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 amb-P-6 ab-2ce, -3a-3, 5 am b¬r+11 a b−2 co, a ̃3 — 14 am b ̃o. Ans. 5 a-3+5ab-2ce. 2. Reduce the polynomial 9 a-3b-2c4-7 ba−3+ (18a-3b-5 an bmc* -3.25) (3 a" bm-a-3b-2c4 +3c-5.25) to its simplest form. Ans. 10a-3b-2c4+11a-3b-8 an bm-2 c*+2.25. 3. Multiply a-m by an. 4. Multiply am by a-". 5. Multiply am by a ̄". Ans. am-n Ans. a-m-n =a−(m+n). 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 8. Find the continued product of 5 a3 b-4, 10 a2 65 c, and 3 а7. Ans. 150 a12 b c. 9. Multiply 13 a-1bc-3 by -4a-3b-6 c2. 13. Divide 14 a-8bc3 d-1e by 2 a b-3 c5 d h. Ans. 7a-9b4c-2d-2eh-1, Ans.-a-b-1 c. 14. Divide -3 am by 2 am+" b c1. Division of Polynomials. 42. Problem. To divide one polynomial by another. 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. 1. Divide 43. EXAMPLES. 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. = 64x6-16 a3 x3+a6 4x2-4 a x + a2 Divisor. 64x6—64 a x5+16 a2 x1 | 16 x1+16 a x3+ 12 a2 x2+4 a3 x+a1 64 a x5 16 a2 x4 16 a3 x3 + a6 = 1st Remainder. a2x2 64 a x5-64 a2 x4 + 16 a3 x3 48 a2 x4 32 a3 x3+a6 =2d Remainder. 48 a2 x4. 16 a3 x3. 16 a3 x3 4 a4 x2 4 a4 x2 0. Ans. 16 x1 + 16 a x3 + 12 a2x2 +4 a3 x + a2 2. Divide b c3-c3 x by c3. 3. Divide a2+2ab+b2 by a+b. 4. Divide a8b4+15 all b5. · 48 a14 b6. 10 a9 b2-aồ b. Ans. b Ans. a+b. 20 a17 b7 by 5. Divide 118 22+81 24 by 1+6z+9z2. 2 a8 b5. Ans. 1-6x+9 z2. 6. Divide 81 a + 16 b12 — 72 a4 b6 by 9 a1+ 12 a2 b3 +466. Ans. 9 a4-12 a2 b3 +4 b6. a2 Ans. x3m+3x2m yn +3 xm y2n +y3n. 8. Divide 1+ a3 n3 by - 1+an. Ans. 1+an+a2 no. 9. Divide 2 a4-13 a3b+ 31 a2 b2-38 a b3 + 24 b4 Ans. a1 + a3 b + 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 targer the absolute value of the exponent the lower the power. Thus a5 x-6— a1x-3+ a3 + a ̄1 x + a−2x2, is arranged according to the ascending powers of x, and according to the descending powers of a. Division of Polynomials. 46. EXAMPLES. 1. Divide a1 + a2-a-2-a-4 by a2-a-2. Ans. a2+1+a-2. 2. Divide 4 a4 b-6+12 a3b-59a2b-4b-2+2a-2 -a-4b2 by 2 a2b-3+3 ab-2-b-1+ a2 b. Ans. 2 a2b-3+3ab2+b-1a-2b. 47. In the course of algebraic investigations, it is often convenient to separate a quantity into its factors. This is done, when one of the factors is known, by dividing by the known factor, and the quotient is the other factor. And when a letter occurs as a factor of all the terms of a quantity, it is a factor of the quantity, and may be taken out as a factor, with an exponent equal to the lowest exponent which it has in any term, and indeed by means of negative exponents any monomial may be taken out as a factor of a quantity. 48. EXAMPLES. 1. Take out 3 a2 b as a factor of 15 a5 b2 + 6 a3 b + 9a2 b2+3 a2 b. Ans. 3 a2b (5a3b+2a+3b+1). 2. Take out am as a factor of 3 am +1+2 am. Ans. am (3a+2). 3. Take out 2a3 b5 c as a factor of 6 a6 b7 c2+6 a b3 c -2ab+2 a2 c. Ans. 2 a3 b5 c (3 a3 b2 c +3 a−2 b3 —a-2b-4c-1+ -1-2-1a-1b-5. a-3b -5 c 4. Take out b as a factor of an-1b-bn. Ans. b (a-1-bn-1). |
