Thus, to find the other factor of x3-y3, we divide by x-y, and the quotient is x2+xy+y2; hence, x3—y3=(x—y)(x2+xy+y2). Similarly, x5—5—(x—y)(x2+x3y+x2y2+xy3+y1). 4th. Any binomial which is the difference of the even powers of two quantities, higher than the second degree, can be separated into at least three factors, one of which is the sum, and another the difference of the quantities. (See Art. 85.) Thus, by Art. 84, aa—ba, is divisible by a+b, and, by Art. 85, it is divisible by a-b; hence it is divisible by both a+b and a—b, and the other factor will be found by dividing by their product. Or, it may be separated into factors, thus, a1-b1=(a2+b2)(a2—b2)=(a2+b2)(a+b)(a—b). 5th. Any binomial which is the sum of the odd powers of two quantities, can be separated into at least two factors, one of which is the sum of the quantities. (Art. 86.) The other factor will be found by dividing the given binomial by this sum. Thus, a3-b3-=(a+b)(a2—ab+b2). Separate the following expressions into their simplest factors ABT. 94: To separate a quadratic trinomial into its factors. A quadratic trinomial is of the form, x2+ax+b, in which the sign of the second term may be either plus or minus. To explain the method of performing this operation, let us examine the relation that exists between two binomial factors and their product. 1. (x+a)(x+b)=x2+(a+b)x+ab. 4. (x—a)(x+b)=x2+(b—a)x—ab. From this we see that any trinomial may be resolved into two factors, when the first term is a square, and the coëfficient of the second term equal to the sum of any two quantities, whose product is equal to the third term. REMARK.-In Equations of the Second Degree (Art. 234), it will be shown how to perform this operation by a direct method; it is, however, a useful exercise for the pupil to do it by inspection, the only difficulty being to find two quantities whose sum is equal to the coefficient of the second term, and product equal to the third term. ART. 95. Examples of binomials and trinomials that may be separated into factors, by first separating the monomial factor, and then applying the principles in Art. 93. Ex. 1. ax3y-axy3=axy(x2—y2)=axy(x+y)(x—y). 2. 3ax2+6axy+3ay2. 3. 2cx2-12cx+18c. 4. 27a-18ax+3ax2. Ans. 3a(x+y)(x+y). Ans. 2c(x-3)(x-3). Ans. 3a(3-x)(3—x). Ans. 3mn(m+n)(m—n). Ans. 22(2+2)(2—z). Ans. 2xy(x2+y2)(x+y)(x—y). 9. 2x+4x270x. Ans. 2x(x+7)(x-5). Ans. 3ab(a-5)(a+4). 10. 3a3b-3a2b-60ab. Solve the following questions by first indicating the operations to be performed, and then canceling the factors common to the dividend and divisor. 11. Multiply 4x-12 by 1-x2, and divide the product by 2+2x. (4x-12)(1-x2) 4(x-3)(1+x)(1—x). 2+2x = =2(x-3)(1-x)= 2(1+x) 8x-6-2x2. 12. Multiply x2+2xy+y2 by x-y, and divide the product by x2-y2. Ans. x+y. 13. Multiply 6am2—6an2 by m+n, and divide the product by 2m2+4mn+2n2. Ans. 3a(m-n). 14. Multiply together 1-c, 1-c2, and 1+c2, and divide the product by 1-2c+c2. Ans. 1+c+c+c3. 15. Multiply together x2+x-2 and x2-x-6, and divide the product by x2+4x+4. Ans. x2-4x+3. 16. Multiply together x3-x2-30x and x2+11x+30, and divide the product by the product of x2-36 and x2+10x+25. Ans. x. GREATEST COMMON DIVISOR. ART. 96. Any quantity that will exactly divide two or more quantities, is called a common divisor, or common measure, of those quantities. Thus, ab is a common divisor of ab2 and abx. REMARK.-Two quantities, like two numbers, often have more than one common divisor. Thus, a2cx and abdx have three common divisors, a, x, and ax. ART. 97. That common divisor of two quantities which is the greatest, both with regard to the coefficients and exponents, is called their greatest common divisor, or greatest common measure. Thus, the greatest common divisor of 6a2bx2 and 9a3cxz is 3a2x. ART. 98. Quantities that have a common divisor are said to be commensurable, and those that have no common divisor are said to be incommensurable. ART. 99. To find the greatest common divisor of two or more monomials. 