It remains now to multiply the first product I. II. by this second product III. IV. :

238. Let us resume the same example, but change the order of it, first multiplying the factors I. and III. and then II. and IV. together.

II. IV. = a + aabb + b2.

Then multiplying the two products I. III. and II. IV.,

239. We shall perform this calculation in a still different manner, first multiplying the Ist. factor by the IVth. and next the II. by the IIId.

It remains to multiply the product I. IV. and II. III.

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 ab= 1; further, a a 9, ab 6, b b = 4. Therefore a a + ab + bb = 19, and a aab+bb = 7. So that the product required is that of 5 × 19 × 1 × 7, which is 665. Now a 729, and b 64, consequently the product required is ab 665, as we have already seen.

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=

=

CHAPTER IV.

Of the Division of Compound Quantities.

241. 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 parentheses, placing two points between the divisor and dividend. If it were required, for example, to divide a + b by cd, we should represent the quotient thus

a + b
c + ď

according

to the former method; and thus, (a + b) : (c + d) according to the latter. Each expression is read a + b divided by c + d.

242. When it is required to divide a compound quantity by a simple one, we divide each term separately. For example;

6a-8b4c, divided by 2, gives 3a-4b+2c;

=

and (a a 2 ab): (a) a 2 b.

In the same manner

(a32aab+3a ab): (a) = aa2ab+3ab;

(4 a ab6aac+8abc): (2 a) = 2ab3ac4bc; (9 a abc-12 abb c + 15 a b c c) : (3 a b c)=3a-4b+5c, &c. 243. If it should happen that a term of the dividend is not divisible by the divisor, the quotient is represented by a fraction, as in the

division of a + b by a, which gives 1+

b

Likewise,

For the same reason, if we divide 2 a+b by 2, we obtain

b

2

a+; and here it may be remarked, that we may write

b

26
3

is the same asb, and the same asb, &c.

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-be by the divisor ab, the quotient must then be such as, when multiplied by the divisor ab, will produce the dividend a c-bc. 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. no less evident, that

It is

(a a + ab): (a + b) = a; (3a a −2 ab): (3a-2 b) = a;

(6aa9ab): (2a-3b) = 3 a, &c.

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 +3ab2bb 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 2 a b + 2 bb. 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 2 a b + 2 bb; consequently a+2b is the quotient required; which, multiplied by the divisor ab, ought to produce the dividend aa+3 ab+2bb. See the whole operation :

a+b) aa + 3 ab + 2 bb (a + 2b

a a + ab

2ab+2bb
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.