63a0 — 33a1b + 77a3b2 — 69a2b3 + 19ab1 — 116 | ± 9a2 — 3ab — 1162 3ab11b5 = 3d new dividend. 9a2b3 + 3ab1 + 116° and the division is exact. 0 + 0 + 0 = the last remainder. Ex. 4. To divide - 72+ 21x426x3-x-7x+20 by -x+3x-4. Changing the signs of the terms of the divisor, etc., it becomes a 3x + 4, then observing that for we must use in finding the terms of the quotient, and when the divisor is multiplied by any quotient-term; and placing the divisor to the right of the dividend, we have the following: x2 7x+201 = x2 3x + 4 ·Ta +21x4 - 26x3 7x21x+28x3 7x3-2x-5 7x = 1st new dividend. 15x+20=2d new dividend. 5x2 + 15x 0÷0 6 3. Divide 3a - 13a2b2 + 146* by 3a2 — 7b2. 6. Divide a + 4a3b + 6a2b2 + 4ab3 + b1 by a2 + 2ab + b2. Ans. a2+2ab+b2. 9. Divide a + a2b2 + b1 by a2 — ab + b2. Ans. a2 + ab + b2. 10. Divide 12a1. 21a3b — 26a2b2 + 30ab3 + 25b1 by 4a2 · 3ab5b2. Ans. 3a2 3ab -563. (7.) It is clear from what was shown in the preliminary remarks on Division, that if we divide any divisor and its corresponding dividend by the same quantity, and then use the quotients thus obtained for a new divisor and dividend, the same quotient will result from the new divisor and dividend as from the given divisor and dividend. Hence, having arranged the divisor and dividend as in Rule I., if we divide each term of the divisor and dividend by the first term of the divisor, we may take the results thus obtained for the new divisor and dividend; and it is easy to see that the first term of the new divisor is always 1; and the first term of the new dividend will be the first term of the quotient. We may divide the new dividend by the new divisor, either by Rule I. or Rule II.; but it will be better to proceed as in the following rule, which is a modification of Rule II. RULE III. 1. Having arranged the terms of the divisor and dividend, as in Rule I., divide each term of the divisor and dividend by the first term of the divisor, and cali the results the new divisor and dividend. 2. Change the signs of all the terms of the new divisor, except that of the first term, which is 1, and call the result the changed divisor. 3. Multiply all the terms of the changed divisor except the first term, by the first term of the new dividend; and put the products successively under the corresponding terms of the new dividend; then the sum of the results in the column that immediately follows the first term of the new dividend will be the second term of the quotient. 4. Multiply the same terms of the changed divisor (as before) by the second term of the quotient, and put the products successively under the corresponding terms of the new dividend and the preceding products; then the sum of the results in the column that immediately follows the second term of the quotient will be the third term of the quotient. 5. Then proceed in the same way as before with the third term of the quotient and the same terms of the changed divisor to get the fourth term of the quotient, and so on. We will now give some examples for the purpose of illustrating the rule. Ex. 1. To divide 2+ 2x + 1 by x + 1. Here, supposing the divisor and dividend to be arranged according to the descending powers of a, we must divide. the divisor and dividend by x, or which is the same, we must multiply each of their terms by x-1, and we shall have (x+1)x-1=1+1= the new divisor, and (a2 + 2x + 1) x¬1 = x + 2 + x-1= the new dividend; and changing the sign of the second term in the divisor, we have 1-1 for the changed divisor. Hence, by the rule, we have the following process. Ex. 2. To divide - 7+ 21a - 26x3- x2 - 7x + 20 by x2 + 3x - 4. Here, in order to divide the dividend and divisor by — x2, the first term of the divisor, we may multiply each of their terms by x-2, and we get 73 — 21x2 + 26x + 1 + 7x−1 202 the new dividend, and 1-3x-1+4x-2= the new divisor, and changing the signs of all the terms of the new divisor, except that of 1, the first term, we have 1 + 3x-1. 42 the changed divisor. Hence, by the rule, we must proceed as follows. 1+3x-1 · 4x-2|7x3-21x2 + 26x + 1 + 7x−1 — 20x-2 212-28x 0+ 0 - 15x-1+20x-2 7x3 +0-2x-5=7x3- 2x-5 the quotient, as required; which agrees with the answer to the same question as found in Ex. 4 under Rule II. Explanation.-If we divide the first term, 7, of the new dividend by 1, the first term of the changed divisor, we shall have, as in Rule II., 73 for the first term of the quotient, which, multiplied by all the terms of the changed divisor, except the first term 1, gives the product (3-1 — 4x ̄2) × 7x=212-28, which is placed under the corresponding terms of the new dividend. Then, by adding - 21a and 212, which are the terms in the column that immediately follows the first term of the quotient, we get 0 for the sum; consequently, the second term of the quotient = 0, which, multiplied by the same terms of the changed divisor, gives 0 for each product, which we put down in the scheme merely for the sake of uniformity. To get the third term of the quotient, we add the quan tities which stand in the third vertical column, and get 26x -28x=2x, the third term of the quotient, since -2x, divided by 1, the first term of the divisor, is 2x; and the product of 2 and the same terms of the changed divisor as before is (3x-1-4x-2) × 2x=6+ 8x-1, which is put under the corresponding terms of the new dividend, then the sum of the quantities in the fourth vertical column is 165 the fourth term of the quotient. And, as before, we get (3x-1-4x-2) × 515x-1 + 20x-2, which, being put under the corresponding terms of the new dividend and the preceding products, the sums of the quantities in each of the vertical columns destroy each other, and the new dividend is exhausted; and of course the division is exact, and the sought quotient is 7-2x-5. It is easy to see that it is useless to divide by 1, the first term of the changed divisor, since any quantity is the same after it is divided by 1 as before; consequently, the quotientterms will be found as stated in the rule in all cases. It is also easy to see that Rule III. is substantially the same as Rule II., when applied to the new divisor and dividend, when we reject the first term, 1, of the changed divisor, which is clearly allowable. Ex. 3.-To divide 15a6ab17a2b2 + 8ab3 — 461 by 3a2 462. The terms of the dividend and divisor are divided by the first term, 3a2, of the divisor, by multiplying them by 3-1a-2 and we shall have 5a2 - 2ab — 17 × 3−12 + 8 × 3−1a-1b3 — 4 x 3-1a-2 and 1 - 4 × 3-1a-273 for the new dividend and divisor; and changing the sign of the second term of the new divisor, we have 1 + 4 x 3-1a-262 for the changed divisor. Hence, by the rule, we must proceed as follows. 1+4x 3-15a2-2ab-17x3-172+8x3-1a-163-4 x 3-1a-2b4 20 x 3-172-8×3-1a-13+4x3-1a-2b+ a-22 5a2-2ab+b2 the exact quotient. For exercise, we shall refer to the examples under Rules I. and II. (8.) DIVISION BY DETACHED COEFFICIENTS. If we arrange the divisor and dividend in the same way, according to the successive powers of some letter (or letters) |