CHAPTER VI. DIVISION. 46. THE object of division is to find out the quantity, called the quotient, by which the divisor must be multiplied so as to produce the dividend. or Division is thus the inverse of multiplication. The above statement may be briefly written quotient x divisor dividend, dividend÷divisor=quotient. It is sometimes better to express this last result as a fraction; thus Example 1. Since the product of 4 and x is 4x, it follows that when 4x is divided by x the quotient is 4, the divisor and dividend the factors common to both, just as in Arithmetic. 47. RULE. To divide one simple expression by another, divide the coefficient of the dividend by that of the divisor, and subtract the index of any letter in the divisor from the index of that letter in the dividend. Example 4. 84a5x312a4x=7a5-4x3-1 =7ax2. Example 5. 77a2x3y1÷7ax2y=11axy3. NOTE. If we apply the rule to divide any power of a letter by the same power of the letter we are led to a curious conclusion. This result will appear somewhat strange to the beginner, but its full significance will be explained in chap. xxx, 48. It is easy to prove that the rule of signs holds for division. ab ax b =b. - aba b. ab÷ -α= -ax-b - b. - ab÷ -α= 49. RULE. To divide a compound expression by a single factor, divide each term separately by that factor. This follows at once from Art. 34. Examples. (1) (9x-12y+32)÷-3=-3x+4y-z. (2) (36a3b2-24a2b5 – 20a4b2)÷4a2b-9ab - 6b4 - 5a2b. (3) (2x2 − 5xy+2x2y3) ÷ − 1x=− 4x+10y – 3xy3. 31. 33. 34. 35. 18. -50y3x3 by -5x3y. 20. x3-3x2+x by x. 28. a3-a2b-a2b2 by a2. -3a2+fab-6ac by -a. 32. 2-3x3y by -32. §x2+xy+x by -fx. -2a5x3+3a2x2 by Ja3x. a2x-abx-facx by jax. 50. To divide one compound expression by another. RULE. 1. Arrange divisor and dividend in ascending or descending powers of some common letter. 2. Divide the term on the left of the dividend by the term on the left of the divisor, and put the result in the quotient. 3. Multiply the WHOLE divisor by this quotient, and put the product under the dividend. 4. Subtract and bring down from the dividend as many terms as may be necessary. Repeat these operations till all the terms from the dividend are brought down. Example 1. Divide x2+11x+30 by x+6. x+6)x2+11x+30 ( divide x2, the first term of the dividend, by x, the first term of the divisor; the quotient is x. Multiply the whole divisor by x, and put the product x2+6x under the dividend. We then have On repeating the process above explained we find that the next term in the quotient is +5. The entire operation is more compactly written as follows: x+6)x2+11x+30 (x+5 The reason for the rule is this: the dividend may be divided into as many parts as may be convenient, and the complete quotient is found by taking the sum of all the partial quotients. Thus x2+11x+30 is divided by the above process into two parts, namely x2+6x, and 5x+30, and each of these is divided by x+6; thus we obtain the complete quotient x+5. Example 2. Divide 24x2 - 65xy+21y2 by 8x-3y. 51. In the examples given hitherto the divisor has been exactly contained in the dividend. It is a good exercise for the beginner to work cases in which the accurate value of the remainder is required. Example. Find the remainder when x4-7x3 +9x2 - 11x+15 is divided by x-5. x-5)x1-7x2+9x2 - 11x+15 (x3 – 2x2 – x − 16 x4-5x3 - 2x3+9x2 |