30. Roots of polynomials. Horner’s method. Basic theorem and its consequences

If is some polynomial, andis a number then the numberobtained by replacing the unknownonis called thevalue of the polynomial at . If, the numberis called aroot of the polynomial (or the equation).

If we divide a polynomial on an arbitrary polynomial of the first degree (or as we say further alinear polynomial) then the remainder will be either some polynomial of the zero degree or zero, i.e. in any case some number .

Theorem 3. The remainder of dividing a polynomial on a linear polynomialis equal to the valueof the polynomialat.

Proof: In fact, let . Taking the values of both parts of this equality at, we obtain:.

Corollary 4. A number is a root of a polynomialiffdivides.

On other hand, if is divided on some polynomial of the first degreethen obviously it also is divided on the polynomial, i.e. on a polynomial of the form. Thus, finding roots of a polynomialis equivalent to finding its linear divisors.

Horner’s method

Let , and let(*)

where .

Comparing the coefficients at the same degrees of in (*), we obtain:

……………

Then . At last. Thus, the coefficients of the quotient and the remainder can be sequentially obtained by calculations of the same type.

Basic theorem and its consequences

Theorem 5 (Basic theorem of algebra of complex numbers) Every polynomial with arbitrary numeric coefficients of which the degree is greater than or equals 1 has at least one root (in general, it is complex).

Let be a polynomial of the^th degree, with arbitrary complex coefficients. The basic theorem allows to assert an existence of a rootfor.

Therefore, .

The coefficients of are also real or complex numbers, and thereforehas a root, and consequently.

Continuing by such way we come after a finite number of steps to a decomposition of the polynomial of the ^th degree into product of linear multipliers:

(1)

The decomposition (1) is unique for a polynomial up to the order of multipliers.

Indeed, if there is another decomposition: (2)

then from (1) and (2) follows the following:

(3)

If the root would be different from allthen substitutinginstead of the unknown in (3) we would be obtained on the left zero and on the right a number differed from zero. Thus, each rootis equal to some rootand conversely. This doesn’t imply coinciding (1) and (2). Indeed, among the rootscan be equal each other. Let for example there areroots equal toand on the other hand there areroots amongthat are equal to. It needs to show that. If for examplethen reducing both parts of (3) on the multiplierwe come to the equality of which the left part contains the multiplier, and the right part doesn’t contain it. It is a contradiction. Thus, the decomposition (1) foris unique.

Joining together identical multipliers, the decomposition (1) can be rewritten as follows:

(4)

where , and allare distinct.

We proved the following important result:

Theorem 6. Every polynomial of the degree, with arbitrary numeric coefficients hasroots if each of the roots is counted as many as its multiplicity.