Question

How do you find all the real and complex roots of `P(x) = 3x^4 + x^3 + 2x^2 - 2`?

Answer 1

It gets rather complicated, but it's doable...

Explanation:

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

You might try looking for rational zeros first, but you would not find any. This is a full blown quartic with `2` Real roots and `2` Complex roots, all irrational.

Normally with a quartic of the form `ax^4+bx^3+cx^2+dx+e`, the first thing you would do would be to substitute something like `t=x+b/(4a)` in order to get a quartic in `t` with no `t^3` term.

In our example, there's an easier method by substituting `t = 1/x` and multiplying by `-2t^4` to get:

`0 = 4t^4-4t^2-2t-6`

`=(2t^2-at+b)(2t^2+at+c)`

`=4t^4+(2b+2c-a^2)t^2+a(b-c)t+bc`

Equating coefficients and rearranging a little, we get:

`{ (b+c = (a^2-4)/2), (b-c=-2/a), (bc=-6) :}`

Then:

`((a^2-4)/2)^2 = (b+c)^2 = (b-c)^2+4bc = (-2/a)^2-24`

Hence:

`(a^2)^3-8(a^2)^2+112(a^2)-16 = 0`

Multiply through by `3^3=27` to get:

`0 = 27(a^2)^3-216(a^2)^2+3024(a^2)-432`

`=(3(a^2)-8)^3 + 816(3(a^2)-8)+6608`

Substitute `w=3a^2-8` to get:

`w^3+816w+6608= 0`

Let `w=u+v` to get:

`u^3+v^3+3(uv+272)(u+v)+6608= 0`

Add the constraint `v = -272/u` to eliminate the term in `(u+v)` and multiply through by `u^3` to get:

`(u^3)^2+6608(u^3)-20123648 = 0`

Since this derivation is symmetric in `u` and `v`, we can use the quadratic formula and hence the two Real roots of this quadratic as `u^3` and `v^3`, to provide a Real value for `w` and hence `a^2`...

`u^3 = (-6608+-sqrt(6608^2+4*20123648))/2`

`=(-6608+-sqrt(124160256))/2`

`=(-6608+-48sqrt(53889))/2`

`=(-6608+-48sqrt(53889))/2`

`=8(-413+-3sqrt(53889))`

Hence:

`3a^2-8 = w = u+v =2(root(3)(-413+3sqrt(53889))+root(3)(-413-3sqrt(53889)))`

Hence, choosing the positive root:

`a = sqrt(2/3(4+root(3)(-413+3sqrt(53889))+root(3)(-413-3sqrt(53889))))`

Then `b = (a^3-4a-4)/(4a)` and `c = (a^3-4a+4)/(4a)`.

Hence quadratics in `t` to solve.

Then the zeros of the original polynomial are the reciprocals of the roots of these quadratics.

The resulting expressions are rather complicated, but tractable.