38. Corollary. When, in example 6 of art. 30, the ex-6 ponent n of a in the divisor is greater than its exponent m in the dividend, the exponent m-n in the quotient is negative; and a negative exponent is thus substituted for the usual fractional form of the quotient. In the same way we should have a b2c3a5 b2 c8 d= = a1 b2 c3 ÷ a5 b2 c8 d1 — a−4 c−5 d−1. Any quotient of monomials may thus be expressed by means of negative exponents without using fraçtional forms, 39. EXAMPLES. 1. Divide 5 a4 b3 c2 d by 15 a b5 c2 d3 12. 40. Corollary. Quantities, thus expressed by means of fractional exponents, may be used in all Ans. 3 a-1. 4 Division of Polynomials. calculations, and may be added, subtracted, multiplied, or divided by the rules already given, the signs being carefully attended to. 1. Find the sum of 7 a-3+9 am b-P-6 ab-2 ce, -3a-3, 5 amb-r+11 ab-2c, a-3-14 am b-P. Ans. 5 a-5 ab-2ce. 2. Reduce the polynomial 9 a−3 b −2 c4 — 7 ba-3 + (18a-3 b — 5 a1 bm + c≈ −3.25) — (3 a′′ bm — a−3 b−2c4 +3c-5.25) to its simplest form. Ans. 10 a-3b-2c4+11a-3b-8 aa bm—2 c2+2.25. 6. Find the continued product of 11 a-2, -2 a−5, 4 ao, and -9 a1. Ans. 792 a6. 7. Find the continued product of 2 a-3, 7 a-9, and _ 3 a. Ans. -42 a-6 42 a6° Ans. 150 a12 b c. 8. Find the continued product of 5 a3 b-4, 10 a2 b5 c, and -3 a1. 9. Multiply-13 a-1bc-3 by -4a-3b-6 c2. 10. Divide a-m by a". Ans. 52a-4b-5c-1. Ans. a m-n =a-(m+n). Ans. am+". Ans. a-m+n=an—m ̧ 13. Divide 14 a-8bc3d-1e by 2 a b-3 c5 d h. Ans. 7a-9b4c-2d-2eh-1. Division of Polynomials. 42. Problem. To divide one polynomial by an other. Solution. The term of the dividend, which contains the highest power of any letter, must be the product of the term of the divisor which contains the highest power of the same letter, multiplied by the term of the quotient which contains the highest power of the same letter. A term of the quotient is consequently obtained by dividing, as in art. 35, the term of the dividend which contains the highest power of any letter by that term of the divisor which contains the highest power of the same letter. But the dividend is the sum of the products of the divisor by all the terms of the quotient; and, therefore, If the product of the divisor by the term just found is subtracted from the dividend, the remainder must be equal to the sum of the products of the divisor by the remaining terms of the quotient, and may be used as a new dividend to obtain another term of the quotient. By pursuing this process until the dividend is entirely exhausted, all the terms of the quotient may be obtained. It facilitates the application of this method to arrange the terms of the dividend and divisor according to the powers of some letter, the term which contains the highest power being placed first, that which contains the next to the highest power being placed next, and so on. Division of Polynomials. 43. EXAMPLES. 1. Divide -16 a3 x3+a6 +64 x6 by 4x2+a2-4ax. Solution. In the following solution the dividend and divisor are arranged according to the powers of the letter z; the divisor is placed at the right of the dividend with the quotient below it. As each term of the quotient is obtained, its product by the divisor is placed below the dividend or remainder from which it is obtained, and is subtracted from this dividend or remainder. a2 4x2-4 a x + a2 64 26-16 a3 x3+a6 Divisor. 64x6—64 a x5-16 a2 x4 | 16 x2+16 a x3+12 a2x2+4 a3 x+a1 64 a x5 16 a2 x4 16 a3 x3 + a6 = 1st Remainder. 64 a x5-64 a2 x4 +16 a3 x3 48 a2 x4 − 32 a3 x3 + ao = 2d Remainder. 48 a2 x4 — 48 a3 x3 + 12 aa x2 16 a3 x3 — 12 a1 x2+a6 = 3d Remainder. 16 a3 x3 16 a1 x2+4 a5 x 4 a1 x2 — 4 a5 x +a6= 4th Remainder. 4 a4 x2. 4 a5 x + a 0. Ans. 16 x2+16 a x3 + 12 a2x2 + 4 a3 x + a1 2. Divide b c3 c3 x by c3. 3. Divide a2+2 a b + b2 by a+b. 4. Divide a8b4+ 15 all b5 48 a14 b6 10 a9 b2- - a6 b. Ans. b. -x. Ans. a+b. 20 a17 b7 by Ans, a2b3-5 a5 b4—2 a8 b5. 5. Divide 118 22+81 24 by 1+6x+9z2. Division of Polynomials. 324m yên t3 xem yên — yên by z3m - 7. Divide 26m 3x2m yn+3xm yê — y3n ̧ Ans. x3m+3x2m yn +3 xm y2n +yзn. 8. Divide 1+ a3 n3 by - 1+an. Ans. 1+an+ a2 n2. 9. Divide 2 a1- 13 a3 b + 31 a2 b2 — 38 a b3 + 24 64 Ans. a+a3b + a2 b2+ a b3 + ba. 44. Corollary. The quotient can be obtained with equal facility by using the terms which contain the lowest powers of a letter instead of those which contain the highest powers. In this case, it is more convenient to place the term containing the lowest power first, and that containing the next lowest next, and so on. This order of terms is called an arrangement according to the ascending powers of the letter; whereas that of the preceding article is called an arrangement according to the descending powers of the letter. 45. Corollary. Negative powers are considered to be lower than positive powers, or than the power zero, and the larger the absolute value of the exponent the lower the power. Thus a5x-6-a4x-3+ a3+a-1x+a-2x2, is arranged according to the ascending powers of x, and according to the descending powers of a. |