Question

How do you find all zeros of the function `f(x)= 3x^3-2x^2+x+2`?

Answer 1

Find Real zero:

`x_1 = 1/9(2+root(3)(-262+9sqrt(849))+root(3)(-262-9sqrt(849)))`

and related non-Real Complex zeros using Cardano's method.

Explanation:

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

Rational root theorem

By the rational root 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 `2` and `q` a divisor of the coefficient `3` of the leading term.

That means that the only possible rational zeros are:

`+-1/3`, `+-2/3`, `+-1`, `+-2`

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

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=3`, `b=-2`, `c=1` and `d=2`, so:

`Delta = 4-12+64-972-216=-1132`

Since `Delta < 0`, this cubic has `1` Real zero and `2` non-Real Complex zeros.

Cardano's method

We can use Cardano's method to find the zeros, but first we need to simplify the form of the cubic by premultiplying by `3^5=243` and making a linear substitution (called a Tschirnhaus transformation).

`243 f(x) = 243(3x^3-2x^2+x+2)`

`= 729x^3-486x^2+243x+486`

`= (9x-2)^3+15(9x-2)+524`

Let `t=9x-2`

We want to solve:

`t^3+15t+524 = 0`

Let `t = u+v`

Then:

`u^3+v^3+3(uv+5)(u+v)+524=0`

Add the constraint:

`v = -5/u`

to eliminate the term in `(u+v)` and get:

`u^3-125/u^3+524 = 0`

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

`(u^3)^2+524(u^3)-125 = 0`

Use the quadratic formula to find roots:

`u^3=(-524+-sqrt(524^2-4(1)(-125)))/2`

`=(-524+-sqrt(274576+500))/2`

`=(-524+-sqrt(275076))/2`

`=(-524+-18sqrt(849))/2`

`=-262+-9sqrt(849)`

Since these roots are Real and the derivation was symmetrical in `u` and `v`, we can use one of the roots for `u^3` and the other for `v^3` to find Real root:

`t_1 = root(3)(-262+9sqrt(849))+root(3)(-262-9sqrt(849))`

and corresponding Complex roots:

`t_2 = omega root(3)(-262+9sqrt(849))+omega^2 root(3)(-262-9sqrt(849))`

`t_3 = omega^2 root(3)(-262+9sqrt(849))+omega root(3)(-262-9sqrt(849))`

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

Hence zeros for the original cubic:

`x_1 = 1/9(2+root(3)(-262+9sqrt(849))+root(3)(-262-9sqrt(849)))`

`x_2 = 1/9(2+omega root(3)(-262+9sqrt(849))+omega^2 root(3)(-262-9sqrt(849)))`

`x_3 = 1/9(2+omega^2 root(3)(-262+9sqrt(849))+omega root(3)(-262-9sqrt(849)))`