CHAPTER V. DIVISION. 32. To divide one Algebraical Quantity by another is to find how many times the second is contained in the first, then to add the result if the divisor be positive, and to subtract it if the divisor be negative. We shall now consider four different cases of Division. 1o. To divide 12a by 4 is to find how often 4 is contained in 12a, and add the result. 2o. To divide - 12a by 4. ... 12a+4 = 3a Now 4 is contained in - 12a, 3a times, for 3a repeated 4 times gives - 12a, and 4 repeated -3a times must be the same thing. ... - 12a÷4 = -3a By similar reasoning it will appear that 12a-4 = -3a -12a-4 = +3α 33. Hence the rule of signs in Division may be thus stated: : If the dividend and the divisor have the same sign, the quotient is positive, if they have different signs the quotient is negative. Ex. 1. Divide 14abc by 7a. Since 7a x2bc = 14abc, it is evident that 14abc÷7a = 2bc. Ex. 2. Divide 9a6 by -9a2. Since 9a2x - a4 = 9a6, we have 9a6-9a2 = — a1. 33. From the last two examples we may deduce the following Rule for the Division of two simple Algebraical Quantities. (a) RULE FOR THE SIGN OF THE QUOTIENT. When the dividend and divisor have each the same sign, the sign of the Quotient will be positive, but when the dividend and divisor have different signs, the sign of the Quotient will be negative. (b) RULE FOR THE COEFFICIENT OF THE QUOTIENT. The Coefficient of the Quotient is found by dividing the Coefficient of the dividend by the Coefficient of the divisor. (c) RULE FOR THE INDEX. The Index of the Quotient is found by subtracting the Index of the divisor from the Index of the dividend. Examples VIII. (1.) Divide 12a2 by 2a; -40a3b2 by -ab; 18ab by 3b; -15b2 by b; 25a3 by -5a2; a3b2c by ab2; -10x3 by -2x2; 100x1y by xy3; 49a5 by -7a2; 120abcxyz by -12xyz; 3a2b3c4 by 3ab3c1. Note. If any of the coefficients be fractional the rules for division of fractions must be applied. (2.) Divide 2x2 by 3x The coefficient of the quotient will be 23 = 2x3 = 3 .*. 2x2÷3x = 3x (3.) Find the following quotients : 10xy÷x; 2x2÷3⁄4x2; 17a2b2÷-fa; fxyz÷2xz; }x1÷Jx2; }a2y2 ÷3y2; ab÷4b; 3b2÷}b; −x1÷5x2; − a2x2y2z2÷2az; - Hab2 ÷b; -13c2d2÷-20cd; -8x2÷-12x; −40a2p2q2÷- Faq2; -abcpqr-acq (4.) Divide a2x2+3abx2 - a3x by ax The division in this case is effected by dividing each of the terms of the dividend by the divisor; the sum of all the separate quotients will be the complete quotient required. Divide 6a2+662 by 2; 7a2-14ab by 7a; x2yz - xy2z+xyz2 by xyz; -4a3+8a2b-4a2c by -4a2; 3a3-a2b by 2a2; 3x2-6x3y+3x2y m2n mn2 - 6xy3 by 12x; m3 + + by 2m; 100m2-10m+1 by 10; 3a1-3a3b+b by b2; 4x3yz - {xy3z+‡xyz3 by (5.) Divide a2-b2 by a-b 2 3 p3+2p2q by p; a++3a3x+a2x2 by 2a2; xyz; 9x3yz − x3yz+¿xyz3 by jxz In this case the process of division is more complicated, and the method to be pursued is similar to that of Long Division in Arithmetic, as will be seen in the following working : a-b) a2-b2 (a+b a2-ab ab-ba * See p. 28. Note.