Question

How do you factor completely `2x^3+2x^2+3x+2` ?

Answer 1

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

where `x_1, x_2, x_3` are derived below...

Explanation:

Given:

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

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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=2`, `b=3`, `c=2` and `d=2`, so we find:

`Delta = 36-64-216-432+432 = -244`

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

`=(2x+1)^3+(2x+1)+6`

`=t^3+t+6`

where `t=(2x+1)`

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Cardano's method

We want to solve:

`t^3+t+6=0`

Let `t=u+v`.

Then:

`u^3+v^3+(3uv+1)(u+v)+6=0`

Add the constraint `v=-1/(3u)` to eliminate the `(u+v)` term and get:

`u^3-1/(27u^3)+6=0`

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

`27(u^3)^2+162(u^3)-1=0`

Use the quadratic formula to find:

`u^3=(-162+-sqrt((162)^2-4(27)(-1)))/(2*27)`

`=(-162+-sqrt(26244+108))/54`

`=(-162+-sqrt(26352))/54`

`=(-162+-12sqrt(183))/54`

`=(-81+-6(183))/27`

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)(-81+6sqrt(183))+root(3)(-81-6sqrt(183)))`

and related Complex roots:

`t_2=1/3(omega root(3)(-81+6sqrt(183))+omega^2 root(3)(-81-6sqrt(183)))`

`t_3=1/3(omega^2 root(3)(-81+6sqrt(183))+omega root(3)(-81-6sqrt(183)))`

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

Now `x=1/2(-1+t)`. So the zeros of our original cubic are:

`x_1 = 1/6(-3+root(3)(-81+6sqrt(183))+root(3)(-81-6sqrt(183)))`

`x_2 = 1/6(-3+omega root(3)(-81+6sqrt(183))+omega^2 root(3)(-81-6sqrt(183)))`

`x_3 = 1/6(-3+omega^2 root(3)(-81+6sqrt(183))+omega root(3)(-81-6sqrt(183)))`