A=BXQ+R, and by dividing by D, we obtain But since we suppose B and R to be divisible by D, and know Q to be an entire quantity, the second part of the equality is entire; hence the first part, to which it is equal, is also entire; that is, A is exactly divisible by D. Hence, 2dly. The greatest common divisor of two polynomials is the same as that which exists between the least polynomial and their remainder after division. These principles being established, let us suppose that it is required to find the greatest common divisor between the two poly. nomials Hence, a-b is the greatest common divisor. We begin by dividing the polynomial of the highest degree by that of the lowest degree; the quotient is, as we see in the above table, a+4b and the remainder is 19ab2-1963. By the second principle, the required common divisor is the same as that which exists between this remainder and the polynomial divisor. But 19ab2-1963 can be put under the form 1962(a—b). Now the factor, 1962, will divide this remainder without dividing a2-5ab+4b2, hence, by the first principle, this factor cannot enter into the greatest common divisor; we may therefore suppress it, and the question is reduced to finding the greatest common divisor between a2-5ab+b' and a-b. Dividing the first of these two polynomials by the second, there is an exact quotient, a-4b; hence a-b is their greatest common divisor, and is consequently the greatest common divisor of the two proposed polynomials. Again, take the same example, and arrange the polynomials with reference to b. 1st. Rem. 2d. Rem. or Hence, b+a, or a-b, is the greatest common divisor. Here we meet with a difficulty in dividing the two polymonials, because the first term of the dividend is not exactly divisible by the first term of the divisor. But if we observe that the co-efficient 4 of this last, is not a factor of all the terms of the polynomial 4b2-5ab+a2, and that therefore, by the first principle, 4 cannot form a part of the greatest common divisor, we can, without affecting this common divisor, introduce this factor into the dividend. This gives -1263+12ab2-4a3b+4a3, and then the division of the first two terms is possible. Effecting this division, the quotient is 3b, and the remainder is — 3ab2 —a2b+4a3. As the exponent of b in this remainder is still equal to that of the divisor, the division may be continued, by multiplying this remainder by 4, in order to render the division of the first term possible. This done, the remainder becomes —12ab2—4a2b+16a3, which divided by 462-5ab+a2, gives the quotient -3a, which should be separated from the first by a comma, having no connexion with it; and the remainder is -19a2b+19a". Placing this last remainder under the form 19a'(-b+a), and suppressing the factor 19a, as forming no part of the common divisor, the question is reduced to finding the greatest common divisor between 46-5ab+a, and -b+a. Dividing the first of these polynomials by the second, we obtain an exact quotient, -4b+a; hence -b+a, or a-b, is the greatest common divisor required. 68. In the above example, as in all those in which the exponent of the principal letter is greater by unity in the dividend than in the divisor, we can abridge the operation by multiplying every term of the dividend by the square of the co-efficient of the first term of the divisor. We may easily conceive that, by this means, the first partial quotient obtained will contain the first power of this co-efficient. Multiplying the divisor by the quotient, and making the reductions with the dividend thus prepared, the result will still contain the coefficient as a factor, and the division can be continued until a remainder is obtained of a lower degree than the divisor, with reference to the principal letter. Take the same example as before, viz. -3b3+3ab2 —a2b+a3 and 463-5ab+a'; and multiply the dividend by the square of 4=16: and we have 1st. Rem. or 2d. Rem. First Operation. -48b3+48ab2 —16a2b + 16a3||4b2—5ab+a2 REMARK 1. When the exponent of the principal letter in the dividend exceeds that of the same letter in the divisor by two, three, &c. units, multiply the dividend by the third, fourth, &c. power of the co-efficient of the first term of the divisor. It is easy to see the reason of this. 2. It might be asked if the suppression of the factors, common to all the terms of one of the remainders, is absolutely necessary, or whether the object is merely to render the operations more simple. Now, it will easily be perceived that the suppression of these factors is necessary; for, if the factor 19a2 was not suppressed in the preceding example, it would be necessary to multiply the whole dividend by this factor, in order to render the first term of the dividend divisible by the first term of the divisor; but then, a factor would be introduced into the dividend which was also contained in the divi. sor; and consequently the required greatest common divisor would be combined with the factor 19a2, which should not form a part of it. 69. For another example, it is proposed to find the greatest common divisor between the two polynomials, a*+3a3b+4a2b2-6ab3+2b1 and 4a2b+2ab2-263, or simply, 2a+ab-b2, since the factor 26 can be suppressed, being a factor of the second polynomial and not of the first. First Operation. 8a*+24a3b+32a3b3-48ab3+16b* || 2a1+ab—b3 Second Operation. Multiply by 2601, the square of 51. 5202a2+2601ab—26016'|| 51a— 29b 1st. Rem. 5202a2-2958ab 102a+1096 +5559ab-260162 5559ab-316162 The exponent of the letter a in the dividend, exceeding that of the same letter in the divisor by two units, we multiply the whole dividend by the cube of 2, or 8. This done, we perform three consecutive divisions, and obtain for the first principal remainder, -51ab3+296*. Suppressing b3 in this remainder, it becomes 51a+296 for a new divisor, or, changing the signs, which is permitted, 51a-29b: the new dividend is 2a2+ab-b3. Multiplying this dividend by the square of 51, or 2601, then effecting the division, we obtain for the second principal remainder, +5606", which proves that the two proposed polynomials are prime with respect to each other, that is, they have not a common factor. In fact it results from the second principle (Art. 67), that the greatest common divisor must be a factor of the remainder of each operation; therefore it should divide the remainder 56062; but this remainder is independent of the principal letter a; hence, if the two polynomials have a common divisor, it must be independent of a, and will consequently be found as a factor in the co-efficients of the different powers of this letter, in each of the proposed polynomials; but it is |