240. It will be proper to illustrate this example by a numerical application. Let us make a 3 and b = 2, we shall have a+b=5 and a-b=1; further, a a = 9, a b = 6, b b = 4. Therefore a a+ab+bb = 19, and a a-ab+bb 7. So that the product required is that of 5 x 19 x 1 x 7, which is 665. Now a 729, and b = 64, consequently the product required is ab6 665, as we have already seen. = = CHAPTER IV. Of the Divison of Compound Quantities. 141. WHEN we wish simply to represent division, we make use of the usual mark of fractions, which is, to write the denominator under the numerator, separating them by a line; or to inclose each quantity between a parenthesis, placing two points between the divisor and dividend. If it were required, for example to divide a + b by c+d, we should represent the quoa+b tient thus c+d' according to the former method; and thus, (a+b): (c+d) according to the latter. Each expression is read ab divided by c+d. 242. When it is required to divide a compound quantity by u simple one, we divide each term separately. For example; 6a-8b4c, divided by 2, gives 3 a4 b+2c; and (a a-2 ab): (a) = a—2b. In the same manner - 2ab+3ab; (a3 — 2a a b + 3 a ab) : (a) = a a — 2 a b + 3 a b ; (4aab-6aac+8abc): (2a)=2ab-3 ac+4bc; (9aabc-12abbc + 15 abec): (3abc) = 3a-4b+5 c, &c. 243. If it should happen that a term of the dividend is not divisible by the divisor, the quotent is represented by a fraction, as in the division of a +b by a, which gives 1 + b b b b a (aaab+bb): (a a) = 1 — + -· a a a Likewise, For the same reason, if we divide 2 a+b by 2, we obtain and here it may be remarked, that we may write b, 244. But when the divisor is itself a compound quantity, division becomes more difficult. Sometimes it occurs where we least expect it; but when it cannot be performed, we must content ourselves with representing the quotient by a fraction, in the manner that we have already described. Let us begin by considering some cases, in which actual division succeeds. 245. Suppose it were required to divide the dividend a c-bc by the divisor a-b, the quotient must then be such as, when multiplied by the divisor ab, will produce the dividend a c-b c. Now it is evident, that this quotient must include c, since without it we could not obtain a c. In order, therefore, to try whether c is the whole quotient, we have only to multiply it by the divisor, and see if that multiplication produces the whole dividend, or only part of it. In the present case, if we multiply a-b by c, we have a c-bc, which is exactly the dividend ; so that c is the whole quotient. It is no less evident, that (a a + a b) : (a + b) = a; (saa— 2 a b) : (3 a — 2 b) = a ; a − s b) = 3 a, &c. (6 aa—9 ab): (2 246. We cannot fail, in this way, to find a part of the quotient ; if, therefore, what we have found, when multiplied by the divisor, does not yet exhaust the dividend, we have only to divide the remainder again by the divisor, in order to obtain a second part of the quotient; and to continue the same method, until we have found the whole quotient. Let us, as an example, divide a a+ 3 a b + 2 bb by a+b; it is evident, in the first place, that the quotient will include the term › a, since otherwise we should not obtain a a. Now, from the multiplication of the divisor a+b by a, arises a a+ab; which quantity being subtracted from the dividend, leaves a remainder 2ab+2bb. This remainder must also be divided by a+b; and it is evident that the quotient of this division must contain the term 2 b. Now 2 b, multiplied by a + b, produces exactly 2ab+ 2bb; consequently a +2b is the quotient required; which, mul tiplied by the divisor a+b, ought to produce the dividend aa + sab + 2b b. See the whole operation : a+b)a a + s ab + 2 b b (a + 2 b aa+ab 2ab+2bb 0. 247. This operation will be facilitated if we choose one of the terms of the divisor to be written first, and then, in arranging the terms of the dividend, begin with the highest powers of that first term of the divisor. This term in the preceding example was a ; the following examples will render the operation more clear. a—b) a3 - 3 aab + 3 a b b — b3 (a a — 2 a b + b b a3-a ab -2aab Sabb — 2a ab+2abb abbb3 aa—2ab+bb) a* — 4 a 3 b + 6 a a b b — 4 a b3 + ba aa-2ab+bb) a* — 2 a3 b + aabb -- -2a3b+5aabb-4ab3 aabb-2 ab3 +b2 aabb-2ab3 +64 0. aa-2ab+4bb) a* +4 a abb + 16 b*( aa+2ab+4bb at — 2 a3 b + 4aabb 2 a3 b + 16 b 2a-4 a abb + 8 abs 4 aabb-8ab3 +16b 4aabb-8ab3 +16 b* 0: Eul. Alg. 10 aa2ab2b b) a* + 4 b* (aa+2ab+2bb a^2 a3 b+2aabb 2a3b-2aabb4b4 2a3b-4aabb +4 abs 2aabb-4ab34b4 2aabb-4abs + 4b4 0. 1-2x+xx) 1-5 x + 10 x x 10 x3+5x4x5 1-3x+3xx-x3) 1-2x+xx Of the Resolution of Fractions into infinite series. 248. WHEN the dividend is not divisible by the divisor, the quotient is expressed, as we have already observed, by a frac tion. Thus, if we have to divide 1 by 1 a, we obtain the fraction 1 This, however, does not prevent us from attempting the a division, according to the rules that have been given, and continuing it as far as we please. We shall not fail to find the true quotient, though under different forms. |