or 4 a3 (2 a3 — b2 + 1) + 2 a3 — b2 +1, 3 (2 a 3 — b2 + 1) (4 a3 + 1). The division is performed at once by suppressing the factor 2 a3 — b2 + 1, equal to the divisor, and the quotient will be 4a3 +1. After a little practice, methods of this kind will readily occur, by which algebraic operations are abridged. By frequent exercise in examples of this kind, the resolution of a quantity into its factors is at length easily performed; and it is often rendered very conspicuous, when, instead of performing the operations represented, they are only indicated. Of algebraic fractions. 47. WHEN we apply the rules of algebraic division to quantities, of which the one is not a factor of the other, we perceive the impossibility of performing it, since in the course of the operation we arrive at a remainder, the first term of which is not divisible by that of the divisor. See an example; 2d. rem. The first term · a b2 + b3 ab2 of the second remainder cannot be divided by a2, the first term of the divisor, so that the process is arrested at this point. We can however, as in arithmetic, annex a b2 + b3 to the quotient ab the fraction a2 + b2 -, having the remainder for the numerator, and the divisor for the denominator; and the quotient will be b3 It is evident, that the division must cease, when we come to a remainder, the first term of which does not contain the letter with reference to which the terms are arranged, or to a power inferior to that of the same letter in the first term of the divisor. 48. When the algebraic division of the two quantities cannot be performed, the expression of the quotient remains indicated under the form of a fraction, having the dividend for the numerator, and the divisor for the denominator; and to abridge it as much as possible, we should see if the dividend and divisor have not common factors, which may be cancelled (38). But when the terms of the fraction are polynomials, the common factors are not so easily found as when they are simple quantities. They are in general to be sought by a method analogous to that, which is given in arithmetic for finding the greatest common divisor of two numbers. We cannot assign the relative magnitudes of algebraic expressions, as we do not give values to the letters which they contain; the denomination of greatest common divisor therefore, applied to these expressions, ought not to be taken altogether in the same sense as in arithmetic. In algebra we are to understand by the greatest common divisor of two expressions, that which contains the most factors in all its terms, or which is of the highest degree (27). Its determination rests, as in arithmetic, upon this principle; Every common divisor to two quantities must divide the remainder after their division. The demonstration given in arithmetic (art. 61) is rendered clearer by employing algebraic symbols. If we represent the common divisor by D, the two quantities proposed might be expressed by the products A D and B D, formed from the common divisor and the factor by which it is multiplied in each of the quantities. This being supposed, if Q stands for the entire quotient, and R for the remainder resulting from the division of AD by B D, we have AD BD Q+R (Arith. 61); dividing now the two members of the equation by D, we obtain and since the first member, which in this case must be composed of the same terms, as the second, is entire, it must follow, that R D is reduced to an expression without a divisor, that is to say, that R is divisible by D. According to this principle, we begin, as in arithmetic, by inquiring whether one of the quantities is not itself the divisor of the other; if the division cannot be exactly performed, we divide the first divisor by the remainder, and so on; and that remainder, which will exactly divide the preceding, will be the greatest common divisor of the two quantites proposed. But it will be necessary in the divisions indicated, to have regard to what belongs to the nature of algebraic quantities. We are not, in the first place, to seek a common divisor of two algebraic quantities, except when they have common letters; • and we must select from them a letter, with reference to which the proposed expressions are to be arranged, and that is to be taken for the dividend in which this letter has the highest exponent the other being the divisor. Let there be the two quantities which are already arranged with reference to the letter a; we take the first for the dividend and the second for the divisor. A difficulty immediately presents itself, which we do not meet with in numbers, and this is, that the first term of the divisor will