4. Factor ac3x+cmx. 5. Factor x3-x1y+xy3—y3. 6. Factor ab'+2a'b'+a*b*. Ans. a'+c(c+m) x. Ans. a'b'(a+b) (a+b). 7. Arrange (x-x)a+(x2+x)(3b-c)-q according to the pow Ans. (a+3b-c)x2—(a—3b+c)x—q. ers of x. 8. Factor a'm-9am3. 9. Factor 8a'-x'. 10. Factor y+243. Ans. am(a-3m) (a'+3m). Ans. (4a3+2ax+x1) (2a—x). Ans. (y'-3y'+9y'-27y+81) (y+3). 11. Find the factors of x-y. Ans. (x2+xy+y3) (x2—xy+y3) (x+y) (x—y). 12. Find the factors of a'-ab'+2abc-ac. Ans. a(a+b−c) (a−b+c). SUBSTITUTION. 96. Substitution, in Algebra, is the process of putting one quantity for another, in any given expression. 1. Substitute y-1 for x, in x+x2-5x-3. x2+x3—5x—3 =y3—2y3—4y+2, Ans. Hence, for substitution we have the following RULE. Perform the same operations upon the substituted quantity as the expression requires to be performed upon the quantity for which the substitution is made. EXAMPLES FOR PRACTICE. 1. Substitute a-b for u in a'+ab +33. Ans. a'-ab+b. Ans. x+2x+1. Ans. x+10x+37x+60x+30. 4. Substitute s+r for x, in x2+ax+b, and arrange the result according to the descending powers of r. Ans. r2+(2s+a)r+s+as+b. 5. What will a+a*b+a3b3+ab+b become, when b=a? Ans. 5a*. 6. What will x+ax2+a2x+a' become, when m+1 is put for x and m-1 for a ? Ans. 4m(m2+1). 7. What will x*+y' become, when a+b is put for x and a-b for y? Ans. 2(a*+6a3l3+b*). 8. What is the value of (x+a+b+c)*+(x—a—b—c)', when a+b+c=s? Ans. 2(x2+10x*s2+5xs*). 9. In x-7x-6 substitute y-2 for x. Ans. y'-6y+5y. 10. In x-2x+3x-7x2+8x-3 substitute y+1 for x. Ans. y'+3y+5y'. 11. If a—b—x, b-c-y, and c-a-z, prove that 2(a—b)*(b—c)3 +2(a—b)2(c—a)2+2(b—c)2(c—a)2=x*+y*+z*. THE GREATEST COMMON DIVISOR. 97. A Common Divisor of two or more quantities is a quantity which will exactly divide each of them. 98. The Greatest Common Divisor of two or more quantities is the greatest quantity that will exactly divide each of them; it is composed of all the common prime factors of the quantities. The term, greatest, in this connection, is used in a qualified sense, and has reference to the degree of a quantity, or of its leading term, not to its algebraic or its arithmetical value. Thus, if x-3 and x2+x+2 are the prime factors common to two or more quantities, then according to the above definition, (x2+4x+2)(x−3)=x3+ x2-10x-6, is the greatest common divisor. But this product is not necessarily greater in value than one of the prime factors. For, if x=4, then we have x2+4x+2=34, and x'+x-10x-6-34. 99. Several quantities are said to be prime to each other when they have no common factor. CASE I. 100. When the given quantities can be factored by inspection. It is evident from (81, 2) that no factor of the greatest common divisor can have an exponent greater than the least with which it enters the given quantities. Hence the following obvious RULE I. Find by inspection, or otherwise, all the different prime factors that are common to the given quantities, and affect each with the least exponent which it has in any of the quantities. II. Multiply together the factors thus obtained, and the product will be the greatest common divisor required. EXAMPLES FOR PRACTICE. 1. Find the greatest common divisor of a'—2a'x'+axʻ, and a‘— 2a*x+a'x'. Factoring, we have a'—2a2x2+a x'=a (a'—2a2x2+x)=a(a-x)' (a+x)' The lowest powers of the common factors are a and (a—x)'; and we have a(a—x)'=a'—2a3x+ax3 the greatest common divisor required. 2. Find the greatest common divisor of 2a2bc', 6ab'c', and 10a3bc2. Ans. 2abc. 3. Find the greatest common divisor of 5x'y'z3, 6x3yz3, and 12x3yz3. Ans. xyz. 4. Find the greatest common divisor of x3-y' and x2—2xy+y3 Ans. x-y. b. What is the greatest common divisor of a'm-l'm and 2ac❜m-2c'bm ? Ans. m(a-b). 6. What is the greatest common divisor of a2x3-3a2x2+a3x and 3axz'—ax3z3—az2 ? Ans. a(x2-3x+1). 7. What is the greatest common divisor of 16x-1, x—4x3, and 1-8x+16x1? Ans. 4x-1. CASE II. 101. When the given quantities can not be factored by inspection. 102. The greatest common divisor is found in this case by a process of decomposing the quantities by division. But in order to deduce a rule for the method, it will be necessary first to establish certain principles relating to exact division. 103. First, suppose A to be a quantity which is exactly divisible by another quantity, D, and let q represent the quotient. Then, A =q D If we now multiply the dividend by m, we shall have, from (84 I), in which qm is entire. it will also divide Am. Thus we have shown that if D divides A, 1. If a quantity will exactly divide one of two quantities, it will divide their product. Again, let A and B represent any two quantities, and S their sum. Now suppose both A and B are exactly divisible by D, and let S A+B=S S q+q=D in which must be entire, because its equal, q+q', is entire. Hence, D 2. If a quantity will exactly divide each of two quantities, it will divide their sum. Finally, let d represent the difference of A and B, and suppose A and B to be divisible by D, q and q' being the quotients, as before. We shall have 3. If a quantity will exactly divide each of two quantities, it will divide their difference. 104. We may now show, by the aid of these principles, wh relation the greatest common divisor of two quantities bears to the parts of these quantities when decomposed by division. Suppose two polynomials to be arranged according to the powers of the same letter, and let A represent the greater and B the less. Then let us divide the greater by the less, the last divisor by the last remainder, and so on, till nothing remains. If we represent the several quotients by q, q', q', etc.; and the remainders by R, R', R", etc., the successive operations will appear as follows: To investigate the mutual relations of A, B, R, and R', we observe that in division the product of the divisor and quotient, plus the remainder, if any, is always equal to the dividend. Hence, from the operations above, we have the three following conditions. Now from the first equation it is evident that R' divides R without remainder; it will therefore divide Rn', (103, 1). And since R' divides both Rq' and itself, it must divide their sum, Rq'+R', or B. (103, 2); consequently, it will divide Bq, (103, 1). Finally, since it divides both Bq and R, it must divide their sum, Bq+R, or A, (103, 2). Hence, I. The last divisor, R', is a common divisor of R, B, and A, or of all the dividends. Again, the dividend minus the product of the divisor and quotient, is always equal to the remainder. Therefore, from the first and second operations above, we have A-Bq=R B-Rq'-K' |