Editing Cubic equation (section)

==Discriminant and nature of the roots==
The nature (real or not, distinct or not) of the [[zero of a function|roots]] of a cubic can be determined without computing them explicitly, by using the [[discriminant]].

===Discriminant===
The [[discriminant]] of a [[polynomial]] is a function of its coefficients that is zero if and only if the polynomial has a [[multiple root]], or, if it is divisible by the square of a non-constant polynomial. In other words, the discriminant is nonzero if and only if the polynomial is [[square-free polynomial|square-free]].

If {{math|''r''{{sub|1}}, ''r''{{sub|2}}, ''r''{{sub|3}}}} are the three [[root of a function|roots]] (not necessarily distinct nor [[real number|real]]) of the cubic <math>ax^3+bx^2+cx+d,</math> then the discriminant is
<math display="block">a^4(r_1-r_2)^2(r_1-r_3)^2(r_2-r_3)^2.</math>

The discriminant of the depressed cubic <math>t^3+pt+q</math> is 
<math display="block">-\left(4\,p^3+27\,q^2\right).</math>

The discriminant of the general cubic <math>ax^3+bx^2+cx+d</math> is 
<math display="block">18\,abcd - 4\,b^3d + b^2c^2 - 4\,ac^3 - 27\,a^2d^2.</math>
It is the product of <math>a^4</math> and the discriminant of the corresponding depressed cubic. Using the formula relating the general cubic and the associated depressed cubic, this implies that the discriminant of the general cubic can be written as 
<math display="block">\frac{4(b^2-3ac)^3-(2b^3-9abc +27 a^2d)^2}{27a^2}.</math>

It follows that one of these two discriminants is zero if and only if the other is also zero, and, if the coefficients are [[real number|real]], the two discriminants have the same sign. In summary, the same information can be deduced from either one of these two discriminants.

To prove the preceding formulas, one can use [[Vieta's formulas]] to express everything as polynomials in {{math|''r''{{sub|1}}, ''r''{{sub|2}}, ''r''{{sub|3}}}}, and {{mvar|a}}. The proof then results in the verification of the equality of two polynomials.

===Nature of the roots===
If the coefficients of a polynomial are [[real number]]s, and its discriminant <math>\Delta</math> is not zero, there are two cases:
* If <math>\Delta>0,</math> the cubic has three distinct real [[zero of a function|roots]]
* If <math>\Delta<0,</math> the cubic has one real root and two non-real [[complex conjugate]] roots.

This can be proved as follows. First, if {{mvar|r}} is a root of a polynomial with real coefficients, then its [[complex conjugate]] is also a root. So the non-real roots, if any, occur as pairs of complex conjugate roots. As a cubic polynomial has three roots (not necessarily distinct) by the [[fundamental theorem of algebra]], at least one root must be real.

As stated above, if {{math|''r''{{sub|1}}, ''r''{{sub|2}}, ''r''{{sub|3}}}} are the three roots of the cubic <math>ax^3+bx^2+cx+d</math>, then the discriminant is 
<math display="block">\Delta=a^4(r_1-r_2)^2(r_1-r_3)^2(r_2-r_3)^2</math>

If the three roots are real and distinct, the discriminant is a product of positive reals, that is <math>\Delta>0.</math>

If only one root, say {{math|''r''{{sub|1}}}}, is real, then {{math|''r''{{sub|2}}}} and {{math|''r''{{sub|3}}}} are complex conjugates, which implies that {{math|''r''{{sub|2}} − ''r''{{sub|3}}}} is a [[purely imaginary number]], and thus that {{math|(''r''{{sub|2}} − ''r''{{sub|3}}){{sup|2}}}} is real and negative. On the other hand, {{math|''r''{{sub|1}} − ''r''{{sub|2}}}} and {{math|''r''{{sub|1}} − ''r''{{sub|3}}}} are complex conjugates, and their product is real and positive.<ref>{{Cite book|last=Pratt|first=Orson|url=https://books.google.com/books?id=--A2AAAAMAAJ&q=roots+of+the+cubic+equation|title=New and Easy Method of Solution of the Cubic and Biquadratic Equations: Embracing Several New Formulas, Greatly Simplifying this Department of Mathematical Science|date=1866|publisher=Longmans, Green, Reader, and Dyer|pages=13|isbn=9781974130924|language=en|quote=...if two roots are imaginary, the product is positive...}}</ref> Thus the discriminant is the product of a single negative number and several positive ones. That is <math>\Delta<0.</math>

===Multiple root===
If the discriminant of a cubic is zero, the cubic has a [[Multiplicity (mathematics)#Multiplicity of a root of a polynomial|multiple root]]. If furthermore its coefficients are real, then all of its roots are real.

The discriminant of the depressed cubic <math>t^3 + p t + q</math> is zero if <math>4p^3 + 27q^2 = 0.</math> If {{mvar|p}} is also zero, then {{math|''p'' {{=}} ''q'' {{=}} 0 }}, and 0 is a triple root of the cubic. If <math>4p^3 + 27q^2 = 0,</math> and {{math|''p'' ≠ 0 }}, then the cubic has a simple root
<math display="block">t_1 = \frac{3q}{p} </math>

and a double root
<math display="block">t_2 = t_3 = -\frac{3q}{2p}.</math>

In other words,
<math display="block">t^3 + p t + q = \left(t - \frac{3q}{p}\right)\left(t + \frac{3q}{2p}\right)^2.</math>

This result can be proved by expanding the latter product or retrieved by solving the rather simple [[system of equations]] resulting from [[Vieta's formulas]].

By using the [[#Depressed cubic|reduction of a depressed cubic]], these results can be extended to the general cubic. This gives: If the discriminant of the cubic <math>ax^3 + b x^2 + c x + d</math> is zero, then

*either, if <math>b^2 = 3ac,</math> the cubic has a triple root <math display="block">x_1 = x_2 = x_3 = -\frac{b}{3a},</math> and <math display="block">ax^3 + bx^2 + cx + d = a\left(x + \frac{b}{3a}\right)^3</math>
*or, if <math>b^2 \ne 3ac,</math> the cubic has a double root <math display="block">x_2 = x_3 = \frac{9ad - bc}{2(b^2 - 3ac)},</math> and a simple root, <math display="block">x_1 = \frac{4abc - 9a^2d - b^3}{a(b^2 - 3ac)}.</math> and thus <math display="block">a x^3 + b x^2 + c x + d = a(x - x_1)(x - x_2)^2.</math>

===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>
-->