a6-a2r4 a® + a3x—a1x2 —a3x3° Here a is a simple factor of the numerator, and a3 is a factor of the denominator; hence a2 is the greatest common measure of these simple factors, which must be reserved to be introduced into the greatest common measure of the other factors of the terms of the proposed fraction; viz.: Therefore, restoring a2, the greatest common measure, is a2(a2-r2). (1) Find the greatest common measure of 2a2r2, 4x2y2, and 6x3y. (2) Find the greatest common measure of the two polynomials a3-a2b +3ab3b3, and a2—5ab+4b2. (3) What is the greatest common measure of x3-ry2 and x2+2xy+y2? (4) Find the greatest common measure of r-ys and x13 —y13. (5) Find the greatest common measure of the polynomials (6) Find the greatest common measure of the polynomials (7) y°-5yz-4yz2+2z3 and 7y°z+10yz2+323. (8) Also of (r+a2x2+a1) and (r1+ax3—a3x—a1). (9) Also of (7a2-23ab+6b2) and (5a3—18ba2+11ab2 —6b3). (10) Also of (5a+10a1b+5a3b) and (a3b+2a2b2+2ab3+b1). (11) Also of (6a3 + 15a1b — 4a3c2 — 10a2bc2) and (9a3b — 27aobc +18bc3). 6abc (12) Also of (aa+y+a¥b®+aab8+bB+3) and (aam+aan+b3m+b3n). (13) Find the g. c. d. of the three quantities a3+3a2b+3ab2+b3, a2+2ab +b2, and a2—b2. PROPOSITION.-A divisor of a polynomial, which is independent of the letter of arrangement of that polynomial, must divide separately each of the multipliers of the different powers of that letter. DEMONSTRATION.-Let Arm+B-1+Cxm-2, &c., be the polynomial, and D the divisor. The quotient must contain every power of the letter of arrangement that the dividend does, since the quotient, multiplied by the divisor, must produce the dividend, and the letter of arrangement is not contained in the divisor. The quotient must, therefore, be of the form A'r+B'xm-1 +C'rm2, &c., multiplying which by the divisor gives DA'x+DB ́ÃTM−1 +DC'r2, &c., the original dividend, the multiplier of each power of x in which is evidently divisible by D. EXAMPLES. m Q. E. D. (1) Find a common divisor, independent of the letter a, of the two quantities b a2—ca2+b2a—c2a+b2—2bc+c2 and b3a3-3b3ca3+3bc2a3 —c3a3+b+a2 —c1a2+b3a—c3a+b3—3b2c+3bc2 —c3. Collecting together in the first of these two quantities the multipliers of a2 and a, and observing that b2-2bc+c is the square of b-c, we have (b−c)a2+(b2-c2)a+ (b −c)3, and from the second by a similar process, (b-c)3a3+(b1-c1) a2+(b3—c3)a+(b−c)3. The multipliers of the different powers of a in the two quantities are, therefore, b-c, b2-c2, (b—c)3, (b—c)3, ba—c1, and b3-c3. The only number which will divide them all is their common divisor b-c, which is, therefore, the answer required. (2) Find the greatest common divisor of (bc)a2-2b (b −c)a+b2(b—c) and Here the common divisor, independent of a, is b-c; suppressing which, we have left the two quantities a-2ba+b2 and (b+c)(a2-b2). Suppressing the factor (b+c) not common to both, we shall find the common divisor of these last two quantities to be a-b, and the greatest common divisor of the two original quantities is (b-c) (a-b) or ab-ac-b2+bc. The success of the process for finding a greatest common divisor depends upon the fact that the quantities being arranged according to the powers of some letter, each division leads to a remainder of a degree inferior to the divisor. When the polynomials contain many terms of the same degree, a precaution is necessary, without which this reduction would not always obtain, which consists in uniting all these terms under a single multiplier. Let there be the two polynomials: I write them thus: A=x2 +yx2+x2 —y2x+2yx—y3+y2 A=x3+(y+1)x2—(y2—2y)x—ys +ys +1)x+y. The first term, a3, not being divisible by (y+1)x2, on account of the factor y+1, I know (Prop. above), that if a quantity is arranged like the preceding, every divisor of this quantity, independent of x, must divide separately the multiplier of each power of z. From this it follows that y+1 has no common factor with B, because, if it had, this factor would be found in y+y--1 and in y; but it is evident that y has no factor common with y+1. We can then multiply A by y+1 without affecting the common divisor sought; and as it would be necessary to multiply again by y+1, we multiply at once by (y+1)2, or y+2y+1.