Question

How do you use the rational roots theorem to find all possible zeros of `f(x)=2x^3+7x^2+8x-8`?

Answer 1

Find `f(x)` has no rational zeros.

Use Cardano's method to find Real zero:

`x_1 = 1/6(-7+root(3)(-593+12sqrt(2442))+root(3)(-593-12sqrt(2442)))`

and related Complex zeros.

Explanation:

`f(x) = 2x^3+7x^2+8x-8`

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 `-8` and `q` a divisor of the coefficient `2` of the leading term. So the only possible rational zeros are:

`+-1/2, +-1, +-2, +-4, +-8`

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

We can find the irrational zeros using Cardano's method.

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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=2`, `b=7`, `c=8` and `d=-8`, so we find:

`Delta = 3136-4096+10976-6912-16128 = -13024`

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=108f(x)=216x^3+756x^2+864x-864`

`=(6x+7)^3-3(6x+7)-1186`

`=t^3-3t-1186`

where `t=(6x+7)`

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Cardano's method

We want to solve:

`t^3-3t-1186=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv-1)(u+v)-1186=0`

Add the constraint `v=1/u` to eliminate the `(u+v)` term and get:

`u^3+1/u^3-1186=0`

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

`(u^3)^2-1186(u^3)+1=0`

Use the quadratic formula to find:

`u^3=(1186+-sqrt((-1186)^2-4(1)(1)))/(2*1)`

`=(-1186+-sqrt(1406596-4))/2`

`=(-1186+-sqrt(1406592))/2`

`=(-1186+-24sqrt(2442))/2`

`=-593+-12sqrt(2442)`

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)(-593+12sqrt(2442))+root(3)((-593-12sqrt(2442))`

and related Complex roots:

`t_2=omega root(3)(-593+12sqrt(2442))+omega^2 root(3)(-593-12sqrt(2442))`

`t_3=omega^2 root(3)(-593+12sqrt(2442))+omega root(3)(-593-12sqrt(2442))`

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

Now `x=1/6(-7+t)`. So the zeros of our original cubic are:

`x_1 = 1/6(-7+root(3)(-593+12sqrt(2442))+root(3)(-593-12sqrt(2442)))`

`x_2 = 1/6(-7+omega root(3)(-593+12sqrt(2442))+omega^2 root(3)(-593-12sqrt(2442)))`

`x_3 = 1/6(-7+omega^2 root(3)(-593+12sqrt(2442))+omega root(3)(-593-12sqrt(2442)))`