multiplied, (Art. 143), as well as the quantities b4c2b2c4 and 63-bac, which comprehend not at all this letter.

The question is therefore reduced to finding the common divisors of the quantities b2-c2 and b-c, and, to verify afterward, if, among these divisors, there be found some, that would also divide b3. bc2 and b2 —bc, ba c2-b2 c1 and b3 —b3c.`

Dividing b2-c2 by b-c, we find an exact quotient b+c; b-c is therefore a common divisor of the quantities b2-c2 and b-c, and it appears that they cannot have any other divisor, because the quantity bc is divisible but by itself and unity. We must therefore try if it would divide the other quantities referred to above, or, which is equally as well, if it would. divide the two proposed quantities; but it will be found to succeed, the quotients coming out exactly,

3

(b+c)a2 + (b2+bc) a3 +b3c2+b2c3 ;

In order to bring these last expressions to the greatest possible degree of simplicity, it is expedient to try if the first be not divisible by b+c; this division being effected, it succeeds, and we have now only to seek the greatest common divisor of these very simple quantities;

a2+ba3+b2c2, and a2 +ba+b2. .

Operating on these, according to the Rule, (Art. 141), we will arrive, after the second division, at a remainder containing the letter a in the first power only; and as this remainder is not the common divisor, hence we may conclude that the letter a does not make a part of the common divisor sought, which is consequently composed but of the factor b-c.

343221B

Ex. 7. Required the greatest common divisor of (d2 — c2) Xa2 + c13-d2 c2 and 4da2 — (2c2+4cd)a+ 2c3.

Arranging these quantities according to d, we have ́(a2 -c2) d2+c1-a2c2, or (a2-c2)d2 — (a2 —c2)c2, and (4a2-4ac) xd-(a-c) x2c2;

2

it is evident, by inspection only, that a2-c2 is a divisor of the first, and a-c of the second. But a2-c2 is divisible by a-c; therefore a -c is a divisor of the two proposed quantities: Dividing both the one and the other by a-c, the quotients will be (a+c)×(d2-c2), and 4ad-2c2;

which, by inspection, are found to have no common divisor, consequently a-c is the greatest common divisor of the proposed quantities.

Ex. 8. Required the greatest common divisor of `y--x1 and y3-y2x—yx2+x3.

4

Ans. y2-x2.

Ex. 9. Required the greatest common divisor of ab and a¤—bo.

Ans. a2-b2. Ex. 10. Required the greatest common divisor of a2+a3b—ab3-b4 and a1+a2b2+b1.

2

Ans. a+ab+b2. Ex.11. Required the greatest common divisor of · a2-2αx+x2 and a3—a2x-ax2+x3. Ans. α-x. Ex. 12. Find the greatest common divisor of 6x3-Eyx2+2y2x and 12x2-15yx+3y2. Ans. x-y. Ex. 13. Find the greatest common divisor of 36b2a6-18b3a5-2762a962a3 and 27b2a5Ans. 962a-9b2a. Ex. 14. Find the greatest common divisor of (c-d)a2+(2bc-2bd)a+(b2c-b2d) and (bc-bd+ c2-cd)a+(b3d+bc-b3c-bcd). Ans. c-d. Ex. 15. Find the greatest common divisor of Imp3 +3np3 q2 —2npq3—2nq* and 2mp3 q3-imp1 ~mp3q+3mpq3.

1863a962a3.

3

Ans. q-p.

Ex. 16. Find the greatest common divisor of x33+9x2+27x-98 and x2+12x-28.

Ans.x-2.

§ III. METHOD OF FINDING THE LEAST COMMON MUL

TIPLE OF TWO OR MORE QUANTITIES.

145. The least common multiple of two or more quantities is the least quantity in which each of them is contained without a remainder. Thus, 20abc is the least common multiple of 5a, 4ac, and 26.

146. The least common multiple of two numbers. or quantities, is equal to their product divided by their greatest common measure, or divisor.

For, let a and b be any two quantities, whose greatest common measure, or divisor, is x, and let a=mx, b=nx; then mnx is a multiple of a by the units in n, and of b by the units in m; consequently it is a common multiple of a and b.

But since x is the greatest common measure of a and b, m and n can have no common divisor; mnx is, therefore, the least common multiple of a and b. Now mx=a, and nx=b; therefore mx Xnxa ×b, ab

(Art. 50), and mnx= (Art. 51). Hence the rule

x

is evident; as for example:

Let the least common multiple of 18 and 12 be required. Their greatest common measure is 6; therefore their least common multiple is:

12 X 18

6

=36.

Every other common multiple of a and b is a multiple of mnx.

Let q be any other common multiple of the two quantities, and, if possible, let mnx be con

tained in q, r times, with a remainder s, which is less than mnx; then q―rmnx=s; and since a and b measure q and rmnx, they measure q-rmnx, or s, (Art. 131); that is, they have a common multiple less than mnc, which is contrary to the supposition.

22

n

To find the least common multiple of three quantities a, b, and c. Let m be the least common multiple of a and b, and ʼn the least common multiple of mande; then is the least common multiple sought. For every common multiple of a and b is a multiple of m, therefore every common multiple of a, b, and c, is a multiple of m and c; also, every multiple of m and c is a multiple of a, b, and c; consequently the least common multiple of m and c is the least common multiple of a, b, and c. Or, in general, let a, b, c, d, &c. be any set of quantities, and let x be the greatest common divisor of a and b ;

ab of and c

then will be the least common multiplier of a and b :

&c.

ab

x

147. Hence the following method for finding the least common multiple of two or more quantities.

RULE.

I. Find the least common multiple of two quantities, by the preceding Article.

II. Find, in like manner, the least common multiple of the result, thus found, and the third quantity.

III. Find also the least common multiple of the last result and the fourth quantity; and proceed in the same manner with this result and the fifth; and so on: the result last found will be the least common multiple of all the quantities.

Ex. 1. Required the least common multiple of a3b2x, acbx2, and abc2 d.

Here, the greatest common measure of a3b2x and acbx2, is abx, and the least common multiple is, a3b2 x Xacbx2 a3b3cx; the greatest

therefore,

abx

common measure of a3b2cx2 and abcd is abc; a3b2 cx2 Xabc2 d

hence,

abc

tiple required.

= a3b2 c2 dx2 = the least mul

Ex. 2. Required the least common multiple of 2a3x, 4ax, and 6x3.

Here, the greatest common measure of 2a x and 4ax2, is 2ax; hence, the least common multiple is 2α2 x × 4αx2 4a2x3; again, the greatest common

Zax

measure of 4a2x2 and 6x3 is 2x2; and therefore

4a2x2 X6x3 =12a2x2= the least common multiple

required.