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The background is about the relation of roots and coefficients, and the symmetric polynomial.

My book directly write (1), can you tell me where and how does this come from?

I know $\prod \left(x_i-x_j\right)$ is the result of Vandermonde Determinant, so is (1) related with Det?

$$\begin{align*}D=\prod _{i<j} \left(x_i-x_j\right)^2\tag{1}\end{align*}$$

$$\begin{align*}x^3+a_1x^2+a_2x+a_3\tag{2}\end{align*}$$

$$\begin{align*}D=a_1^2a_2^2-4a_2^3-4a_1^3a_3-27a_3^2+18a_1a_2a_3.\tag{3}\end{align*}$$

Can you show me how to make (1) becomes (3)?

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The discriminant $D$ is a symmetric version of the Vandermonde determinant.

One reason this symmetric version is important is that every symmetric polynomial function of the roots of a polynomial can be written as a polynomial in the elementary symmetric functions, which are the coefficients of the original polynomial. There is even an algorithm for finding this expression, but in this case it can be found manually. It is probably easier to use the expression of the discriminant as a resultant, but that requires a theorem.

The discriminant is important because it gives you information about the number of real roots. Since it can be expressed in terms of the coefficients, this information can be obtained without knowing the roots.

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  • $\begingroup$ See people.math.carleton.ca/~williams/ant/ch6-solns/ch6-qu15.pdf for the detailed computation. $\endgroup$ – lhf Aug 2 '13 at 11:46
  • $\begingroup$ The discriminant is important for all sorts of reasons. $\endgroup$ – Lubin Aug 2 '13 at 12:17
  • $\begingroup$ Why $D=\prod _{i<j} \left(x_i-x_j\right)^2$=0, means the monic polynomial has multiple root over C? $\endgroup$ – User19912312 Aug 11 '13 at 14:17

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