6a10a3b+23a2 b2-15ab3+2166a-16a+b+36aba—40a2b3 +38ab*-14ba-b 6a-10ab+23a3b2—15a2b3+21ab* 6a+b+13a3b2-25a2b3+17ab1-14b5 3ab2ab+2ab+ 7b3 = the remainder. And taking the last divisor for a dividend, and the remainder, after the useless factor b2 has been rejected, for a divisor, we get 3a3-2a2b+2ab+7b3 6a*—10a3b+23a2b2-15ab3 +21b1|2a-2b the remainder. Then, taking the last divisor for a dividend, and the last remainder for a divisor, after the factor 562 has been rejected, we have 3a2-5ab7b23a3 — 2ab+2ab2 + 763 | a + b 3a3-5ab7ab2 3a2b5ab2 + 763 3ab5ab2 + 768 0+0+0 the remainder. Hence, the last divisor 3a-5ab + 7b3 is the required compound divisor, because the last remainder is 0. If we now multiply the compound by the simple divisor, we have (3a2 - 5ab + 7b3) × 3ab2 = 9a3b2 — 15a2b3 + 21ab for the required greatest common divisor of the given quantities. Ex. 9.-To find the greatest common divisor of 6(a + b)1⁄23 − (2a + 2b — 9c + 9d)x2 — 3(c — d − e)x — e, and 2(a + b)x* +(2a + 2b+3c- 3d)x+(2a + 26 + 3c 3d + e)x2 + (3c3d + e)x + e. By taking 9 times the second quantity for a dividend, and the first quantity for the corresponding divisor, and using the vertical bar for the parenthesis, we get = the remainder. Rejecting the useless factor 13 in the 2a 26 3c remainder, it becomes x2 + _3 a x + e = 2 (a + b) ∞0o + 3 (cd)x+e, which will be found, on trial, to divide the last divisor, and of course it is the required divisor of the given quantities. Ex. 10. To find the greatest common divisor of a - ab + a2b3 — b3, a3 + a3b2 + a2b3 + b3, and a + b3. The greatest common divisor of the first two quantities is a+b3, and the greatest common divisor of a3 + b3 and a3 + b3 is a+b; consequently, a + b is the required divisor. (35.) It may be proper here to notice that there are some particular methods by which the greatest common divisor can often be found with more facility than by the general rule. 1. One of these methods consists in resolving the quantities into factors, and then taking the product of the factors that are common to them for their greatest common divisor. Thus, if we take the quantities 3a2a1 and 4a 2a2 3a+1, given in Bonnycastle's Algebra, Case I. of Fractions, since the quantities are both reduced to 0 by putting a = 1, we infer that they involve a - 1 as a common factor; and dividing each of them by a 1, we easily find that we shall have 3a2 · 2a − 1 = (3a + 1) ( a − 1), 4a3 2a2 3a+1 = (4a2 + 2a - 1) (a− 1), which show that a-1 is the greatest common divisor. · Again, in the same case, Bonnycastle requires the greatest common divisor of 7a2 - 23ab + 6b2 and 5a3 - 18a2b+ 11ab2 - 6b3. Since the quantities are both reduced to 0 by putting a = 3b, it follows that they have a 36 for a common factor. And it is easy to put the expressions in the forms, 7a3 23ab + 6b2 (7a — 2b) (a — 3b), 5a3 18ab11ab2 -6b3 (5a2 - 3ab + 262) × (a — 36), which show a 36 to be the greatest common divisor. = - As this method has been heretofore used, we shall not prosecute it any further. 2. Another method is deduced from the circumstance, that any factor which is common to any two quantities is common to any number of times one of them increased or diminished by any number of times the other. Thus, in the same case from Bonnycastle, it is required to find the greatest common divisor of a2 — ab 262 and a2. 3ab + 262. It is easy to see that, if we take the sum of the quantities, we shall have 2a (a — 2b), and if we take their difference we have 26 (a — 26); it hence follows that a 26 is the sought divisor. Again, in Case II. of Bonnycastle, the greatest common · divisor of the quantities 6a2+7ax-3x2 and 6a2 + 11ax + 32 is required. Here, proceeding in the same way as before, we get 2a + 3x for the sought divisor. EXAMPLES FOR EXERCISE. 1. Find the greatest common divisor of 1143 and 1791. Ans. 9. 2. Find the greatest common divisor of 637, 889, and 4949. Ans. 7. 3. Find the greatest common divisor of x2 - 2xy + y2 and x2 — y3. 4. Find the greatest common divisor of 33 +2y and 3- 2x2y + 3xy2 - 2y3. Ans. x 2x2y y. 3xy Ans. 3x 2y. 5. Find the greatest common measure of a — b3 and a ab + ab3 — bε. Ans. a2-b3. 6. Find the greatest common measure of 21 - 69×3 + + 4x3 + 3x2 Ans. x + 1. 7. Give the greatest common measure of 2x +2x+1 and 8x + 12x2 + 6x + 2. 8. Give the greatest common divisor of 182+4523y — 8zy2 — 20y3 and 5421 + 135z3y — 16zy3 — 40y*. Ans. 62+11zy - 10y3. 9. Determine the greatest common divisor of a1 + ab + ab3 + b1, a2 + ab ab1- b3, and a + ab - ab® — b3. Ans. (a + b)3. 10. Determine the greatest common divisor of x3 + 2x2 4x + 1, x + 5x — 7x3 11x+2x+10, and 3x3 5. 11. Find the greatest common divisor of (y+1)x3 — mx2 − (y2 + 1)x+m, and (p-1)(y2+1)x2 +[n(y2−1)—m(p−1)]x - mn, by using the vertical bar for the parenthesis. OF THE LEAST COMMON MULTIPLE. The least common multiple of two or more numbers, or quantities of the same kind, is the least number or quantity which can be divided by each of them without a remainder. If we use am and bm to stand for two numbers or quantities of the same kind, having m for their greatest common divisor, it is easy to see that their least common multiple must be the product of a, b, and m, or abm. For the least common multiple must involve a and m as factors, in order to be divisible by am, and it must involve band m as factors in order that it may be divisible by bm. Since abm = am xbm am m bm = xbm= × am, we easily m m find the least common multiple of two or more numbers, or quantities of the same kind, by the following RULE. Divide the product of any two of the numbers or quantities by their greatest common divisor; or divide one of them by their greatest common divisor, and multiply the quotient by the other, and the result is the least common multiple of the two numbers or quantities. Then the least common multiple of one of the remaining numbers or quantities and the least common multiple just found, is clearly the least common multiple of the three numbers or quantities used. If there are more than three numbers or quantities, we proceed in the same way to find the least common multiple of one of the remaining numbers or quantities, and of the least common multiple of the three just found; and so on, for any number of quantities. EXAMPLES. Ex. 1.-To find the least common multiple of 36 and 48. Since 12 is the greatest common divisor of 36 and 48, we have 36 12 3, and 3 x 48 = 144 the required least common multiple. = = Ex. 2. To find the least common multiple of 36, 48, and 33. By the first example, 144 is the least common multiple of 36 and 48; and since 3 is the greatest common divisor of 144 and 33, we have 33311, and by the rule 144 × 11 = 1584, is the least common multiple required. Ex. 3.-To find the least common multiple of a2 - b2 and a3 + b3. It is easy to see that a + b is the greatest common divisor |