Question

How do you use the rational roots theorem to find all possible zeros of `F(X) = 6x^4 + 2x^3 - 6x^2 + 3x - 5`?

Answer 1

`F(x)` has rational zero `x=1`, another Real zero:

`x_1 = 1/18(-8+root(3)(2510+54sqrt(2153))+root(3)(2510-54sqrt(2153)))`

and two related Complex zeros.

Explanation:

`F(x) = 6x^4+2x^3-6x^2+3x-5`

By the rational root theorem, any rational zeros of `F(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-5` and `q` a divisor of the coefficient `6` of the leading term.

That means that the only possible rational zeros are:

`+-1/6, +1/3, +-1/2, +-5/6, +-1, +-5/3, +-5/2, +-5`

Notice that the sum of the coefficients of `F(x)` is zero, hence:

`F(1) = 6+2-6+3-5=0`

So `x=1` is a zero and `(x-1)` a factor:

`6x^4+2x^3-6x^2+3x-5`

`=(x-1)(6x^3+8x^2+2x+5)`

None of the remaining possible rational zeros work, so let's focus on the cubic:

`f(x) = 6x^3+8x^2+2x+5`

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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=6`, `b=8`, `c=2` and `d=5`, so we find:

`Delta = 256-192-10240-24300+8640 = -25836`

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=972f(x)=5832x^3+7776x^2+1944x+4860`

`=(18x+8)^3-84(18x+8)+5020`

`=t^3-84t+5020`

where `t=(18x+8)`

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Cardano's method

We want to solve:

`t^3-84t+5020=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv-28)(u+v)+5020=0`

Add the constraint `v=28/u` to eliminate the `(u+v)` term and get:

`u^3+21952/u^3+5020=0`

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

`(u^3)^2+5020(u^3)+21952=0`

Use the quadratic formula to find:

`u^3=(-5020+-sqrt((5020)^2-4(1)(21952)))/(2*1)`

`=(5020+-sqrt(25200400-87808))/2`

`=(5020+-sqrt(25112592))/2`

`=(5020+-108sqrt(2153))/2`

`=2510+-54sqrt(2153)`

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)(2510+54sqrt(2153))+root(3)(2510-54sqrt(2153))`

and related Complex roots:

`t_2=omega root(3)(2510+54sqrt(2153))+omega^2 root(3)(2510-54sqrt(2153))`

`t_3=omega^2 root(3)(2510+54sqrt(2153))+omega root(3)(2510-54sqrt(2153))`

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

Now `x=1/18(-8+t)`. So the roots of our original cubic are:

`x_1 = 1/18(-8+root(3)(2510+54sqrt(2153))+root(3)(2510-54sqrt(2153)))`

`x_2 = 1/18(-8+omega root(3)(2510+54sqrt(2153))+omega^2 root(3)(2510-54sqrt(2153)))`

`x_3 = 1/18(-8+omega^2 root(3)(2510+54sqrt(2153))+omega root(3)(2510-54sqrt(2153)))`