Question

How do you factor `5x^3 -3x^2 +16x-6`?

Answer 1

Use Cardano's method to find:

`5x^3-3x^2+16x-6 = 5(x-x_1)(x-x_2)(x-x_3)`

where:

`x_n = 1/15(3+omega^(n-1)root(3)(972+15sqrt(58983))+omega^(1-n)root(3)(972-15sqrt(58983)))`

Explanation:

Given:

`f(x) = 5x^3-3x^2+16x-6`

We can find the zeros `x_1`, `x_2` and `x_3`

Then:

`f(x) = 5(x-x_1)(x-x_2)(x-x_3)`

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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=5`, `b=-3`, `c=16` and `d=-6`, so we find:

`Delta = 2304-81920-648-24300+25920 = -78644`

Since `Delta < 0` this cubic has `1` Real zero and `2` non-Real Complex zeros, which are Complex conjugates of one another.

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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=25f(x)=125x^3-75x^2+400x-150`

`=(5x-1)^3+77(5x-1)-72`

`=t^3+77t-72`

where `t=(5x-1)`

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Cardano's method

We want to solve:

`t^3+77t-72=0`

Let `t=u+v`.

Then:

`u^3+v^3+(3uv+77)(u+v)-72=0`

Add the constraint `v=-77/(3u)` to eliminate the `(u+v)` term and get:

`u^3-456533/(27u^3)-72=0`

Multiply through by `27u^3` and rearrange slightly to get:

`27(u^3)^2-1944(u^3)-456533=0`

Use the quadratic formula to find:

`u^3=(1944+-sqrt((-1944)^2-4(27)(-456533)))/(2*27)`

`=(1944+-sqrt(3779136+49305564))/54`

`=(1944+-sqrt(53084700))/54`

`=(1944+-sqrt(30^2*58983))/54`

`=(972+-15sqrt(58983))/3^3`

Since this is Real and the derivation is symmetric in `u` and `v`, we can use one of these roots for `u^3` and the other for `v^3` to find Real root:

`t_1=1/3(root(3)(972+15sqrt(58983))+root(3)(972-15sqrt(58983)))`

and related Complex roots:

`t_2=1/3(omega root(3)(972+15sqrt(58983))+omega^2 root(3)(972-15sqrt(58983)))`

`t_3=1/3(omega^2 root(3)(972+15sqrt(58983))+omega root(3)(972-15sqrt(58983)))`

where `omega=-1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.

Now `x=1/5(1+t)`. So the roots of our original cubic are:

`x_n = 1/15(3+omega^(n-1)root(3)(972+15sqrt(58983))+omega^(1-n)root(3)(972-15sqrt(58983)))`