not exactly divide the first term of the dividend, on account of the factors 4 and b in the one, which are not in the other. But the letter b being common to all the terms of the divisor and not to those of the dividend, it follows (40) that b is a factor of the divisor, and that it is not of the dividend. Now every divisor common to two quantities can consist only of factors which are common to the one and to the other; if then there be such a divisor with respect to the two quantities proposed, it is to be looked for among the factors of the quantity 4 a2 — 5 a b + b2, which remains of the quantity 4a2b-5 a b2+b3, after suppressing b; so that the question reduces itself to finding the greatest common divisor of the two quantities 3 a3-3a2b+ab2 —b3, 4a25ab+b2. For the same reason that we may cancel in one of the proposed quantities the factor b which is not in the other, we may likewise introduce into this a new factor, provided it is not a factor of the first. By this step, the greatest common divisor, which can consist only of terms common to both, will not be affected. Availing myself of this principle, I multiply the quantity 3a3 3 a2 b + a b2 —b3 by 4, which is not a factor of the quantity 4 a25ab+b2, in order to render the first term of the one divisible by the first term of the other. I shall thus have for the dividend the quantity 12 as 12 a2 b+4 a b2 — 4 b3, for the divisor the quantity 4a25ab+b2, and the quotient will be 3a. Multiplying the divisor by this quotient, and subtracting the product from the dividend, I have for a remainder a quantity which, according to the principle stated at the commencement of this article, must have with 4 a2-5 ab+b2, the same greatest common divisor as the first. Profiting by the remarks made above, I suppress the factor b, common to all the terms of this remainder, and multiply it by 4 in order to render the first term divisible by that of the divisor; I have then for a dividend the quantity 12 a2+4 ab- 16 b2, and for a divisor the quantity 4a2-5ab+b2; and the quotient thence arising is 3. Multiplying the divisor by the quotient, and subtracting the product from the dividend, we obtain the remainder 19ab19b2, and the question is reduced to finding the greatest common divisor to this quantity, and 4 a2 5 a b + b2. But the letter a, with reference to which the division is made, not being in the remainder, except of the first degree, while it is of the second degree in the divisor, it is this which must be taken for the dividend, and the remainder must be made the divisor. Before beginning this new division I expunge from the divisor 19 ab-19 b2, the factor 19 b common to both the terms, and which is not a factor of the dividend; I have then for a dividend the quantity and for a divisor 4a25ab+b2, b. The division leaves no remainder; so that ab is the greatest common divisor required. By retracing these steps, we may prove à posteriori, that the quantity a-b must exactly divide the two quantities proposed, and that it is the most compounded of those which will do it. In dividing by ab the two quantities proposed 3 a3-3a2 bab2 — b3, 4 a2 b — 5 a b2 + b3, we resolve them as follows; (sa2 + b2) (a - b), (4 ab — b2) (a - b). 49. When the quantity, which we take for a divisor, contains several terms having the letter, with reference to which the arrangement is made, of the same degree, there are precautions to be used, without which the operation would not terminate. See an example of this. Let there be the quantities a2 b+ ac2 — d3, ab—ac+d2; if we make the preparation as for common division ab-ac+d2 d3 a Rem. a2c+a c2 — a d2 — d3, by dividing first a2 b by a b, we have for the quotient a; multiplying the divisor by this quotient, and subtracting the products from the dividend, the remainder will contain a new term, in which a will be of the second degree, namely, a2 c, arising from the product of ac by a. Thus no progress has been made; for by taking the remainder a2 c + ac2 a d2 for a dividend, and multipling by b to render the division possible by ab, we have a2bc + abc2abd2bd3 - a2bc+a2 c2acd2 rem. a2 c2 + a b c2 — a c d2 — abd2 —bd3, ab-ac+d2 ac and the term a c produces still a term a2 c2, in which a is of the second degree. To avoid this inconvenience, it must be observed, that the divisor a bac+d2=a (b—c) + d2, by uniting the terms ab-acin one; and for the sake of shortening the operation, making b―c=m, we have for a divisor a m+d2; but then the whole dividend must be multiplied by the factor m, to make a new dividend, the first term of which may be divided by am, the first term of the divisor; the operation then becomes |