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 DIVISION OF SIMPLE EXPRESSIONS. 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. 84a5x3÷12a4x=7a5-4x3-1 =7ax2. Example 5. 77a2x3y1÷7ax3y=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 Chapter xxxi. 48. It is easy to prove that the rule of signs holds for division. Examples. (1) 6ab÷2a=36. (2) - 15xy ÷ 3x=- 5y. (3) - 21a2b3÷– 7a2b2=3b. (4) 45ab2x1÷- 9a3bx2= - 5a3bx2. DIVISION OF A COMPOUND EXPRESSION BY A SIMPLE EXPRESSION. 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+3z)÷-- 3=-3x+4y-z. (2) (36a3b2 - 24a2b5 – 20a4b2)÷-4a2b=9ab - 6b± − 5a2b. (3) (2x2 - 5xy+3⁄4x2y3) ÷ − {x=− 4x+10y − 3xy3. 31. 33. 34. 20. x3-3x2+x by x. 22. 10x7-8x6+3x1 by x3. 28. a3-a2b-a2b2 by a2. -3a2+2ab-6ac by -3a. 32. x2-3x3y by -3x3y2. - §x2+§xy+10x by −5x. -2a5x3+a1x1 by Za3x. DIVISION OF COMPOUND EXPRESSIONS. 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)x+11x+30 (x+5 5x+30 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+6a, 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. 17. 12a2-7ax-12x2 by 3a - 4x. 18. 15a2+17ax - 4x2 by 3a+4x. 19. 12a2-11ac -36c2 by 4a - 9c. 20. 9a2+6ac-35c2 by 3a+7c. 21. 60x2 - 4xy - 45y2 by 10x - 9y. 22. -4xy-15y2+96x2 by 12x - 5y. 23. 7x3+96x2- 28x by 7x - 2. 24. 100x3-3x-13x2 by 3+25x. 25. 27x3+9x2-3x-10 by 3x-2. 26. 16a3-46a2+39a-9 by 8a-3. 27. 15+3α-7a2-4a3 by 5-4a. 28. 16-96x+216x2 - 216x3+81x4 by 2-3x. 51. The process of Art. 50 is applicable to cases in which the divisor consists of more than two terms. Example. Divide 6x5 -- x1+4x3 – 5x2 - x - 15 by 2x2 −x +3. x2+4x3- 5x2- x −15 (3x3 + x2 - 2x – 5 6x53x4+9x3 2x45x35x2 2x4x3+ 3x2 - 4x3- 8x2 - x -4x3+ 2x2-6x - 10x2+5x-15 - 10x2+5x-15 52. Sometimes it will be found convenient to arrange the expressions in ascending powers of some common letter. Example. Divide 2a3+10-16a - 39a2+15a4 by 2-4a-5a2. 2-4a-5a2)10-16a - 39a2+2a3 + 15aa (5+2q − 3a2 |