Question

`2x^3+4x^2-13x+6` Can you factorise this please?

Answer 1

`"There exists no easy factorization here. Only a general method"`
`"for solving a cubic equation can help us here."`

Explanation:

`"We could apply a method based on the substitution of Vieta."`
`"Dividing by the first coefficient yields :"`
`x^3 + 2 x^2 - (13/2) x + 3 = 0`
`"Substituting "x=y+p" in "x^3+ax^2+bx+c" yields :"`
`y^3 + (3p+a) y^2 + (3p^2+2ap+b) y + p^3+ap^2+bp+c = 0`
`"if we take "3p+a=0" or "p=-a/3", the first coefficient"` `"becomes zero, and we get :"`
`=> y^3 - (47/6) y + (214/27) = 0`
`"(with "p = -2/3")"`
`"Substituting "y=qz" in "y^3 + b y + c = 0", yields :"`
`z^3 + b z / q^2 + c / q^3 = 0`
`"if we take "q = sqrt(|b|/3)", the coefficient of z becomes"`
`"3 or -3, and we get :"`
`"(here "q = 1.61589329")"`
`=> z^3 - 3 z + 1.87850338 = 0`
`"Substituting "z = t + 1/t", yields :"`
`=> t^3 + 1/t^3 + 1.87850338 = 0`
`"Substituting "u = t^3", yields the quadratic equation :"`
`=> u^2 + 1.87850338 u + 1 = 0`
`"The roots of the quadratic equation are complex."`
`"This means that we have 3 real roots in our cubic equation."`
`"A root of this quadratic equation is "`
`u=-0.93925169 + 0.34322917 i`

`"Substituting the variables back, yields :"`
`t = root3(u) = 1.0*(cos(-0.93041329)+i sin(-0.93041329))`
`= 0.59750263 - 0.80186695 i.`
`=> z = 1.19500526 + i 0.0.`
`=> y = 1.93100097 + i 0.0.`
`=> x = 1.26433430`
`"The other roots can be found by dividing and solving the"` `"remaining quadratic equation."`
`"The other roots are real : -3.87643981 and 0.61210551."`

Answer 2

`2x^3+4x^2-13x+6 = 2(x-x_0)(x-x_1)(x-x_2)`

where:
`x_n = 1/6(-4+2sqrt(94) cos(1/3 cos^(-1)(-214/2209 sqrt(94))+(2npi)/3))`

Explanation:

Given:

`2x^3+4x^2-13x+6`

Note that this does factorise much more easily if there is a typo in the question.

For example:

`2x^3+4x^2-color(red)(12)x+6 = 2(x-1)(x^2+3x-6) = ...`

`2x^3+4x^2-13x+color(red)(7) = (x-1)(2x^2+6x-7) = ...`

If the cubic is correct in the given form, then we can find its zeros and factors as follows:

`f(x) = 2x^3+4x^2-13x+6`

Discriminant

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example, `a=2`, `b=4`, `c=-13` and `d=6`, so we find:

`Delta = 2704+17576-1536-3888-11232 = 3624`

Since `Delta > 0` this cubic has `3` Real zeros.

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+432x^2-1404x+648`

`=(6x+4)^3-282(6x+4)+1712`

`=t^3-282t+1712`

where `t=(6x+4)`

Trigonometric substitution

Since `f(x)` has `3` real zeros, Cardano's method and similar will result in expressions involving irreducible cube roots of complex numbers. My preference in such circumstances is to use a trigonometric substitution instead.

Put:

`t = k cos theta`

where `k = sqrt(4/3 * 282) = 2sqrt(94)`

Then:

`0 = t^3-282t+1712`

`color(white)(0) = k^3 cos^3 theta - 282k cos theta+1712`

`color(white)(0) = 94k(4 cos^3 theta - 3 cos theta)+1712`

`color(white)(0) = 94k cos 3 theta+1712`

So:

`cos 3 theta = -1712/(94 k) = -1712/(188 sqrt(94)) = -(1712sqrt(94))/(188*94) = -214/2209 sqrt(94)`

So:

`3 theta = +-cos^(-1)(-214/2209 sqrt(94))+2npi`

So:

`theta = +- 1/3cos^(-1)(-214/2209 sqrt(94))+(2npi)/3`

So:

`cos theta = cos(1/3 cos^(-1)(-214/2209 sqrt(94))+(2npi)/3)`

Which givens `3` distinct zeros of the cubic in `t`:

`t_n = k cos theta = 2sqrt(94) cos(1/3 cos^(-1)(-214/2209 sqrt(94))+(2npi)/3)" "` for `n = 0, 1, 2`

Then:

`x = 1/6(t-4)`

So the three zeros of the given cubic are:

`x_n = 1/6(-4+2sqrt(94) cos(1/3 cos^(-1)(-214/2209 sqrt(94))+(2npi)/3))`

with approximate values:

`x_0 ~~ 1.2643`

`x_1 ~~ -3.8764`

`x_2 ~~ 0.61211`