Question

How do you use the rational roots theorem to find all possible zeros of `f(x) = x^3 - 10x^2 + 9x - 24`?

Answer 1

Find `f(x)` has no rational zeros, but we can use Cardano's method to find Real zero:

`x_1 = 1/3(10+root(3)(-919+18sqrt(1406))+root(3)(-919-18sqrt(1406)))`

and related Complex zeros.

Explanation:

`f(x) = x^3-10x^2+9x-24`

By the rational roots 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 `-24` and `q` a divisor of the coefficient `1` of the leading term. That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-24`

In addition, using Descartes' rule of signs we find that `f(x)` has `1` or `3` positive Real zeros and no negative Real zeros. So the only possible rational zeros are:

`1, 2, 3, 4, 6, 8, 12, 24`

None of these work, so `f(x)` has no rational zeros.

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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=-10`, `c=9` and `d=-24`, so we find:

`Delta = 8100-2916-96000-15552+38880 = -67488`

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^3-270x^2+243x-648`

`=(3x-10)^3-219(3x-10)-1838`

`=t^3-219t-1838`

where `t=(3x-10)`

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Cardano's method

We want to solve:

`t^3-219t-1838=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv-73)(u+v)-1838=0`

Add the constraint `v=73/u` to eliminate the `(u+v)` term and get:

`u^3+389017/u^3-1838=0`

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

`(u^3)^2-1838(u^3)+389017=0`

Use the quadratic formula to find:

`u^3=(1838+-sqrt((-1838)^2-4(1)(389017)))/(2*1)`

`=(-1838+-sqrt(3378244-1556068))/2`

`=(-1838+-sqrt(1822176))/2`

`=(-1838+-36sqrt(1406))/2`

`=-919+-18sqrt(1406)`

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)(-919+18sqrt(1406))+root(3)(-919-18sqrt(1406))`

and related Complex roots:

`t_2=omega root(3)(-919+18sqrt(1406))+omega^2 root(3)(-919-18sqrt(1406))`

`t_3=omega^2 root(3)(-919+18sqrt(1406))+omega root(3)(-919-18sqrt(1406))`

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

Now `x=1/3(10+t)`. So the zeros of our original cubic are:

`x_1 = 1/3(10+root(3)(-919+18sqrt(1406))+root(3)(-919-18sqrt(1406)))`

`x_2 = 1/3(10+omega root(3)(-919+18sqrt(1406))+omega^2 root(3)(-919-18sqrt(1406)))`

`x_3 = 1/3(10+omega^2 root(3)(-919+18sqrt(1406))+omega root(3)(-919-18sqrt(1406)))`