# Resultant

of two polynomials and

The element of the field defined by the formula:

 (1)

where is the splitting field of the polynomial (cf. Splitting field of a polynomial), and are the roots (cf. Root) of the polynomials

and

respectively. If , then the polynomials have a common root if and only if the resultant equals zero. The following equality holds:

The resultant can be written in either of the following ways:

 (2)
 (3)

The expressions (1)–(3) are inconvenient for computing the resultant, since they contain the roots of the polynomials. Using the coefficients of the polynomials, the resultant can be expressed in the form of the following determinant of order :

 (4)

This determinant contains in the first rows the coefficients of the polynomial , and in the last rows the coefficients of the polynomial , and in the free spaces there are zeros.

The resultant of two polynomials and with numerical coefficients can be represented in the form of a determinant of order (or ). For this one has to find the remainders from the division of by , . Let these be

Then

The discriminant of the polynomial

can be expressed by the resultant of the polynomial and its derivative in the following way:

## Application to solving a system of equations.

Let there be given a system of two algebraic equations with coefficients from a field :

 (5)

The polynomials and are written as polynomials in :

and according to formula (4) the resultant of these polynomials (as polynomials in ) is calculated. This yields a polynomial that depends only on :

One says that the polynomial is obtained by eliminating from the polynomials and . If and is a solution of the system (5), then , and, conversely, if , then either the polynomials or have a common root (which must be looked for among the roots of their greatest common divisor), or . Solving system (5) is thereby reduced to the computation of the roots of the polynomial and of the common roots of the polynomials and in one indeterminate.

By analogy, systems of equations with any number of unknowns can be solved; however, this problem leads to extremely cumbersome calculations (see also Elimination theory).

#### References

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