Question

How do you factor `2x^3+2x^2-8x+8`?

Answer 1

`2x^3+2x^2-8x+8 = 2(x^3+x^2-4x+4) = 2(x-x_1)(x-x_2)(x-x_3)`

where `x_1, x_2` and `x_3` are found below...

Explanation:

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

Note that if either of the last two signs were inverted then this cubic would factor by grouping. As it is, it is somewhat more complicated...

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

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Descriminant

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=1`, `b=1`, `c=-4` and `d=4`, so we find:

`Delta = 16+256-16-432-288 = -464`

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=27f(x)=27x^3+27x^2-108x+108`

`=(3x+1)^3-39(3x+1)+146`

`=t^3-39t+146`

where `t=(3x+1)`

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Cardano's method

We want to solve:

`t^3-39t+146=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv-13)(u+v)+146=0`

Add the constraint `v=13/u` to eliminate the `(u+v)` term and get:

`u^3+2197/u^3+146=0`

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

`(u^3)^2+146(u^3)+2197=0`

Use the quadratic formula to find:

`u^3=(-146+-sqrt((146)^2-4(1)(2197)))/(2*1)`

`=(146+-sqrt(21316-8788))/2`

`=(146+-sqrt(12528))/2`

`=(146+-12sqrt(87))/2`

`=73+-6sqrt(87)`

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=root(3)(73+6sqrt(87))+root(3)(73-6sqrt(87))`

and related Complex roots:

`t_2=omega root(3)(73+6sqrt(87))+omega^2 root(3)(73-6sqrt(87))`

`t_3=omega^2 root(3)(73+6sqrt(87))+omega root(3)(73-6sqrt(87))`

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

Now `x=1/3(-1+t)`. So the zeros of our original cubic are:

`x_1 = 1/3(-1+root(3)(73+6sqrt(87))+root(3)(73-6sqrt(87)))`

`x_2 = 1/3(-1+omega root(3)(73+6sqrt(87))+omega^2 root(3)(73-6sqrt(87)))`

`x_3 = 1/3(-1+omega^2 root(3)(73+6sqrt(87))+omega root(3)(73-6sqrt(87)))`