9. (3+) (3})=. 16. (4+1) (4—})=. The student should be drilled upon examples like those apар pended to the preceding theorems until he can produce the results mentally with as great facility as he could read them if exhibited upon paper. The utility of these theorems will be the more apparent, the more complicated the expressions to which they are applied. Frequent examples of their application will be seen hereafter. (63.) The same theorems will enable us to resolve many complicated expressions into their factors. 1. Resolve a2+4ab+4b' into its factors. Ans. (a+2b) (a+2b). 2. Resolve a'-6ab+9b' into its factors. 3. Resolve 9a2-24ab+166' into its factors. 4. Resolve a-b' into three factors. 5. Resolve a"-b" into its factors. 6. Resolve a-b into four factors. 7. Resolve 25a*-60a'b'+366° into its factors. 8. Resolve n2+2n+1 into its factors. 9. Resolve 4m2n2-4mn+1 into its factors. 10. Resolve 49a*b*-168a3b'+144a'b' into its factors. 11. Resolve n3+2n2+n into three factors. 12. Resolve 1- into two factors. 25 13. Resolve 4- into two factors. MULTIPLICATION BY DETACHED COEFFICIENTS. (64.) The coefficients of a product depend simply upon the coefficients of the two factors, and not upon the literal parts of the terms. Hence we may obtain the coefficients of the product by multiplying the coefficients of the multiplicand severally by the coefficients of the multiplier. To these coefficients the proper letters may afterward be annexed. This will be best understood from a few examples. Thus, take the first example of Art. 52, to multiply a+b by a+b. which is the same result as before ol tained. Ex. 2. Multiply 3a2+4ax-5x2 by 2a2-6ax+4x2 Coefficients of multiplicand 3+ 4-5 2- 6+ 4 6+ 8-10 -18-24+30 +12+16-20 Coefficients of the product 6-10-22+46-20 It may seem difficult in this case to supply the letters; but a little consideration will render it perfectly plain. Thus, 3a2x2a' is equal to 6a; hence a' is the proper letter to be attached to the first coefficient. For the same reason, x is the proper letter to be attached to the last coefficient. Moreover. we see that both the proposed polynomials are homogeneous, and of the second degree. Hence the product must be homogeneous, and of the fourth degree. The powers of a must decrease successively by unity, beginning with the first term, while those of x increase by unity. Hence the required product is 6a1-10a3x-22a2x2+46ax3 — 20x1. Ex. 3. Multiply x3+x3y+xy2+y3 by x—y. If we should proceed with this example precisely in the same manner as with the preceding, we should commit an error by attempting to unite terms which are dissimilar. The reason is, that the multiplicand does not contain the usual complete series of powers of a. The term containing the second power. of a is wanting. This does not render the method inapplicable, but it is necessary to preserve dissimilar terms distinct from each other; and since, while we are are operating on the coefficients, we have not the advantage of the letters to indicate what are similar terms, we supply the place of the defi cient term by a cipher. The operation will then proceed with entire regularity. 2+0-3+ 5 2-5 4+ 0-6+10 -10-0+15-25 4-10-6+25-25 Hence the product is 4a-10a3b-6a2b2+25ab3—25b*. Ex. 6. Multiply 2a-3ab'+5b' by 2a-5b'. Here there is a term in each polynomial to be supplied by a cipher. The preceding examples are intended to lead the student to consider the properties of coefficients by themselves, and prepare him for some investigations which are to follow, particularly in Section XX. The beginner, however, in attempting to apply the method, must be cautious not to unite dissimilar terms. MISCELLANEOUS EXAMPLES. 1. Multiply 15a+86-2c+7d by 4a-3b-5c+11d. Ans. Ans. 2. Multiply 3—3x3y3+2y3 by 4x3-3x2y2—6y3. 3. Multiply 3x2+8xy-5 by 4x2—7xy+9. Ans. Ans. 4. Multiply 6a-5ax-3x2 by 4a2—3ax+2x3. 5. Multiply 5x+xy+4y2 by x2-2xy+3y2. Ans. 6. Multiply x3-x3y+xy2 by x2—xy3—y3. Ans. 7. Multiply 7a'x'-3a'x3-5x' by 4a'x'-2a'x3-3x‘. 8. Multiply a*b*—5a3b'+3b° by 8a'b'+3a'b-2b3. Ans. Ans. Ans. 9. Multiply 6a-4ab2-8c3 by 7a3-5ab'—3c'. SECTION V. DIVISION. (65.) The object of division in Algebra is the same as in Arithmetic, viz., The product of two factors being given, and one of the factors, to find the other factor. The dividend is the product of the divisor and quotient, the divisor is the given factor, and the quotient is the factor required to be found. CASE I. (66.) When the divisor and dividend are both monomials. Suppose we have 63 to be divided by 7. We must find such a factor as, multiplied by 7, will give exactly 63. We perceive that 9 is such a number, and therefore 9 is the quotient obtained when we divide 63 by 7. Also, if we have to divide ab by a, it is evident that the quotient will be b; for a multiplied by b gives the dividend ab. So, also, 12mn divided by 3m gives 4n; for 3m multiplied by 4n makes 12mn. Suppose we have a to be divided by a'. We must find a number which, multiplied by a2, will produce a'. We perceive that a is such a number; for, according to Art. 50, we multiply a' by a', by adding the exponents 2 and 3, making 5. That is, the exponent 3 of the quotient is found by subtracting 2, the exponent of the divisor, from 5, the exponent of the dividend. Hence the following RULE OF EXPONENTS IN DIVISION. In order to divide quantities expressed by different powers of the same letter, subtract the exponent of the divisor from the exponent of the dividend. Let it be required to divide 35a' by 5a'. We must find a quantity which, multiplied by 5a', will produce 35a. Such a quantity is 7a'; for, according to Arts. 49 and 50, 7a X 5a is equal to 35a. Therefore, 35a' divided by 5a2 gives for a quotient 7a; that is, we have divided 35, the coefficient of the dividend, by 5, the coefficient of the divisor, and have subtracted the exponent of the divisor from the exponent of the dividend. (67.) Hence, for the division of monomials, we have the following RULE. 1. Divide the coefficient of the dividend by the coefficient of the divisor. 2. Subtract the exponent of each letter in the divisor from the exponent of the same letter in the dividend. EXAMPLES 1. Divide 20x3 by 4x. 8 Divide 150a'b'cd3 by 30a3b3ď. Ans. 5x2. (68.) The rule given in Art. 66 conducts, in some cases, to negative cxponents. Thus, let it be required to divide a3 by a3. We are directed to subtract the exponent of the divisor from the exponent of the dividend. We thus obtain |