Question

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

Answer 1

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

`x_1 = 1/6(-8+root(3)(-424+54sqrt(58))+root(3)(-424-54sqrt(58)))`

and related Complex zeros.

Explanation:

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

By the rational roots theorem, any rational zeros of `f(x)` can be expressed 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.

That means that the only possible rational zeros are:

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

None of these work, so we need other means to find the 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=2`, `b=8`, `c=7` and `d=-8`, so we find:

`Delta = 3136-2744+16384-6912-16128 = -6264`

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+864x^2+756x-864`

`=(6x+8)^3-66(6x+8)-848`

`=t^3-66t-848`

where `t=(6x+8)`

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Cardano's method

We want to solve:

`t^3-66t-848=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv-22)(u+v)-848=0`

Add the constraint `v=22/u` to eliminate the `(u+v)` term and get:

`u^3+10648/u^3-848=0`

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

`(u^3)^2-848(u^3)+10648=0`

Use the quadratic formula to find:

`u^3=(848+-sqrt((-848)^2-4(1)(10648)))/(2*1)`

`=(-848+-sqrt(719104-42592))/2`

`=(-848+-sqrt(676512))/2`

`=(-848+-108sqrt(58))/2`

`=-424+-54sqrt(58)`

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)(-424+54sqrt(58))+root(3)(-424-54sqrt(58))`

and related Complex roots:

`t_2=omega root(3)(-424+54sqrt(58))+omega^2 root(3)(-424-54sqrt(58))`

`t_3=omega^2 root(3)(-424+54sqrt(58))+omega root(3)(-424-54sqrt(58))`

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

Now `x=1/6(-8+t)`. So the zeros of our original cubic are:

`x_1 = 1/6(-8+root(3)(-424+54sqrt(58))+root(3)(-424-54sqrt(58)))`

`x_2 = 1/6(-8+omega root(3)(-424+54sqrt(58))+omega^2 root(3)(-424-54sqrt(58)))`

`x_3 = 1/6(-8+omega^2 root(3)(-424+54sqrt(58))+omega root(3)(-424-54sqrt(58)))`