47. A factor which does not contain any factor common to both A and B may be rejected at any stage of the process. Let the operation stand thus: where neither m nor n contains any factor common to A and B. It will be an exercise for the student to show that D is the G.C.M. of A and B. 48. A factor, which has no factor that the divisor has, may be introduced into the dividend at any stage of the process. The operation may stand thus B)mA(p, where m has no factor that B has ; CnBg, where n has no factor that C has; 0 As in Arts. 46, 47, it may be easily shown that D is the G.C.M. Both the above principles are made use of in working out Ex. 3 Art. 45. 49. When a common factor can be found by inspection, it is advisable to strike it out of the given expressions. Then, having found by the ordinary process the G.C.M. of the resulting quantities, we must multiply the G.C.M. so found by the rejected factor. Thus, 4 x is common to the quantities 4x3- 20 x2 + 24 x; and 4+ 16x2 84 20. Rejecting it, we get x2 5 x+6, and x2 + 4 x whose G. C.M. is easily found to be x 2. Multiplying by 4 x, we find the required G.C.M. to be 4x2-8x. 50. By a little ingenuity on the part of the student in breaking up the given expressions into factors, the ordinary and often tedious process of finding the G.C.M. may be avoided. The limits of our space will allow us only one example. Ex. Find the G.C.M. of 3x23 + 4x2 15 x3 + 47 x2 + 13 x 12. 10x+3, and 1 as a factor (Art. 30), The other factor may 3 x2 + 7 x2 1) + 7 x (x - 1) - 3(x-1) 3x+4x2 - 10x + 3 = = 3 x3 Now, 3x2+ 7 x 3 is not further resolvable, and x is evidently (Art. 30) not a factor of 15 a3 + 34 x2 + 13 x 12. It is, therefore, very probable that 3x2 + 7x - 3 is the G.C.M. required. We may test it thus— 15x3 +47 x2+13x-12= 15x3 + 35x2 - 15x+12x2 + 28x-12 =5x (3x2+7x-3)+4(3x2+7x-3) =(5x + 4) (3x2+7x-3). Hence, 3x2+ 7 x 3 is the G.C.M. required. G.C.M. of Three or More Quantities. 51. The G.C.M. of three or more quantities may be found thus RULE. Find the G.C.M. of any two of the quantities, then the G.C.M. of the G.C.M. so found and a third quantity, and The last found G.C.M. will be the G.C.M. required. 2. 23 + 6 x2 + 11 x + 6 and x3 + 5 x2 + 7 x + 3. 3. 2x3 + 10 x2 18x 90 and 3 x3 + 16 x2 4. x2 + (a + b x + ab and x2 + (a + c) x + ac. 8 x3 + 31 x2 + 15 x and 24 x2 + 22 x3 + 17 x2 + 5 x. 10. 15 a3 + a2b 3 ab2 + 2 b3 and 54 a2b2 24 64. (c + 1) x2 + (c + 1) x 3 x 1 and 4x + 2 x3 c2 and 9 ac + 2 a2 e2x + x2 c and ex3 1. (f − g) x2 + 9, 8 x2 + 20 x2 + 51x a, bxt (b + d) x3 + (c + e) x + e, and (c + 1) x5 + (ď + 2) x2 (c + 2) x2. b3+ c3 + 3 abc and a2 b2 + c2 + 2 ac. 52. When two or more algebraical expressions are arranged according to the powers of some letter, the expression of lowest dimensions which is divisible by each of the given expressions is called the L.C.M. 53. The L.C.M. of monomials and of expressions whose factors are apparent may be found by inspection. Ex. 1. Find the L.C.M. of ab, ac, ad, bc, bd, cd. If we form an expression, whose elementary factors contain each of the elementary factors of the given quantities, we shall evidently have a common multiple; and if no ele mentary factor of this expression is of a higher power than the highest power of the same factor in the given quantities, we shall get the L.C.M. Hence, the required L.C.M. = abcd. 54. The L.C.M. of two quantities is found by dividing their product by their G.C.M. Let a and b be the two quantities, and d the G.C.M.; And suppose a = pd and b = qd. It is evident that p and q contain no common factor. Hence pq is the L.C.M. of p and q ; and, therefore, no expression of lower dimensions than pqd can possibly be divisible by pd and qd. Now pqd = Hence pqd is the L.C.M. of pd and qd, or of a and b. pd × qd ÷ d = a × b÷d, and hence the rule: 55. To find the L.C.M. of three or more quantities. RULE. Find the L.C.M. of two of the quantities, then the L.C.M. of the expression thus obtained and a third quantity, and so on. The last expression so found is the L.C.M. required. We shall prove this rule in the case of three quantities. Let a, b, c be the quantities, and m be the L.C.M. of a and b. Then the L.C.M. of m and c is the L.C.M. required. For every common multiple of m and c is a common multiple of a, b, c. And every common multiple of a and b Hence, must contain the m, their least common multiple. every common multiple of a, b, c must be a common multiple of m and c, and the converse is also true. L.C.M. of m and c is the L.C.M. of a, b, c. Find the L.C.M. of Ex. XIII. 1. axy2, 3 a2x3y, 4 a3y3, 6 x2y2. Hence, the 7. 6x2 + 37 x + 56, 8 x2 + 38x + 35, 12x2 + 47 x + 40. 8. 5(x+1), 6 (x2 + 1), 7 (23 + 1). 9. x2 + a2x2 + a2, a2x2 + a3x + a*, ax2 a2x + a3. 10. x2 + (a + b) x + ab, x2 + (b + c) x + bc. 1 + x, 1 + x2, 1 + 11. 1 X, 3 ab (a 1, 6 x 2 a2b2 + + a3 (α + c) x + ac, x2 + x3, 1 + x3. 5 x + 8x3 + b2, a1 + 4a3b + 6a2b2 + 1. 4 ab3 + b2, 7x+5, 4x + 6 x2 + 12 x 3 x2 16, 16 x + 20. 20. 28-y3, 24 - 2x2y2 + y2, x3 + x2y + xy2 + y3. 21.x + αx2 + a2x2 + a3x2 + a1x + a5 and x5 a2x3 a2x2 + a*x 22. a2 + b2 c2 |