(16 cos^3 theta-12 cos theta)-4`

`color(white)(0) = 16(4 cos^3 theta-3 cos theta)-4`

`color(white)(0) = 16cos 3 theta-4`

Hence:

`cos 3 theta = 1/4`

So:

`3 theta = +-cos^(-1)(1/4)+2npi" "` for some integer `n`

So:

`theta = +-1/3cos^(-1)(1/4)+(2npi)/3`

So:

`cos theta = cos(+-1/3 cos^(-1)(1/4) + (2npi)/3)`

So:

`t_n = k cos theta = 4cos(1/3 cos^(-1)(1/4)+(2npi)/3)`

(( we can discard the `+-` since `cos(-alpha) = cos(alpha)` for any `alpha` ))

This last formula provides `3` distinct values `t_n`, for `n = 0, 1, 2`.

Then `x = 1+t/2`

Hence zeros of our original cubic:

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

`x_0 ~~ 2.8100379`

`x_1 ~~ -0.6417835`

`x_2 ~~ 0.8317456`

Then `f(x)` can be factorised as:

`-2x^3+6x^2-3 = -2(x-x_0)(x-x_1)(x-x_2)`