Question

How do you find all the real and complex roots of `x^3+x^2+x+2`?

Answer 1

Use Cardano's method to find Real zero:

`x_1 = -1/3(1+root(3)((47+3sqrt(249))/2)+root(3)((47-3sqrt(249))/2))`

and related Complex zeros.

Explanation:

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

By the rational root theorem, any zeros of `f(x)` must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `2` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-2`

In addition, all of the coefficients of `f(x)` are positive, so any Real zeros it has will be negative ones. That leaves:

`-1`, `-2`

We find:

`f(-1) = -1+1-1+2 = 3`

`f(-2) = -8+4-2+2 = -4`

So `f(x)` has no rational zeros and has an irrational zero in `(-2, -1)`.

The discriminant `Delta` of a cubic `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=b=c=1`, `d=2` and we find:

`Delta = 1-4-8-108+36 = -83 < 0`

Since `Delta < 0`, we can deduce that `f(x)` has one Real zero and a Complex conjugate pair of non-Real zeros.

To simplify the cubic to have no square term, first multiply by `3^3 = 27` to reduce the arithmetic involving fractions.

`27f(x) = 27x^3+27x^2+27x+54`

`=(3x+1)^3 + 6(3x+1) + 47`

Let `t=3x+1`

We want to solve:

`t^3+6t+47=0`

Using Cardano's method, let `t = u+v`

`u^3+v^3+3(uv+2)(u+v)+47 = 0`

Let `v = -2/u` to eliminate the term in `(u+v)`

`u^3-8/u^3+47 = 0`

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

`(u^3)^2+47(u^3)-8 = 0`

Use the quadratic formula to find:

`u^3=(-47+-sqrt(47^2-4(1)(-8)))/2`

`=(-47+-sqrt(2209+32))/2`

`=(-47+-sqrt(2241))/2`

`=(-47+-3sqrt(249))/2`

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

`t_1 = -root(3)((47+3sqrt(249))/2)-root(3)((47-3sqrt(249))/2)`

and Complex roots:

`t_2 = -omega root(3)((47+3sqrt(249))/2)-omega^2 root(3)((47-3sqrt(249))/2)`

`t_3 = -omega^2 root(3)((47+3sqrt(249))/2)-omega root(3)((47-3sqrt(249))/2)`

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

Then `x = 1/3(t-1)`, so the zeros of the original cubic are:

`x_1 = -1/3(1+root(3)((47+3sqrt(249))/2)+root(3)((47-3sqrt(249))/2))`

`x_2 = -1/3(1+omega root(3)((47+3sqrt(249))/2)+omega^2 root(3)((47-3sqrt(249))/2))`

`x_3 = -1/3(1+omega^2 root(3)((47+3sqrt(249))/2)+omega root(3)((47-3sqrt(249))/2))`

Answer 2

`((x-0.1766)^2+1.2028^2)(x+1.3532) approx x^3+x^2+x +2`

Explanation:

A polynomial with real coefficients and with an odd maximum degree has at least a real root.
Having this in mind we propose for the polynomial a structure such as

`x^3+x^2+x+2=((x-a)^2+b^2)(x-c)`
Here we are supposing that the two other roots are complex conjugate. Equating the coefficients we have

#{ (2 + a^2 c + b^2 c=0),( 1 - a^2 - b^2 - 2 a c=0), (1 + 2 a + c=0) :} #

Handling the first and the second equations eliminating `a^2+b^2` we get the reduced system

# { (2 + c - 2 a c^2 = 0), (1 + 2 a + c = 0) :} #

plotting those equations we can observe that there is an intersection approximately for `a = 0.1, b = -1.0`

This coarse initial "solution" will be conveniently approximated using correction formulas. The approximation formulas can be obtained substituting for `a->a+delta_a` and `b->b+delta_b` and considering that `delta_a^2 < abs(delta_a)` and `delta_b^2 < abs(delta_b)` resulting in

#{ (2 + c - 2 a c^2 - 2 c^2 delta_a + delta_c - 4 a c delta_c = 0), (1 + 2 a + c + 2 delta_a + delta_c =0):}#

Solving for `delta_a,delta_b`

#{(delta_a = -((-1 + 2 a - 4 a c - 8 a^2 c - 2 a c^2)/( 2 (1 - 4 a c + c^2)))), (delta_c = -1 - 2 a - c + (-1 + 2 a - 4 a c - 8 a^2 c - 2 a c^2)/( 1 - 4 a c + c^2)) :}#

substituting the initial values we get
`delta_a=0.108333,delta_b=-0.416667`
once more now with `a = 0.208333,b=-1.41667`
`delta_a = -0.0301958,delta_c =0.0603957`
obtaining after three iterations
`a =0.1766, c = -1.3532` within an acceptable error.
The calculation of `b` follows without more problems giving
`b = -1.2028`
The final result is
`((x-0.1766)^2+1.2028^2)(x+1.3532) approx x^3+x^2+x +2`