Question

How do you find the zeros of `x^5 - 5x^4 - x^3 + x^2 + 4 = 0`?

Answer 1

Zeros:

`x_1 = 1`

`x_2 = -1`

`x_3 = 1/3(5+root(3)(179+12sqrt(114))+root(3)(179-12sqrt(114)))`

and two related Complex zeros.

Explanation:

`x^5-5x^4-x^3+x^2+4 = 0`

First note that the sum of the coefficients is zero.

That is:

`1-5-1+1+4 = 0`

So `x_1=1` is a zero and `(x-1)` a factor:

`x^5-5x^4-x^3+x^2+4 = (x-1)(x^4-4x^3-5x^2-4x-4)`

The sum of the coefficients of the remaining quartic with signs reversed on terms of odd degree is zero.

That is:

`1+4-5+4-4 = 0`

So `x_2=-1` is a zero and `(x+1)` a factor:

`x^4-4x^3-5x^2-4x-4=(x+1)(x^3-5x^2-4)`

By the rational root theorem, any rational zeros of the remaining cubic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-4` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational zeros are:

`+-1, +-2, +-4`

None of these work, so the cubic only has irrational zeros.

Let: `f(x) = x^3-5x^2-4`

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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=1`, `b=-5`, `c=0` and `d=-4`, so we find:

`Delta = 0+0-2000-432+0 = -2432`

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^3-135x^2-108`

`=(3x-5)^3-75(3x-5)-358`

`=t^3-75t-358`

where `t=(3x-5)`

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Cardano's method

We want to solve:

`t^3-75t-358=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv-25)(u+v)-358=0`

Add the constraint `v=25/u` to eliminate the `(u+v)` term and get:

`u^3+15625/u^3-358=0`

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

`(u^3)^2-358(u^3)+15625=0`

Use the quadratic formula to find:

`u^3=(358+-sqrt((-358)^2-4(1)(15625)))/(2*1)`

`=(-358+-sqrt(128164-62500))/2`

`=(-358+-sqrt(65664))/2`

`=(-358+-24sqrt(114))/2`

`=-119+-12sqrt(114)`

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_3=root(3)(-119+12sqrt(114))+root(3)(-119-12sqrt(114))`

and related Complex roots:

`t_4=omega root(3)(-119+12sqrt(114))+omega^2 root(3)(-119-12sqrt(114))`

`t_5=omega^2 root(3)(-119+12sqrt(114))+omega root(3)(-119-12sqrt(114))`

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

Now `x=1/3(5+t)`. So the zeros of `f(x)` are:

`x_3 = 1/3(5+root(3)(-119+12sqrt(114))+root(3)(-119-12sqrt(114)))`

`x_4 = 1/3(5+omega root(3)(-119+12sqrt(114))+omega^2 root(3)(-119-12sqrt(114)))`

`x_5 = 1/3(5+omega^2 root(3)(-119+12sqrt(114))+omega root(3)(-119-12sqrt(114)))`