3 n. 42. Divide 12 b c d by 3 b c x. 47. Divide 7 a b by 3 a. 3 a. 8 a b c by 3 a b c. SECTION III. Compound Quantities. 1. Divide a b + a c by a. ANS. b + c. We have here a compound quantity, a b + a c, to be divided by a simple quantity, a. We first divide a b by a, and the quotient is b; we then divide a cby a, and the quotient is c: that is, we divide each term of the compound quantity separately. 2. Divide 12 a c+9bc by 3 c. ANS. 4a+3b. 3. Divide 18 a x + 15-21 a b by 3 a. 4. Divide 3 a b c + 12 a b x-9 a b by 3 a b. 5. Divide 10 a x 15 x by 5 x. 9. Divide 35 d n + 14 d x by 7 d. 10. Divide 12 aby+6 abx-18 bm +24b by 6 b. 11. Divide 16 a 8 12. Divide 15 a b 12 y20 a dx+m by 4. 25 b x + 10 am by 10 a. 13. Divide 28 a b x-14 bx-49 b x y by 7 b x. 20. Divide 17 12 a + a b by 3 a. 21. Divide a + b by x + y. In this example, both the divisor and dividend are compound quantities; but, as all their terms are dissimilar, it is evident that the division can only be represented, and not actually performed. 22. Divide a c + b c by a + b. a + b) a c + b c ( c In this example, we divide the first term of the dividend, a c, by the first term of the divisor, a, and obtain c for the quotient. We then multiply the whole divisor, a + b, by c, to ascertain whether it be the whole quotient, or only a part of it; and the product is a c+ b c, that is, the dividend. 23. Divide bx + cx by b + c. 24. Divide a a + a b by a + b. 25. Divide 3 a a-2 a b by 3 a — 2 b. 26. Divide 12 a a + 6 a b by 4 a + 2 b. 27. Divide 18 aaa+ 6 a ab 12 a a x by 6 a 28. Divide 14 a b + 21 a c + 49 a x by 2 b + 3 c +7x. 29. Divide b b + 3 b c + 2 cc by b + c. b + c ) b b + 3 b c + 2 c c ( b + 2 c. In this example, as in the preceding, we divide the first term of the dividend by the first term of the divisor. The quotient of b b divided by b, is b. We then multiply the whole divisor, b + c, by b, and obtain the product b b + b c, which is not equal to the dividend. We are, therefore, certain that b is not the whole quotient. This product is then subtracted from the dividend, and the remainder is 2 b c + 2 cc, which must also be divided. We begin as before, and divide the first term of this remainder by the first term of the divisor; that is, 2 = 2 c, which is the b c b second term of the quotient. We next multiply the divisor, bc, by 2 c, and the product is equal to the remainder of the dividend. The whole quotient is + 2 c, which, being multiplied by the divisor, will reproduce the dividend. 30. Divide a a+2ab+bb by a + b. αα a b a b + b b * b b by a + b. a a − b b ( a — b. This example is like the former, excepting that it contains negative quantities. We obtain the second dividend, -ab-bb, by subtracting a a + a b from bb; and the sign of the second term of the quotient must be —, because the signs of the divisor, +a, and of the dividend, — a b, are not alike. It is obvious that the first term of the quotient is too large; for (a + b) a = a a + a b, which is evidently a larger quantity than the dividend, a a—b b, whatbe the values of a and b. ever may yyy (4 x x + 2 x y + y y. 4 x x y SECTION IV. General Rule for Division. From the foregoing examples and observations, we ́derive the following general RULE for Division in Algebra, when both the divisor and dividend are compound quantities: Divide the first term of the dividend by the first term of the divisor, for the first term of the quotient. Multiply the whole divisor by this term, and subtract the product from the dividend. Divide the first term of the remainder by the first term of the divisor, for the second term of the quotient. Multiply the whole divisor by this second term, and subtract the product from the remainder. Continue this series of operations as long as the nature of the question may require. 1. Divide a a + ab + ac + 5 a + 5 b + 5 c 3 m m x y 2 m m x 3. Divide b b b + c c c by b + c. 4. Divide 6 m m +mx y + 2 m x 5. Divide x x x 3 z by x + 1. 6. Divide a + b m x x y by 2 m y. + x x + x x y + x y + 3x z + a + x. |