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

##### 2 Answers

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:

By the rational root theorem, any zeros of

That means that the only possible rational zeros are:

#+-1# ,#+-2#

In addition, all of the coefficients of

#-1# ,#-2#

We find:

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

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

So

The discriminant

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

In our example

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

Since

To simplify the cubic to have no square term, first multiply by

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

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

Let

We want to solve:

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

Using Cardano's method, let

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

Let

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

Multiply through by

#(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

#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

Then

#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))#

#### 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

Here we are supposing that the two other roots are complex conjugate. Equating the coefficients we have

Handling the first and the second equations eliminating

plotting those equations we can observe that there is an intersection approximately for

This coarse initial "solution" will be conveniently approximated using correction formulas. The approximation formulas can be obtained substituting for

Solving for

substituting the initial values we get

once more now with

obtaining after three iterations

The calculation of

The final result is