Editing Cubic equation (section)

===Characteristic 2 and 3===
The above results are valid when the coefficients belong to a [[field (mathematics)|field]] of [[characteristic (algebra)|characteristic]] other than 2 or 3, but must be modified for characteristic 2 or 3, because of the involved divisions by 2 and 3.

The reduction to a depressed cubic works for characteristic 2, but not for characteristic 3. However, in both cases, it is simpler to establish and state the results for the general cubic. The main tool for that is the fact that a multiple root is a common root of the polynomial and its [[formal derivative]]. In these characteristics, if the derivative is not a constant, it is a linear polynomial in characteristic 3, and is the square of a linear polynomial in characteristic 2. Therefore, for either characteristic 2 or 3, the derivative has only one root. This allows computing the multiple root, and the third root can be deduced from the sum of the roots, which is provided by [[Vieta's formulas]].

A difference with other characteristics is that, in characteristic 2, the formula for a double root involves a square root, and, in characteristic 3, the formula for a triple root involves a cube root.
<!-- This is a unfinished extension of the section for giving explicitly the roots. This has been commented, because this seems too detailed for the importance of the case.

====Characteristic 2====
In characteristic 2, the derivative of the cubic <math>ax^3+bx^2+cx+d</math> is 
<math display="block">3bx^2+c=bx^2+c.</math>
As this polynomial is a square (at least over some [[field extension]] of the field of the coefficients), one gets: 
* if the discriminant is 0, and {{math|1=''b'' = 0}}, one has also {{math|1=''d'' = 0}}, and thus the factorization <math display="block">ax^3+bx^2+cx+d=x(ax^2+b);</math> that is, the simple root is 0, and the double root is the square root of {{math|''b''/''a''}} (remember that {{math|1=1 = –1}} in characteristic two);
* if the discriminant is zero, and {{math|1=''b'' ≠ 0}}, one has the factorization <math display="block">ax^3+bx^2+cx+d=(ax+b)(x^2+c/b);</math>
* there is a triple root if and only <math>c=b^2/a</math> and <math>d=b^3/a^2.</math>

====Characteristic 3====
In characteristic 3, the derivative of the cubic <math>ax^3+bx^2+cx+d</math> is 
<math display="block">2bx+c=c-bx.</math>

If the discriminant is zero and {{math|''b'' ≠ 0}}, the multiple root is this <math>c/b.</math> This gives the factorization 
<math display="block">ax^3+bx^2+cx+d=(ax+b-ac/b)(x-c/b)^2.</math>
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