a3+2a2x+3ax2 +4x3 | a2+ax+x2

Here, the last remainder is found to be the simple quantity 7x2; we may therefore conclude that the given quantities do not admit of any divisor whatever.

143. When the quantity which is taken for the divisor contains many terms where the letter, according to which we have arranged, has the same exponent; then every successive remainder becomes more complicated than the preceding one; in this case, Analysts make use of various artifices which can be only learned by experience.

Ex. 5. Required the greatest common divisor of a2b+ac2-d3 and ab-ac+d2.

rem.

a2c+ac2-ad2-d3 | Partial quot. a

Dividing at first a2b by ab, we find for the quotient, a; multiplying the divisor by this quotient, and subtracting the product from the dividend, the remainder contains a new term, ac, arising from the product of -ac by a.

By proceeding after this manner there will be no progress made in the operation; for, taking a2c+ac2 -ad2-d3 for a dividend, and multiplying it by b, to render possible the divisor by ab, we will have

Dividend.

a2bc-abc2-abd2 -bd3
a2bc-a2c2+acd2

Divisor. ab-ac+d2

Partial quot.

rem. a2c2+abc2-acd2-abd2 -bd3 ac

and the term ac will still reproduce a term a c2, in which the exponent of a is 2.

To avoid this inconveniency, we must observe that the divisor ab-ac+d2=a(b-c)+d2, reuniting the terms ab-ac into one, and putting, to abridge the calculations, b-c=m; we will have for the divisor am+d2; it is necessary to multiply all the dividend a2b+ac2-d3 by the factor m, for the purpose of finding a new dividend whose first term would be divisible by the quantity am forming the first term of the divisor; the operation will become,

Dividend.

a2 bm+ac2m-d3m

a2 bm+abd2

Divisor. am+d2

Partial quot.

1st rem. ac2 m-abd2 d3m | ab+c2

By the first operation, the terms involving a are taken away from the dividend, and there remain no terms involving a except in the first power. In order to make them disappear, we will at first divide the term ac2m by am, and it gives for the quotient c2; multiplying the divisor by the quotient, and subtracting the product from the dividend, we will have the second remainder; taking this second remainder for a new dividend, and cancelling in it the factor d2, which is not a factor of the divisor, it will become

-ab-c2-dm; multiplying by m, we shall have

Dividend.

— abm — c2m-dm2
-abm-bd2

rem. +bd2-c2m-dm2

Divisor.

am+da

Partial quot. b

The remainder, bd2-cam-dm2, of this last division does not contain the letter a; it follows, then, that if there exist between the two proposed quantities a common divisor, it must be independent of the letter a.

Having arrived at this point, we cannot continue the division with respect to the letter a; but observing that if there be a common divisor, independent of a, of the two quantities bd2-c2m-dm2 and am+d2, it may divide separately the two parts am and d of the divisor; for, in general, if à quantity be arranged according to the powers of the letter a, every term of this quantity, independent of a, must divide separately the quantities by which the different powers of this letter are multiplied.

In order to be convinced of what has just been said, it is sufficient to observe, that in this case each of the proposed quantities should be the product of

a quantity dependent on a, and of a common divisor which does not at all depend on it. Now, if we have, for example, the expression

Aa1+Ba3+Ca2+Da+E,

in which the letters A, B, C, D, E, designate any quantities whatever, independent of a, and if we multiply it by a quantity M, also independent of a, the product,

MAa+MBa3+MCa2+MDa+ME, arranged according to a, will still contain the same powers of a as before; but the coefficient of each of these powers will be a multiple of M.

This being admitted, if we substitute for m the quantity (b-c), which this letter represents, we shall have the quantities

bd2-c2(b-e)-c(b—c)2,
a(b-c)+d2;

now it is plain that b-c and d2 have no common factor whatever: therefore the two proposed quantities have not a common divisor.

144. The greatest common divisor of two quantities may sometimes be obtained without having recourse to the general Rule: Some of the methods that are used by Analysis for this purpose, will be exemplified by the following Examples.

Ex. 6. Required the greatest common divisor of a2b2+a3b3+bac2 — aa c2 —a3bc2 -b2 c1 and a3b+ ab2+b3-a2c-abc-b2c.

After having arranged these quantities according to the powers of the letter a, we shall have

(b2 — c2 )α1 + (b 3 —bc2) a3 +b1c2 —b2 c*,

(b−c)a2 + (b2 —bc)a+b3 — b2 c ; it may at first be observed, that if they admit of a common divisor, which should be independent of the letter a, it must divide separately each of the quantities by which the different powers of a are

multiplied, (Art. 143), as well as the quantities bc2b2c and 63-63c, which comprehend not at all this letter.

The question is therefore reduced to finding the common divisors of the quantities b2c2 and b—c, and, to verify afterward, if, among these divisors, there be found some that would also divide b3bc and bbc, b4c2-b2 c1 and b3-b3c.

Dividing b2-c2 by b-c, we find an exact quotient b+c; b-c is therefore a common divisor of the quantities b2 —c2 and b-c, and it appears that they cannot have any other divisor, because the quantity b-c is divisible but by itself and unity. We must therefore try if it would divide the other quantities referred to above, or, which is equally as well, if it would divide the two proposed quantities; but it will be found to succeed, the quotients coming out exactly,

3

3

(b+c)a1+(b2+bc)a3 +-b3c2+b2 c3 ;

In order to bring these last expressions to the greatest possible degree of simplicity, it is expedient to try if the first be not divisible by b+c; this division being effected, it succeeds, and we have now only to seek the greatest common divisor of these very simple quantities;

a+ba3 +b2c3, and a2+ba+b2.

Operating on these, according to the Rule. (Art. 141), we will arrive, after the second division, at a remainder containing the letter a in the first power only; and as this remainder is not the common divisor, hence we may conclude that the letter a does not make a part of the common divisor sought. which is consequently composed but of the factor bc.