-The rules to be observed in the division of quantities consisting of more than one term are 1o. To arrange both the divisor and dividend according to the powers of the same letter (a in the example); then to divide the first term of the dividend by the first term of the divisor, place the result in the quotient and multiply the divisor by it; subtract and proceed similarly with the next line, and so on to the end. (6.) Divide a3-6a2+12a-8 by a- -2 a-2) a3-6a2+12a-8 (a2-4a+4 a3-2a2 -4a2+12a 4a-8 4a-8 (7.) Divide 4x-x2+x1-4 by x2-2+x Here the dividend and divisor must be arranged according to powers of x, as follows: x2+x-2)x-x2+4x-4(x2−x+2 (8.) Divide x2+x−2 by x+2; and divide 3x3-3x2-x-1 by 3x2-1 (9.) Divide 4a3. -a by 2a-1; x2-16 by x-4 (10.) Divide x2-1 by x-1; x3+8 by x+2 (11.) Divide 3x1 — 6x3 – 2x2+5x-12 by 3x2-3x+4 (12.) Divide 14x1 — 65x3+141x2-116x+44 by 2x2-7x+11 (13.) Divide a3- 8b3 by a2+2ab+4b2 (14.) Divide a3+32 by a1 − 2a3+4a2 - 8a+ 16 (15.) Divide a*x*—y1 by ax-y; 100a2-b2 by 10a+b (16.) Divide x+x2y2+y* by x2− xy+y2 (17.) Divide 3x2y2z2+14xyz − 5 by 3xyz-1 (18.) Divide 6a2b2 — abcd - 12c2d2 by 3ab+4cd (19.) Divide 8a3-27 by 9-12a+4a2 (20.) Divide a2+2ab+b2 - c2 'by a+b+c; by a+b-c; by b+c-a; and by a+c-b (21.) Divide a3— a2b— a2c+ab2+ ac2 — b3 — bc2 — b2c-c3 by a2+b2+c2 (22.) Divide 8a-16a+12a2 by 4a2 (23.) Divide 6x2yz — 12xy2z+18xyz2 by 6xyz (24.) Divide a-poq° by a3+a1pq+a3p2q2+a2p3q3+ap*q‘+p3q3 с (25.) Divide 3p-4p*q+3p2r-7pq-7pqr+12qr3+12r+12pr2 by The division in this case does not terminate; the quotient is x2-x+1; and there is a remainder -2. The proper way to write the answer is Here it is plain that the division will never terminate. In fact the quotient is an endless series or terms, four of which have been found; it is easily seen that the next 4 terms are 2ax8, 2x10, 2ax12, 2ax14. When the division is carried on to 4 terms only, the result is (29.) Divide 3 by 1-x; 6+a by 1-a, both to 4 terms. (30.) Divide a2x2 by a-ax to 6 terms. (31.) Divide 1+x2 by 1−x+x2 to 4 terms. (32.) Find the quotient to 4 terms of each of the following divisions : 1+x3÷1−x3; x7 ÷x-1; ao÷a-3; 1÷1+a; (x+a)*÷x-a; (x − x)* ÷x−u; (1+x)3÷(1−x)2; x6+x3+1÷x2+x+1. (34.) Divide a1+}a3b−7 a2b2+}ab3-3 b1 by ja2+}ab-b2 } a2+}ab-b2) } a*+}a3b− Za2b2+}ab3 — } b1 (a2+} b2 In Ex. 33 and 34, it will be observed that the fractional coefficients must be added, subtracted, multiplied and divided by the Rules for vulgar fractions. In other respects the working is the same as for whole numbers. (35.) Divide x2−1 by x−1; x2-} by x+}; a2−ža+f by a−}; a2+ax m by fa; 3m3-m2+ by 2m; a1-1 by a2+1; ja3+b3 by a2-ab+b2. 3 (36.) Find the following quotients :— 3a3-3 by ja2+}a+}; {x2−}xy+{y2÷}x−}y; (37.) Divide 1 by x++; x+§ by i̟+x; 1+22 by 1+1+x by 1-2 (39.) Divide 23− (a+b+c) x2+(ab+ac+bc)x− abc by x-c -x cx2 23. |