1. Let it be required to find the greatest common divisor of the two monomials, 14a3cx and 21a2bx. By separating each quantity into its prime factors, we have 14a3cx=7×2×aaacx, and 21a2bx=7×3×aabx. By examining these quantities we find that 7, aa or a2, and x, are the only factors common to both; hence, both the quantities can be exactly divided by either of these factors, or by their product, 7a2x, and by no other quantity whatever; therefore, 7a2x is their greatest common divisor. This gives the following RULE FOR FINDING THE GREATEST COMMON DIVISOR OF TWO OR MORE MONOMIALS. ·Resolve the quantities into their prime factors; then the product of those factors that are common to all the terms, will form their greatest common divisor. NOTE. -The greatest common divisor of the literal parts of the quantities may generally be found most easily by inspection, by taking each letter that is common to two or more of the quantities, with its least exponent. 2. Find the greatest common divisor of 6a2xy, 9a3x3, and 15a^x1y3. OPERATION. 6a2xy =3X2a2xy 9a3x3 3x3a3x3 15a3x1y3=3×5a1x1y3 Here we find that 3 is the only numerical factor, and a and x the only letters common to all the quantities. The least powers of a and x, in either of the quantities, are a2 and x; hence, the greatest common divisor is 3a2x. Find the greatest common divisor of the following quantities. ART. 100. Previous to investigating the rule for finding the greatest common divisor of two polynomials, it is necessary to demonstrate the following PROPOSITION. ·Any common divisor of two quantities, will always exactly divide their remainder after division. Let AD and BD be either two monomials, or polynomials, of which D is a common divisor, and let AD be greater than BD. Divide AD by BD, and if BD is not contained an exact number of times in AD, suppose it is contained Q times with a remainder, which may be called R. Then, since the remainder is found, by subtract BD)AD(Q BDQ AD-BDQ-R ing the product of the divisor by the quotient from the dividend, we have, R=AD-BDQ. Dividing both sides by D, we get R D =A—BQ; but A and BQ are each entire quantities, therefore R D' their difference, must be an entire quantity. Hence, any common divisor of two quantities (and of course the greatest common divisor), will always exactly divide their remainder after division. REMARK. In the preceding demonstration it is assumed that the pupil understands the following axioms: First. If two equal quantities be divided by the same quantity the quotients will be equal. Second. The difference of two entire quantities is also an entire quantity. ART. 101. Let it be required to find the greatest common divisor of two polynomials, A and B, of which A is the greater. If we divide A by B, and there is no remainder, B is, evidently, the greatest common divisor, since it can have no divisor greater than itself. Divide A by B, and call the quotient Q, then if there is a remainder R, it is evidently less than either of the quantities A and B ; and by the preceding theorem it is also exactly divisible by the greatest common divisor; hence, the greatest common divisor must divide A, B, and R, and cannot be greater than R. But if R will exactly divide B, it will also B)A(Q BQ A-BQ-R, 1st Rem. R)B(Q' RQ' B-RQ' R', 2d Rem. A=BQ+R Since the dividend is B=RQ'+R' equal to the product of the divisor by the quotient, plus the remainder. exactly divide A, since A=BQ+R, and therefore will be the greatest common divisor sought. Suppose, however, that when we divide R into B, to ascertain if it will exactly divide it, we find that the quotient is Q', with a remainder R'. Now, it has been shown that whatever exactly divides two quantities, will divide their remainder after division (Art. 100); and since the greatest common divisor of A and B, has been shown to divide B and R, it must also divide their remainder |