Why discriminant of a polynomial is so special?

Question

Why is $\Delta = b^2 - 4ac$ so special that it deserves a name of its own ? I saw it only once in quadratic formula and nowhere else.

Does higher polynomials also have discriminant ? if so then is there a general formula for it in terms of coefficients ?

Answer 1

It was mentioned already that the discriminant of a polynomial is zero if and only if the polynomial has multiple roots (i.e. it has some factor of $(x-a)^n$ for $n>1$).

The discriminant is defined for higher polynomials (there is a formula on wikipedia in terms of the polynomial and its derivative). Perhaps an easier to explain definition is as follows: given a polynomial $$p(x)=\prod_{i=1}^d(x-\alpha_i)$$ (recall from the fundamental theorem of algebra than any polynomial with real coefficients [and leading coefficient 1] can be written in this form, with $\alpha_i$ complex). Then we say $$\text{disc}(p(x))=\prod_{i<j}(\alpha_i-\alpha_j)^2$$ where the product is taken over all pairs $(i,j)$ with $1\le i<j\le d$. (Simple exercise - verify this definition agrees with the discrimiant in the quadratic formula.)

In abstract algebra (particularly in Galois Theory), we can do a bit more with the discriminant than just check that no roots are the same. Consider a polynomial with rational coefficients, $$p(x)=\sum_{i=0}^d a_ix^i=a_dx^d+\dots+a_1x+a_0$$ for some $a_i\in\mathbb{Q}$. In some cases, we can factor such polynomials into linear factors with coefficients in $\mathbb{Q}$: $x^2-2x+1=(x-1)^2$. In other cases, we cannot: $x^2+1=(x+i)(x-i)$. In this case, the polynomial factors into linear polynomials with coefficients in $\mathbb{C}$. Both $\mathbb{C}$ and $\mathbb{Q}$ are sets called fields - essentially they are sets where you can do multiplication, division (except by $0$), addition, and subtraction. There are many other examples of fields, for instance $\mathbb{R}$. Notice that $\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$. We could add a lot of other fields into this chain of inclusions. There are also fields between $\mathbb{Q}$ and $\mathbb{C}$ that do not include $\mathbb{R}$ or have $\mathbb{R}$ as a subset. One of the fundamental questions of Galois theory is: "What is the smallest field over which we can write a polynomial as a product of linear factors?" This smallest field is called the splitting field. It turns out that the discriminant of a polynomial can be used to compute some interesting properties of splitting fields of polynomials.

As a quick note, the splitting field of $x^2+1$ viewed as a polynomial with rational coefficients is the set $\{a+bi:a,b\in\mathbb{Q}\}\subset\mathbb{C}$. This is usually denoted by $\mathbb{Q}(i)$ (or $\mathbb{Q}[i]$ if we are not emphasizing the fact that it is a field). Also, notice that $\mathbb{R}\not\subset\mathbb{Q}(i)$ and $\mathbb{Q}(i)\not\subset\mathbb{R}$.

I am not sure what your background is, so I will leave it at this for now, but feel free to ask if you want me to expand on more things. Also, I am by no means an expert. Hopefully someone with more experience could talk more about what discriminants are useful for.

Answer 2

Higher polynomials also have discrimants which are much more complicated to calculate.

Discriminant gives you an idea about the roots of the polynomial without having to calculate the roots (and above degree 4, you just cannot algebraically).

You can think it like what is the determinant to a matrix is what discrimant is to a polynomial.

See https://en.wikipedia.org/wiki/Discriminant

Edit: I couldn't find the general formula for discrimant of an n'th degree polynomial on the Wikipedia page but it exists and is known to mathematicians; it is simply a sum of some combinations of coefficients. Edit 2: Someone provided it.

Answer 3

Here's a typical quadratic: $$ p(x) = ax^2+bx+c. $$ We all know this quadratic can be factorised into the form $$ p(x) = a(x-r_1)(x-r_2), $$ where $r_1$ and $r_2$ are two (not necessarily real) numbers that are called the roots: in particular, $p(r_i)=0$. Multiplying this form out gives $$ p(x) = ax^2 -a(r_1+r_2)x + ar_1r_2. $$ This is supposed to be the same polynomial as our original $p$, so the coefficients must be the same. Hence we find $$ r_1+r_2 = -\frac{b}{a} \qquad r_1r_2 = \frac{c}{a}; $$ these are called Vieta's formulae.

Bearing this in mind, what is this function $\Delta_p = b^2-4ac$ in terms of the roots? $$ \Delta_p = b^2-4ac = a^2(r_1+r_2)^2 - 4a^2r_1r_2 = a^2(r_1^2+2r_1r_2+r_2^2 - 4 r_1r_2) = a^2(r_1-r_2)^2. $$ So the discriminant of the quadratic $p(x)$ is the square of the difference of its roots: it is positive if $p(x)$ two distinct real roots, zero if the roots are equal (we say that $p$ has a double root), and negative if the difference is a multiple of $i$, which occurs when $r_1$ and $r_2$ are complex.

But quadratics are easy. Can we find a similar thing that tells us when the cubic $$ q(x) = ax^3+bx^2+cx+d = a(x-r_1)(x-r_2)(x-r_3) $$ has less than three distinct roots (so, for example, $r_1=r_3$ and it factorises to $a(x-r_1)^2(x-r_2)$)?

The answer is yes: we can define $$\Delta_q = a^6 (r_1-r_2)^2(r_1-r_3)^2(r_2-r_3)^2,$$ which is obviously zero if (and indeed, only if) two of the $r_i$s are the same: each bracket is the square of the difference of two of the roots, and the expression contains each possible pair. But, you say, this is useless: if we knew the roots, we'd know if they were different! And you'd be correct. So the real question is whether this quantity $\Delta_q$ can be expressed in terms information we do have about $q$, namely the coefficients? The answer turns out to be yes, because

• $\Delta_q$ is a symmetric polynomial function of the roots (swapping the subscripts on the $r$s does not change its value),
• we have corresponding Vieta formulae for the coefficients of a cubic in terms of its roots,
• and there is a theorem that says that any symmetric polynomial can be written in terms of the polynomials that appear in Vieta's formulae.

The fairly gruesome expression we find is $$ \Delta_q = b^2 c^2 - 4 a c^3 - 4 b^3 d + 18 a b c d - 27 a^2 d^2. $$ It is rather nicer if we look at $q(y-b/(3a))/a$, which has no $y^2$ term: it is of the form $$ y^3+Py+Q, $$ which is known as a depressed cubic. This has discriminant $-4P^2-27Q^3$.

Exactly the same may be done for the quartic and higher: since there is no general formula for the root of a degree-5 or higher polynomial, the discriminant is therefore more useful. Although you can probably imagine how unpleasant the expression in terms of the coefficients becomes for a polynomial of high degree if the cubic case looks like this!

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In the case of the cubic, one can determine a bit more about the roots from the value of the discriminant. A cubic with real coefficients always has a real root (it has different sign when $x$ is large and positive from when $x$ is very negative, so it must cross the $x$-axis somewhere).

• If $\Delta_q = 0$, there are at least two equal $r$s.
• If $\Delta_q < 0$, the other two roots must be complex—there is only one real root.
• If $\Delta_q > 0$, there are three distinct real roots.

One can prove these simply by considering all the possible combinations in the factorisation $q(x) =a(x-\alpha)(x^2+2bx+c)$, where $\alpha$ is the real root and the quadratic has $0$, $1$ or $2$ real roots.

Above a cubic, there are too many factors in the discriminant for it to classify the roots particularly: $x^4+1$ has positive discriminant and no real roots, for example.