a Here the quantities are already arranged according to the powers of the letter a; the first is taken for a dividend, and the second for a divisor: In the first place, the factor b is found in every term of the divisor and not in every term of the dividend ; therefore, the divisor is divided by the factor b, and the result is 4a- 5ab +62; but the first term of this result will not divide exactly that of the divilend, on account of the factor 4, which is not in the dividend; the dividend is therefore multiplied by 4 in order to render the division of their first terms complete: Now, the dividend 12a’ – 12a-5 +4abo ---4b3 is divided by the divisor 4a - 5ab+ 6?, and the partial quotient is 3a. Multiplying the divisor by this quotient, and subtracting the product from the dividend, the remainder is 3a"btab 463, a quantity which, according to (Art. 135), must still have with 4a --5ab-+-69 the same greatest common divisor as the first. Suppressing the factor b, common to all the terms of the remainder, or, which is the same, dividing the remainder by b, and multiplying the result by 1, to render possible the division of its first term by that of the divisor, we have then for the dividend the quantity 1 2a +-4ab--166", and for the divisor the quantity 4a2--5ab +-62; the partial quotient is 3. Multiplying the divisor by the quotient, and subtracting the product from the dividend, the remainder is 19ab-196", and the question is now reduced to finding the greatest common divisor of 19ab-1962 and 4a---5ab-+-62. But the letter a, according to which the division has been performed, being of the second degree in the divisor, and only of the first in the remainder ; it is necessary therefore to take the last divisor for a new dividend, and the remainder for a new divisor. Having, at the commencement of this new division, divided the divisor 19ab-1962 by the factor 19b, common to all its terms, and which is not at all common to those of the dividend ; therefore, the dividend is 4a3--5ab -+-62, the divisor a-b, and the quotient 40-6: The operation is completed, because nothing remains; and consequently, (Art 135), a-b is the ; greatest common divisor sought. If we divide the two proposed quantities by a-b, the quotients will be 3a2 +62 and 4ab-62: Whence, the two given quantities are thus decomposed as follows: (3a® +62)x(a-6), (4ab-b?) (a - b). Ex. 2. Required the greatest common divisor of 394---2--1 and 403 --2a2 --3a ti, : In the above operation, the remainder --110 +11 is divided by -11, (its greatest simple divisor witli a negative sign), so as to make the leading term positive : Or, which is the same, if any of the divisors, in the course of the operation, become negative, they may have their signs changed, or be taken affirmatively, without altering the truth of the result; thus, in the above operation, changing the signs of -lla+11, it becomes 11a-11, and dividing 110–11 by its greatest simple divisor 11, we have a-1, as before. Therefore a -1 is the greatest common divisor sought, and the two given quantities may be readily decomposed thus; (3a- +1)+(a-1), (4Q2 + 2a-1)+(a−1), Ex. 3. Required the greatest common divisor of ao --63, a3 + 2ab + 2ab? +63 and at taoba +64. In the first place, the greatest common divisor of 03 — 62 and a3 + 2a*b+ 2ab2 +63, is aa + ab +62, which is found thus ; Dividend. Divisor. a3 + 2a2b+ 2ab2 -+-b 3 a3-63 -63 Partial quot. 1 aa tab -+-62 a3 -a2b-ab% -63 Hence, the greatest common divisor of ai-63 and a3 +2a2b+ 2ab +63, is atab+b; and the greatest common divisor of a' tab +62 and a't ab2 +-64, is found to be a' tab+ba, thus; Dividend. Divisor. a* taab? +-64 ao tab-t-b2 au+a3b+a2b2 Consequently a” tab +62 is the greatest common divisor which was required; and dividing each of the given quantities by this divisor, we will thus decompose them as follows: (a-6)(a2 + ab +6*),(a+b)(ao tab-64),(až -abt 62)(atab +62). 142. It has been remarked (Art. 136), that if the last divisor be unity, and the remainder nothing; then the fraction is already in its lowest terms; this observation is applicable to numbers, and as in algebraic quantities, the greatest simple divisor may be readily found by inspection. Now, it only remains to discover, if compound algebraic quantities can admit of a compound divisor. If, by proceeding according to the Rule (Art. 141), no compound divisor can be found, that is, if the last remainder be only a simple quantity; we may conclude the case proposed does not admit of any, but is already in its lowest terms. Ex. 4. Required the greatest common divisor of eta tax toxa and a3 + 2a2x +3ax? +-4x3. It is plain by inspection that they do not admit of any simple divisor; then the operation according to the rule will stand thus; |