* In this manner we arrive at the remainder R=(−y+—y3+y2)x—yb—y++y3. Before passing to the second division, it is necessary to suppress in R the factors common to the multipliers of the powers of x. But the two parts of R are evidently divisible by —y1—y3+y, and after this simplification there remains x+y. We can then take x+y for a divisor, and as the division is effected exactly, it follows that the common divisor sought is x+y. The process is not always so easy. To develop the general method to be pursued in such cases, let us consider the polynomials A and B, which contain two letters, x and y. Take first the greatest monomial common divisor of the terms of A; let a be this divisor, and A' the quotient of A by a: we shall have A=aA'. Arrange A according to the pow ers of x, taking care to collect all the terms containing a same power of this letter, and let us suppose, for example, that we have A'=Lx2+Mr+N. All the factors of A', independent of x, must be factors of the quantities L, M, N, which multiply the different powers of x. These quantities containing only the single letter y, it will be easy to find their greatest common divisor; let us name this divisor a', and the quotient of A' by a', A"; we shall have A'=a'A", and, consequently, A=ad' A". a will be the product of the monomial factors of A, a' the product of the polynomial factors which do not contain x, and A" the product of the factors which contain x. Let us effect the same decomposition of the polynomial B, and let B=33'B". Then I determine the greatest common divisor of the monomials a and ẞ, as well as that of the polynomials a′ and 3′, which contain only the letter y; and if I can also find that of the polynomials A" and B", which contain y and x, I shall have three quantities, the product of which will be the greatest common divisor of A and B. But I say that we can find the greatest common divisor of the quantities A" and B", in subjecting them to the same calculus as the preceding examples. It is clear, indeed, that, these quantities having no longer either monomial factors or polynomial independent of x, it will be proper to multiply the partial dividends of the first division by the polynomial which is placed before the highest power of x in the divisor, and that we shall thus arrive at a remainder of a degree less in x than the divisor. It will be easy to take from this remainder all the monomial factors which it contains, as well as the polynomial factors independent of x, and then proceed to a second division, taking for a divisor this remainder simplified. We operate as in the first; then we pass to the third, and continuing always in this manner, we are sure of arriving finally at a remainder zero, or independent of x. In the first case the quantities A" and B" have, for greatest common divisor, the divisor of the last division. We have thus seen that the finding of a common divisor, when the polynomials contain two letters, depends upon finding it when they contain one; so the case where they contain three depends upon that where they contain two, and so on, whatever be the number of letters. There is, therefore, no case in which the common divisor can not be found by the above rules. THE LEAST COMMON MULTIPLE. 32. We have already defined a multiple of a quantity to be any quantity that contains it exactly; and a common multiple of two or more quantities is a quantity that contains each of them exactly. Let N be the dividend, D the divisor, c the coefficient of the first term of the divisor. Multiplying by the square of this coefficient, the dividend becomes c2N. The first term of the quotient will contain the first power of c. Multiplying the whole divisor by this term of the quotient, every term of the product will contain the first power of c, and the whole product may be represented by cP. Subtracting this from the dividend, the remainder is c N-cP, every term of which contains c, and, therefore, its first term is ready for division without multiplying again by c. The least common multiple, of two or more quantities, is, therefore, the least quantity that contains each of them exactly. N.B. The least common multiple must, from its nature, contain all the factors in either of the quantities. 