1. Find the G. C. D. of 17 pq, 34 p2q, and 51 p3q3.

2. Find the G. C. D. of 3x2-6x+3, 6 x2+6x-12, and 12x2 - 12.

Represent the G. C. D. as the product of all the common prime factors, of the following:

103y-60y+5.xy and 5-5.xy3-100 y.

4.

5.

6.

7.

8.

abc (a2 — b2) (a2 - c2)

9.

10.

11.

12.

13.

14.

15.

P2x - Py+2 lmx — 2 lmy + m2x - m2y and Fx — l3y — m2x + m2y.

108. The G. C. D. of Two Algebraic Expressions. As in Arithmetic, the following is the rule for finding the G. C. D. of two algebraic expressions:

Let A and B be the two expressions; arrange A and B with respect to the descending powers of some common letter and suppose that the exponent of the highest power of that letter in A is equal to or greater than the exponent of the highest power of that letter in B. Divide

A by B; make the remainder a new divisor and divide B by it. Proceed in this way until there is no remainder; then the last divisor will be the G. C. D. required.

109. EXAMPLE. - Find the G. C. D. of 817 and 1763; also of x2-4x+3 and 4x3.

In Arithmetic

9x2 15x+18.

The division is continued until the first term of the remainder is of a lower degree than that of the divisor.

110. The truth of the rule given in 108 depends upon the following principles:

1. If P divide A, then it will divide mA. For suppose that P is contained in A, r times, then ArP and mA = mrP. P is a factor of mA and therefore a divisor of mA.

Any divisor of an expression is also a divisor of any multiple of that expression.

=

2. If P divide A and B, then it will divide mA ±n B. For, since P divides A and B we may assume that A rP and B = $P (by 1), then mA±nB = mrP±nsP = (mr±ns) P; .. P divides mAnB by definition of division.

Hence, any common divisor of two expressions is a divisor of the sum or the difference of any multiples of the expressions.

These two principles make it possible to prove the rule given in 108.

111. Let A and B be the two expressions. Suppose that they are arranged according to the descending powers of some common letter, and that the exponent of the highest power of that letter in A is equal to or greater than the highest power of that letter in B. Divide A by B; let a be the quotient and C the remainder. Divide

B) A (a
a B

B by C; let b be the quotient and D the remainder.

C) B (b

bC

D) C (c c D

Divide C by

D; let c be the quotient with no remainder. The laws of subtraction give the following results:

(1) AaB+C;

(2) B=bC+D;

(3) C−cD.

To prove that D is a common divisor of A and B. D is a divisor of C since by (3) C = cD; i. e., D is one of the factors of C; hence by 1, 110, D is a divisor of bC and .. of bC+D, by 2, 8110; or by (2), B = b (cD) + D = (be + 1) D ... D divides B. Since D divides B it divides a B, and divides a B+C by 2, 8110, that is, divides A. Hence D is a divisor of A and B.

To show that D is the greatest common divisor of A and B. We have from (1), (2), and (3),

(1') A — a B = C;

(2')

B-bCD.

By 110, 2, any expression which divides A and B divides A-aB, that is, divides C, (1'); thus an expression which divides A and B is a divisor of B and C. Similarly, from (2'), every expression which divides B and C is a divisor of C and D. That is, every expression which divides A and B is a divisor of D. But no expression of a higher degree than D can divide D. Therefore D is the G. C. D. required.

112. In order to avoid fractional coefficients in the quotients in the operations of finding the G. C. D., (1) certain factors may be rejected which do not form a part of the G. C. D. required; and (2) a factor of a certain kind may also be introduced at any stage

of the process.

EXAMPLE Find the G. C. D. of x 2x2-6x3 +4.c2 + 13x+6 and 310+15x+8.

Thus,

Before proceeding farther divide the new divisor by 2 and multiply the new dividend by 3. Then continue the operation thus:

3x- 6 x 18 x3 12x2 + 39x+18 | 3x1 + 4x3 — 6x2 – 12 x — 5 3x+4x2-6x3-12x2.

5x

10x1 12x + 24x2+44x+18

Divide the last remainder by 2 and multiply the quotient by 3. Then we have:

Dividing the last remainder by 2 and continuing the operation, thus: 3x+4x3 6x2 12.x 5 | x3 +3x2 +3x+1

-

3x+9x3 + 9x2 + 3x

Hence +3x2+3x+1 is the G. C. D. required.

113. The factor 2 was omitted from the first remainder in accordance with rule 1, of 112. The justification of this rule will be now given.

Suppose that it is desired to find the G. C. D. of A and B; and that at any time in the process P and Q are respectively dividend and divisor. Let Qm S where m does not have a factor which P has: m may be rejected; that is the process may be continued with P and S instead of P and Q.

For it has been shown (2111) that A and B have the same common divisor that P and Q have.

But any common divisor of P and S is a common divisor of P and Q and is therefore a common divisor of A and B.

Any common divisor of P and Q is a common divisor of P and m S. But m has no factor which P has. Hence any common divisor of P and Q is a common divisor of P and S. Therefore any common measure of A and B is a common measure of P and S. Thus A and B have just the same common divisors which P and S have; the fact which it was desired to prove.

114. By 112, rule 2, a factor of a certain kind may be introduced at any stage of the process. Thus, in the example in 112, after the second division we divided the remainder - 10 4 · 12 x3 + 24 x +44 +18 by 2 and then multiplied the quotient by 3, which is not a factor of the divisor 3.c1+4x3-6.c2-12x-5. This rule is formed as follows: Given the problem, to find the G. C. D. of A and B. At any time in the process suppose that the expressions L and M are to be respectively the dividend and the divisor. Let P = nL, where n has no factor which M has; then ʼn may be introduced without affecting the G. C. D.; that is, instead of continuing the division with L and M we may continue it with P and M.

By 111, A and B have exactly the same common divisor as L and M have. Moreover, any common divisor of L and M is a common divisor of P and M; so that any common divisor of A and B is a common divisor of P and M.

Any common divisor of P and M is a common divisor of nL and M (since P = nL); but n does not contain a factor of M. There fore the common divisor of P and M is a common divisor of L and M and is therefore a common divisor of A and B.

Thus it is evident that A and B have exactly the same common divisors as P and M have.

115. As has been shown, certain factors may be removed from either dividend or divisor, or introduced into either; in practice factors are usually removed from divisors and introduced into dividends. These factors are, as a rule, numerical factors. The results of 113, 114 show that these operations may be made at any time in the process; for example, at the beginning.

=

bL.

116. Suppose that A and B have a common factor L, which is readily seen. Let A = aL and B Then I will be a common factor of A and B and a part of the G. C. D. Find the G. C. D. of a and b and multiply it by L and the product will be the G. C. D. of A and B. For example,

Find the G. C. D. of

2a1 +3 a3x - 92 and 6 a'r 17 a3x2 + 14 a2x3 — 3 ax1.

Arrange the quantities with respect to the descending powers of a and factor them, thus,

a2 (2a+3 ax-9x)

and ar(6 a3- 17 a2x + 14 ax2 — 3x3).

Set aside the factor a common to both expressions as a part of