Question

What are the zeros of `f(x) = x^3-140x^2+7984x-107584`?

Answer 1

Use Cardano's method to find Real zero:

`x_1 = 1/3(140+8 root(3)(1628+12sqrt(20589))+8 root(3)(1628-12sqrt(20589)))`

and two related Complex zeros.

Explanation:

Given:

`f(x) = x^3-140x^2+7984x-107584`

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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=1`, `b=-140`, `c=7984` and `d=-107584`, so we find:

`Delta = 1249387417600-2035736559616-1180841984000-312506560512+2164555653120 = -115142033408`

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=27f(x)=27x^3-3780x^2+215568x-2904768`

`=(3x-140)^3+13056(3x-140)+1667072`

`=t^3+13056t+1667072`

where `t=(3x-140)`

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Cardano's method

We want to solve:

`t^3+13056t+1667072=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv+4352)(u+v)+1667072=0`

Add the constraint `v=-4352/u` to eliminate the `(u+v)` term and get:

`u^3-82426462208/u^3+1667072=0`

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

`(u^3)^2+1667072(u^3)-82426462208=0`

Use the quadratic formula to find:

`u^3=(-1667072+-sqrt((1667072)^2-4(1)(-82426462208)))/(2*1)`

`=(1667072+-sqrt(2779129053184+329705848832))/2`

`=(1667072+-sqrt(3108834902016))/2`

`=(1667072+-12288sqrt(20589))/2`

`=833536+-6144sqrt(20589)`

`=8^3(1628+-12sqrt(20589))`

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=8 root(3)(1628+12sqrt(20589))+8 root(3)(1628-12sqrt(20589))`

and related Complex roots:

`t_2=8 omega root(3)(1628+12sqrt(20589))+8 omega^2 root(3)(1628-12sqrt(20589))`

`t_3=8 omega^2 root(3)(1628+12sqrt(20589))+8 omega root(3)(1628-12sqrt(20589))`

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

Now `x=1/3(140+t)`. So the roots of our original cubic are:

`x_1 = 1/3(140+8 root(3)(1628+12sqrt(20589))+8 root(3)(1628-12sqrt(20589)))`

`x_2 = 1/3(140+8 omega root(3)(1628+12sqrt(20589))+8 omega^2 root(3)(1628-12sqrt(20589)))`

`x_3 = 1/3(140+8 omega^2 root(3)(1628+12sqrt(20589))+8 omega root(3)(1628-12sqrt(20589)))`

Answer 2

`x ~~ 18.9, x ~~ 60 + 45i, and x ~~ 60 - 45i`

Explanation:

I used the Cubic Equation calculator