But in cases like this last, where the two numerators, or the denominators, have factors common to both, the conclusion will become more neat by first casting off such common factors. Thus casting away ab out of the two numerators, 6a and x out of both the denominators, we have to be 56 5 divided by -; whereof the quotient is 18a : in the 3 256 When either the divisor or the dividend is a whole quantity (instead of a fraction) it may be reduced to the form of a fraction, by writing an unit or 1 under it. since, by casting off the factors common to the dividend and divisor (as directed in the rule) it is plain that we take like parts of those quantities: therefore the quotient arising by dividing the one part by the other, will be the same as that arising by dividing one whole by the other. As to rule 3°, wherein it is asserted that A C AD = it is evident that AD and BC are equimultiples of the AD vided by BC; which, by notation, is BC as was to be shown. The grounds of the note subjoined to this rule are these: by casting away all factors common 4° Surd quantities under the same radical sign, are divided by one another like rational quantities, only the quotient must stand under the given radical sign. Thus, the quotient of✔ab by ✔b is ✔a: 5°. Different powers, or roots of the same quantity are divided one by another, by subtracting the exponent of the divisor from that of the dividend, and placing the remainder as an exponent to the quantity given. But it must be observed, that the exponents here understood are those defined in p. 5; where all roots are represented as fractional powers. It will likewise be proper to remark further, that, when the exponent of the divisor is greater than that of the dividend, the quotient will have a negative exponent, or, which comes to the same thing, the result will be a fraction, whereof the numerator is an unit, and the denominator the same quantity with its exponent changed to an affirmative one. Thus a divided by a gives a3: And a + divided by a + z3 gives a + z to the two numerators we take equal parts of the quantities; and by throwing off the factors common to both denominators, we take equimultiples of those parts. The two preceding rules, being nothing more than the converse of the 4th and 5th rules in multiplication, are demonstrated in them: though perhaps the case, in rule 5, where the exponent comes out negative, may stand in need of a more particular explanation. Accord 6°. A compound quantity is divided by a simple one, by dividing every term thereof by the given divisor. Thus, sab) Sabc+12abx-9aab (c + 4x-3a: also,-5ac) 15a2bc-12acy2+5ad (—sab+ and so of others. 12yy 5 dd : 7°. But if the divisor, as well as the dividend, be a compound quantity, let the terms of both quantities be disposed in order, according to the dimensions of some letter in them, as shall be judged most expedient, so that those terms may stand first wherein the highest power of that letter is involved, and those next where the next highest power is involved, and so on this being done, seek how many times the first term of the divisor is contained in the first term of the dividend, which, when found, place in the quotient, (as in division in vulgar arithmetic) and then multiply the whole divisor thereby, subtracting the product from the respective terms of the dividend; to the remainder, bring down, with their proper signs, as many of the next following terms of the dividend as are requisite for the next operation; seeking again how often the first term of the divisor is contained in the first term of the remainder, which also write down in your quotient, and proceed as before, repeating the operation till all the terms of the dividend are exhausted, and you have nothing remaining. ing to the said rule, the quotient of x3 divided by xs was asserted to be x2, or Now that this is the true 1 value is evident; because 1 and a being like parts of a3 and 25 (which arise by dividing by x3) their quotient will consequently be the same with that of the quantities themselves. F Thus, if it were required to divide a3 + 5a3x + 5ax2 +x3 by a + x (where the several terms are disposed according to the dimensions of the letter a) I first write down the divisor and dividend, in the manner below, with a crooked line between them, as in the division of whole numbers; then I say, how often is a contained in a3, or what is the quotient of a3 by a; the answer is a2, which I write down in the quotient, and multiply the whole divisor, a+x, thereby, and there arises as+ax; which subtracted from the two first terms of the dividend, leaves 4a2x; to this remainder I bring down +5ax, the next term of the dividend, and then seek again how many times a is contained in 4ax; the answer is 4ax, which I also put down in the quotient, and by it multiply the whole divisor, and there arises 4a2x+4ax2, which subtracted from 4a2x +5ax, leaves ax, to which I bring down a3, the last term of the dividend, and seek how many times a is contained in ax2, which I find to be a2; this I therefore also write down in the quotient, and by it multiply the whole divisor; and then, having subtracted the product from ax2+x3, find there is nothing remains; whence I conclude, that the required quotient is truly expressed by a2+4ax+x2. See the operation. a+x) a3+5a2x+5ax2+x3 ( a2+4ax+x2 4a2x+5αx2 4a2x+4αx2 ax2+x3 0 0 In the same manner, if it be proposed to divide a3— 5ax+10a3x2-10a2x2 + 5ax1a—x3 by a2—2ax + x2, the quotient will come out a3-Sa2x+зax2-x, as will appear from the process. a2—2áx+x2 ) a3-5a*x+10a3x3-10a2x2+5ax^—x3 ( a3 as_2a+x+ So likewise, if ass be divided by a-x, the quotient will be a+a3x+a2x2+ax3+4; as by the work will appear. a—x ) a3 —x3 ( aa+a3x+a2x2+ax2+xa a5-a1x Moreover, if it were required to divide a-sa1x2+ 3a2xx by a3-3a2x+Sax2-3, the process will stand thus: a3-Ba2x+\ao-3a1x2+3a2x2x2 ('a3+Sa2x+Sax2+203 sax2-x3 a—3a3x +3aa1x2—a3x3 +Sax-6a4x2+ a3x2+3a2x +Sa3x-9a4x2+9a3x3-3a2x4 |