237. When we have more than two quantities to multiply together, it will easily be understood that, after having multiplied two of them together, we must then multiply that product by one of those which remain, and so on. It is indifferent what order is observed in these multiplications.

Let it be proposed, for example, to find the value, or product, of the four following factors, viz.

I.

II.

III.

IV.

(a+b) (aa+ab+bb) (a−b) (aa —ab+bb).

We will first multiply the factors I. and II.

II. aa+ab+b b

I. a+b

a3 + a ab + abb

+aab+abb + b 3

I. II. a3+2a a b + 2 a b b + b3.

Next let us multiply the factors III. and IV.

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

a3 + 2a ab+2abb + b3 I. II.

a32aab+2abb-b3

III. IV.

-2a5b-4a4bb-4a3b32aabs

2a4bb4a3 b3 +4aab +2ab5

And this is the product required.

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.

I. a + b

III. a-
.b

aa+ab

-ab-bb

I. II. — a a —bb.

II. IV. = a* + a abb+ba.

Then multiplying the two products I. III. and II. IV.
II. IV. = a + a abb + b1

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

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 a-b=1; further, a a = 9, ab=6, bb = 4. Therefore a a+ab+bb19, and a a-ab+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 ab6665, as we have already seen.

CHAPTER IV.

Of the Divison of Compound Quantities.

; or

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 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 quo

a+b

tient thus c+d' according to the former (a+b): (c+d) according to the latter. read a+b divided by c+d.

method; and thus,

Each expression is

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 3 a-4b+2c;

and (aa-2 ab): (a) = a 2 b.

In the same manner

(a32aab +3 a ab): (a) = a a—2ab+sab;

(4 aab-6aac+8abc): (2a)=2ab3ac+4bc; (9aabc-12abbc + 15 ab cc): (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

- •

a

Likewise,

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

b

a +; and here it may be remarked, that we may write 16,

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 ab, the quotient must then be such as, when multiplied by the divisor a―b, 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 ab by c, we have a c-b c, which is exactly the dividend; so that c is the whole quotient. It is no less evident, that

(aa+ab): (a+b) = a; (3 aa— 2 a b) : (3 a — 2 b) = a ;

(6aa9ab): (2 as b) = 3a, &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 + s ab + 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 2 a b + 2 bb; consequently a + 2 b is the quotient required; which, mul

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