Question

What is the fastest and easiest method for solving cubic and quartic equations (without a polynomial calculator)?

Answer 1

It depends...

Explanation:

If the cubic or quartic (or any degree polynomial for that matter) has rational roots, then the rational roots theorem may be the quickest way to find them.

Descartes' Rule of Signs can also help to identify whether a polynomial equation has positive or negative roots, so help narrow down the search.

For a cubic equation, it may be helpful to evaluate the discriminant:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

• If `Delta = 0` then the cubic has a repeated root.

• If `Delta < 0` then the cubic has one real root and two non-real complex roots.

• If `Delta > 0` then the cubic has three real roots.

If `Delta = 0` then the cubic shares a factor with its derivative, so you should be able to find their common factor by calculating the polynomial GCF.

Otherwise, it is probably helpful to use a Tschirnhaus transformation to derive a depressed cubic with no squared term before proceeding further.

If a cubic has one real root and two non-real ones, then I would recommend Cardano's method.

If it has three real roots then I would recommend using a trigonometric substitution instead.

For quartics, you can get a depressed quartic with no cube term by a substitution like `t = x+b/(4a)`.

If the resulting quartic also has no linear term then it is a quadratic in `x^2`. You can either solve that as a quadratic and take square roots, or use a factorisation of the form:

`(x^2-ax+b)(x^2+ax+b) = x^4+(2b-a^2)x^2+b^2`

From this you can find quadratic factors to solve.

If the resulting quartic does have a linear term, then it can be factored in the form:

`(x^2-ax+b)(x^2+ax+c) = x^4+(b+c-a^2)x^2+a(b-c)x+bc`

Equating coefficients and using `(b+c)^2 = (b-c)^2+4bc`, you can derive a cubic in `a^2`. Hence you can find possible values for `a`, `b` and `c`. Then find the zeros of the quadratic factors.

There are other special cases, but that roughly covers it.