G. C. D. and omit the remaining factors, a from the first and x from the second, because they will not affect the G. C. D. (112) and proceed as follows: Omit 40x2, because it is not a factor of the last divisor, and continue 117. Similarly, if at any stage of the operation a divisor and a dividend have a common divisor, it may be set aside and the operation continued with the remaining factors. The G. C. D. will be found by multiplying the final divisor by the factor set aside (107). Suppose that it is desired to find the G. C. D. of three expressions, A, B, and C. Find the G. C. D. of A and B; let D be this G. C. D. of D and C will be the G. C. D. of G. C. D. Then the For any divisor of D and C is a divisor of A, B, and C, since D or each of its factors is a divisor of A and B; and likewise any divisor of A, B, and C is a divisor of D and C, since the G. C. D., D, of A and B must be divisible by any factor common to A and B, and the divisors of A, B, and C, are by hypothesis divisors of C. Therefore, the G. C. D. of D and C is the G. C. D. of A, B, and C. Similarly, the G. C. D. of four algebraic expressions, A, B, C, D, will be found by finding the G. C. D. of any twe, A and B for example, then of the other two, C and D, and finally the G. C. D. of these two G. C. D's just found will be the G. C. D. of A, B, C, and D. 3926 x EXAMPLE. Find the G. C. D. of x1138. 40, and x3 10.r2 + 31 x 30. Thus, 24, x3- 9x2+26x-24 |x3-10x2+31x―30 x3-10 x2+31x-30 | +x2-5x+6 1 118. In accordance with the rule derived in 111 and the results deduced in the paragraphs immediately following it we have the more general rule for finding the G. C. D. of two or more integral polynomials. 1. Remove the simple factors of the given expressions and reserve the factors common to all of the expressions as a part of the G. C. D. sought. 2. Arrange the remaining factors of each quantity in the order of the descending powers of a common letter. That expression which is of the lowest degree is to be taken for the divisor; or if both are of the same degree, that whose first term has the smallest coefficient. 3. Continue each division until the degree of the remainder is less than that of the divisor. 4. If the last remainder of any division is found to contain a factor which is not a common factor of the divisor, this factor is to be removed; the resulting expression to be used as the next divisor, and the last divisor as the next dividend. 5. If at any stage of the division a factor is discovered which is common to a dividend and a divisor it may be removed and the division continued; but it will be a part of the G. C. D. 6. A dividend whose first term is not exactly divisible by the first term of the divisor may be multiplied by such an expression as will make it divisible. EXERCISE XXVII Find the G. C. D. of: The choice of method is left to the student. 1. x1−5x3+5x+5x-6 and x3-3x2-6x+8. 2. 25-20+30 +19x-30 and 4+5x+5x2-5x-6. 3. x+4x3-9x16x+20 and a3-2x2- 23x+60. 4. 16x86 x2 - 176x+105 and -19x+128xo - 356x+336. 6. 1035x-50x +24 and 3-8+17x-10. 7. x1-5x3+5 x2+5x-6 and x+2x3-13x2 - 14x + 24. 8. -5ac5a2x2+5a3x-6at and x+4ax2+a2x-6 a3. 9. x+2ax3-3a2x2 — 8 a3x — 4 aa and xa — aa. 10. -3a+2a2x2+2 a3x2—3ax+-a and x-4 ax3+3a2x2+4a3x-4 a1. 11. 2-3x-5x+15x+4x-12 and -7x3+6x+28x-40. 12. 2-1412 x3+49 x2-84x+36 and x6-14x+49x-36. 13. 2-14x+49-36 and -2x-11x+40x44x+16. 14. x+3x-5x3-27 x2 - 32x-12 and x − 6 x1+9x3 +4 2o 15. +3x-174-39x3+88 x2 + 108 x x+3x-23 x3-51x2+94x+120. - 144 and 16. x1 — yx3 — 7 y2x2+y3x+6y1 and 2x3-9yx2+7 y2x+6y3. 12x. 17. 2x4x3-14x2y2 — 5.xy3+6y and 2x-x3y—14 x2y2+19 xy3—6 y1. -13.67 x — 175x3 +244 x2-172x+48 and 18. x − 2 x3 — 20x1 +10.x3 +79 x2 −8x-60 and 19. 24x-50x2 +59xa — 60 x3 +36x2 — 10x+1 and 24x+14x+11x+12x3-12x2-2x+1 and 20. x-12x+48 x1 — 46 x3 — 153 x2 +378 x - 216 and x3 — 39 x3 — 34 x2 + 252 x +360. 21. x8 — 46 xo — 36 xo + 609 x1+828x3 — 1516x2-1440x+1600 and x1 — 5 x6 — 42x2+182x+497x3—1533x2 - 1800x+2700 and x+5x3-25x145x3-16x+500x+400. 119. Least Common Multiple.-In Arithmetic the Least Common Multiple of two or more whole numbers is the smallest number which each of the numbers will exactly divide. This term is also used in Algebra, and the sense in which it is used in this subject will be understood by the following definition: The least common multiple of two or more expressions which are arranged in the order of the descending powers of some common letter is the expression in the lowest degree of that letter which is exactly divisible by each of them. 120. Any expression which another will exactly divide is a multiple of it. 121. If two expressions have no common factor their least common multiple will be their product, since it will contain each of them. For brevity, let L. C. M. be used for Least Common Multiple. 122. When the Expressions can be Readily Factored. If the expressions can be readily factored their L. C. M. will be an expression which is the product of every factor of each expression taken the greatest number of times it occurs in any one of the given expressions. 1. Find the L. C. M. of 18 ax2, 90 ay2, 12 axy. 2. Find the L. C. M. of x2y — xy2, 3 x(x − y)2, 4 y(x − y)3. The rule can be more simply stated: The L. C. M. of two or more expressions which can be readily factored is the product of each factor taken the greatest number of times it occurs in any one of the expressions. 123. When the Quantities are not Readily Factored.-The L. C. M. of two or more quantities can be found by finding their G. C. D. Suppose that D is the G. C. D. of A and B; then Since D is G. C. D. of A and B, a and b have no common factor, .'. L. C. M. of a and b = - ab. Hence the L. C. M. of A and B or of aD and D is ab D. The L. C. M. of two expressions can be found by dividing their product by their G. C. D.; or by a method which is usually more simple; by dividing one of the quantities by their G. C. D. and multiplying this quotient by the other. EXAMPLE. Find the L. C. M. of 20x1 + x2. 1, .. 5-1 is G. C. D. required. 20x1 + x2 1 - 5x214 x2 + 1. Hence, L. C. M. required is (4x2 + 1)(25 .x1 + 5 x3 − x − 1). 124. To find the L. C. M. of three quantities, A, B, and C; find the L. C. M., say M, of A and B; then the L. C. M. of M and C is the L. C. M. required, say M'. For M will contain each factor of A and B the greatest number of times it occurs in either of them; and M' will contain each factor of |
