CASE I. A Monomial by a Monomial. 80.-Ex. 1. Divide 18ab by 6a. 18ab÷6a=3b SOLUTION. Dividing the dividend 18ab by 6 and a, the factors of the divisor, by rejecting those factors from the dividend, we have the quotient 36; for 36 is such a quantity as, multiplied by the divisor 6a, will give the dividend 18ab. 2. Divide 8ab by 4a2. SOLUTION. Dividing the dividend 8a5b by 4 and a2, factors of the divisor, by rejecting those factors from the dividend, we have the quotient 2a3b. 3. Divide -6a2b by 2a. 4. Divide 12a5b3 by −3a2b. 8a3b÷4a2 = 2a3b Ans. - Sab. Ans. - 4a3b2. 81. Rule for Dividing one Monomial by another.—Divide the numerical coefficient of the dividend by the coefficient of the divisor; annex to this quotient the letters of the dividend, giving each an exponent equal to its exponent in the dividend less the exponent in the divisor, and make the result positive when the dividend and divisor have like signs, and negative when they have unlike signs. When the dividend and divisor contain an equal literal factor, it is cancelled, and does not appear in the quotient. 1. Divide 12abc by 3c. PROBLEMS. 2. Divide - 12abc by -4ab. Ans. 4ab. 17. Divide 8am-1b"c3 by -2am+1b3on. Ans. -4a2n-2,3-n CASE II. A Polynomial by a Monomial. 82.-Ex. 1. Divide 6a2b2 - 4a2b+2a2c by 2a. SOLUTION. Dividing the first term of the dividend by 2a, we have 362; dividing the 2a)6ab2-4a2b+2a2c 3b2-2ab+ac second term by 2a, we have -2ab; and dividing the third term by 2a, we have ac. Uniting these results by their proper signs, we have, as the entire quotient, 362-2ab+ac. 2. Divide a2+ac by a. 3. Divide 6a - 3a2x2 by 3a2. Ans. a+c. Ans. 2a2-ax2. 83. Rule for Dividing a Polynomial by a Monomial.-Divide each term of the dividend by the divisor, and connect the several results. 7. Divide 15a2bc – 10a3b*c3y2+5a2b3ď2 by — 5abc. 8. Divide 4a2b - 6ab2 by - 2ab. 84.-Ex. 1. Divide x3 + 2x2 - 3x by x2+3x. x2+3x)x3 + 2x2 — 3x( x − 1 SOLUTION. The divisor and dividend for convenience are arranged according to the powers of x, beginning with the highest power. Since 23, the first term of the dividend, must equal the product of x2, the first term of the divisor, by the first term of the quotient, we divide x3 by x2, and have x for the first term of the quotient. x times the whole divisor is x3+3x2, which subtracted from the whole dividend leaves x2 3x. Since -x2, the first term of this new dividend, must equal the product of x2, the first term of the divisor, by the second term of the quotient, we divide -2 by x2, and have -1 for the second term of the quotient. - 1 time the whole divisor is - x2-3x, which subtracted from the last dividend leaves no remainder. Hence, the quotient is x-1. 2. Divide a2+2ab+b2 by a+b. 3. Divide a2 - 2ab+b2 by a − b. Ans. a+b. Ans. a-b. 85. Rule for Dividing a Polynomial by a Polynomial.-Arrange the divisor and dividend according to the powers of one of their letters. Divide the first term of the dividend by the first term of the divisor, and write the result for the first term of the quotient; multiply the whole divisor by it, and subtract the product from the dividend. Consider the remainder as a new dividend, find the second term of the quotient in like manner as before, and thus continue till the first term of the divisor is not contained in the dividend. If there be at last a remainder, write it, with the divisor under it, as a fractional part of the quotient. 14. Divide 7x3- 24x2+58x - 21 by 7x-3. Ans. a - 2x. Ans. x-8. Ans. x2-3x+7. Ans. x3-x2y+xy3 — y3 + 2y+ x + y Zero and Negative Exponents, and Reciprocals. 86. The Reciprocal of a quantity is the quotient arising from the division of 1 by that quantity. 87. A Zero Exponent, or o, is used to preserve the trace of a letter, which might otherwise disappear in the process of division. Thus, the quotient of x3y2 divided by a3y may be written xy. |