33. To find the least common multiple of two quantities. (1) Divide the product of the two proposed quantities by their greatest common measure, and the quotient is the least common multiple of these quantities; or divide one of the quantities by their greatest common measure, and multiply the quotient by the other. Let a and b be two quantities, d their greatest common measure, and m their least common multiple; then let a=hd, and b=kd; and since d is the greatest common measure, h and k can have no common factor, and hence their least common multiple is hk; therefore, hkd is the least common multiple of hd and kd; hence, (2) Also, the least common multiple is composed of the highest powers of all the factors which enter into the given quantities. EXAMPLES. (1) Find the least common multiple of 2a'r and 8a3r3. or, by (2), the two quantities being 2ax and 23a3r3, 23a3r3 is the l. c. m.; because 23 is the highest power of 2, a3 the highest power of a, and 3 the highest power of x, in either of the given quantities. or, (2) Find the least common multiple of 4x(x2-y3) and 12x3(x3-y3). Here d=4x2(x-y), and, therefore, we have ab 4x2(x2-y2) × 12x3(x3 — y3) -=12x3(x+y) (x3 —y3) ; m=d= 4x2(x-y) m=12x+12xy—12x1y3—12x3y1; or, the two given quantities being 2x2(x+y)(x-y) and 22.3x3(x-y)(x2+xy +y), the l. c. m. is 22.3x3(x+y)(x−y)(x2+xy+y2). (3) Find the least common multiple of x2+2xy+y2 and x3—xy2. Here d=x+y, and, therefore, we get =x(x+y) (x2—y2)= least common multiple or, the two given quantities being (x+y)3 and x(x+y) (x-y), the l. c. m. is x(x+y)2(x—y). (4) What is the least common multiple of r1—5x3+9x2-7x+2, and x-6x+8x-3? By the process for finding the greatest common measure, we find ..m= =(x-2)(x-6x2+8x-3) =x5—2x1—6.x3+20x2-19x+6, the least common multiple. F (5) Find the least common multiple of a2—2ab+b2 and aa—ba. (6) Find the least common multiple of a2-b2 and a3+b3. (7) Find the least common multiple of x2-y2 and xa3 —y3. (8) Find the least common multiple of y—8y+7 and y2+7y—8. (5) (a-b) (a1-b1). (6) (a—b) (a3+b3). ANSWERS. (7) (x+y) (x3-y3). 34. Every common multiple of two quantities, a and b, is a multiple of m, their least common multiple. For let m' be a common multiple of a and b, then, because m' is greater than m, if we suppose that m' is not a multiple of m, we have, as in the annexed scheme, Now the remainder k is always less than m the divisor; hence, since a and b measure m and m', it is evident that a and b measure m'-hm, or, by (2), k ; therefore, k is a common multiple of a and b, and it has been proved to be less than m, the least common multiple, which is absurd; hence the supposition that m' is not a multiple of m is false, or m' is a multiple of m. 35. To find the least common multiple of three or more quantities. Let a, b, c, d, &c., be the proposed quantities; Then, since every multiple of a and b is a multiple of m, their least common multiple (34), the quantity sought, x, is a multiple of m; but x also is by hypothesis a multiple of c; therefore, x is a multiple of both c and m, and, therefore, it is (34) a multiple of m'; but x is a multiple of d and m', and, therefore, of m'; hence x can not be less than m", and, therefore, m" is the least common multiple. EXAMPLES. (1) Find the least common multiple of 2a2, 4a3b2, and 6ab3. Here, taking 2a and 4a3b2, we find d=2a2, and, therefore, Again, taking m, or 4a3b2, and 6ab3, we find d=2ab2; hence Or, the three given quantities being 2a2, 22a3b2, and 2.3ab3, the l. c. m., by 33. (2), is 22.3a3b3. (2) Find the least common multiple of a-x, a2—x2, and a3 — x3. Taking a-x and a-x, we have d-a-x; and hence |
