(6.) The rule of Division may be separated into three cases. CASE I. When the dividend and divisor are monomials or simple quantities. RULE. Divide the coefficient of the dividend by that of the divisor, and to the quotient annex the quotient obtained by dividing the literal part of the dividend by that of the divisor; and observe that like signs give + and unlike signs - for the sign of the quotient. It is often most convenient to place the divisor under the dividend in the form of a fraction, and then to erase all the factors that are common to both terms of the fraction, and the result will be the sought quotient; observing the rule of signs. But perhaps the most expeditious method, in most cases, is to change the signs of the exponents of the factors of the divisor, and then to proceed as in Multiplication. Thus, to divide 45a3b3 3 by 9a2b3x; according to the first part of the rule, 45 divided by 9 gives 5 for the quotient, and a divided by a gives a for the quotient, b3 ÷ b3 = b2, x3÷x = x2, ... 5abc is the sought quotient; and it is positive, because the dividend and divisor are both positive. The same example, when done by the last part of the rule, gives 45a3b÷ 9a2b3x = 9 × 5a3b3x3 × 9-1a-b-3⁄4x-1 = 5 × 9ab5ab2, since 9° 1. = = As another example, we shall divide 29abe by 42x2yz, according to the second part of the rule. Here we have 29abc ÷ 42x+yzs 29abe 42x+yz' the sign being, because the dividend and divisor have unlike signs; and it is plain that the fraction can not be reduced to lower terms, since it has not explicitly any factors that are common to its numerator and denominator. a2x For another example, we shall divide ax by 4 2 ανα a 3 , by in 4 4 verting the fraction (according to what has been proved to be allowable), and for the letters proceeding according to the last part of the rule; by multiplying the coefficients, and adding the exponents of the same letter, we get the quotient When the divisor is a monomial, and the dividend a compound or polynomial quantity RULE. Divide each term of the dividend by the previous rule for monomials, then the sum of the quotients, when added according to their proper signs, will be the complete quotient required. Thus, to divide 6x-8xy by 2x2, we have 6x1 ÷ 2x2 = 3x2, 8xy 2x2-4y2, consequently there results 3 4y for the sought quotient. Also, to divide 48ay-60a'x'y by — 12a2x3, we proceed as follows, viz., 48a2xy - 60a1x3y 144axy+288ax 144ax3y3 + 288a3x - 60ax3y 144axy 24a, as required. Again, to divide 40abc- Sabg+ 12y by Sab, we have 40abe-Sabg+12y2 = 5c2 − g + Sab 3y2 as required. 2ab' EXAMPLES. 1. Divide 14x2yz — 49x3y2≈3 by — 7xyz, or execute what 121ab1o — 132a3b3y° + 11a1b3c 11ab 3. Execute 4325abg — 4*z°a*b*g* — 4oz°a3b3g3 4. Execute - 4825ab 6m3n3p* — 24m2g2r2z2 3m CASE III. When the divisor and dividend are compound quantities. RULE I. 1. Place the divisor at the left or right of the dividend, and draw a right or curved line between them, as in Arithmetic, being careful to arrange the divisor and dividend the same way, either according to the ascending or descending powers of a common letter. 2. By the rule for the division of monomials, divide the first or left-hand term of the dividend by the first or lefthand term of the divisor, and the result will be the first term of the quotient, placing the quotient to the right of the dividend, with a right or curved line between them, if the divisor is to the left of the dividend; but if the divisor is to the right of the dividend, the quotient may be put below the divisor, with a right-line between them. 3. Multiply the divisor by the first term of the quotient, and subtract the terms of the product from the corresponding terms of the dividend. Then to the remainder bring down as many of the next successive terms of the dividend as may be necessary to form a new dividend; divide the first term of the new dividend by the first term of the divisor, and the result will be the second term of the quotient; multiply the divisor by the second term of the quotient, and subtract the terms of the product from the corresponding terms of the first new dividend. 4. Then to the remainder bring down as many of the remaining next successive terms of the given dividend as may be necessary to form a second new dividend. 5. From the second new dividend we proceed in the same way as before, to get the third term of the quotient, and so on, until all the terms of the given dividend have been brought down. If the last remainder is 0, the division is exact; but if the remainder is different from 0, the division is not exact, and we signify that the remainder is to be divided by the divisor, by writing the divisor under the remainder, with a right-line between them, after the manner of expressing a fraction; so that when the division is inexact, the quotient will generally be of a mixed form, being partly integral and partly fractional. The reason of this rule is evident from the consideration that it is the reverse of the rule of Multiplication. And it is easy to see from what has been shown in Multiplication, if the first term of the arranged dividend can not be exactly divided by the first term of the arranged divisor, or if the last term of the arranged dividend can not be exactly divided by the last term of the arranged divisor, that the dividend can not be exactly divided by the divisor; observing that the divisor and dividend are supposed to be of integral forms. We will now illustrate our rule by a few examples. Ex. 1.-To divide 7x 9+5 3x2 by x-1. Arranging the dividend and divisor according to the descending powers of x, we get (by putting the divisor to the left of the dividend) the following process. 0+0=0= the last remainder. Explanation.-Having arranged the terms of the dividend and divisor, we divide the first term, 5a3, of the dividend by the first term, x, of the divisor, and get 5 for the result, which we put to the right of the dividend (with a curvedline between it and the dividend) for the first term of the quotient. Multiplying the divisor by 5a, we get 5a3- 5 for the product, which we put under 5a-3a2, the corresponding terms of the dividend, and then subtracting 523 5a from 5a-3a, we get 2x for the remainder, to which, bringing down the next successive term, +7x, of the given dividend, we get 2x+7a for the first new dividend. We now divide 22, the first term of the first new dividend, by the first term, x, of the divisor, and get + 2x for the result, which we put for the second term of the quotient. Multiplying the divisor by the second term, 2x, of the quotient, we get 2x22x, which, being written under the first new dividend and subtracted from it, gives 9x for the remainder, to which, bringing down the remaining term, 9, of the given dividend, we get 9x9 for the second new dividend. Dividing 9x, the first term of the second new dividend, by a, the first term of the divisor, we get 9 for the result, which is the third term of the quotient. Multiplying the divisor by 9, we get 9x9, which, being put under the second new dividend and subtracted, gives 0 for the remainder; consequently, since all the terms of the given dividend have been brought down, and since the last remainder equals 0, the division is exact. We shall now put the divisor to the right of the dividend, and divide the last term of the dividend by the last term of the divisor, and so on for the successive dividends; that is, we shall proceed with the division from right to left, as in the following scheme: 5203 3x2+7x 9x 1= the divisor. 9x 9 2x + 2x2 2x -- 5x2 -- 5x+2x+9= the quotient. 0 + 0 = naught = the last remainder. We see that the results are the same as by the preceding |