a. If the addendum or multiplicand, m, is added to itself 2, 3 times, the sums of the repetitions m+m, m+m+m, to m' terms are indicated by the combinations of symbols; m+m+m m' 2×m or 2m, 3×m or 3m, m'×m or m'm or m'm. The numeral factor or coefficient is always prefixed to the literal factor; but the product of two literal factors m, m', is written, indifferently, m'm or mm'. In effect it is proved (Part I. Art. 71) that the product of any given factors expresses the sum of a certain number of units, and that this sum is not made greater or less by mere changes of arrangement. b. Let the literal factors m, m' have numeral coefficients, and let the multiplicand be 6m and the multiplier 4m'; the product is indicated by the expression 6m x 4m', or by 6xmx4xm', or by 6×4xmxm' (Part I. Art. 71), or, since 6×4=24, by 24 mm'. c. If m=m', the product mm is written m2; and by the same convention the product of 3, 4 . . . . n factors equal to m are written respectively, m3, m4 m". If m3 or mmm is multiplied by m3 or mmmmm, the product m3×m3= mmm × mmmmm, or mmmmmmmm, or m3, is obtained by adding together 3 and 5, the exponents of the multiplicand and multiplier, and affecting the root m with an exponent equal to the sum. And if m or mmm ...to e repetitions of m is multiplied by me or mmmm to e' repetitions of m, the product is, in like manner, indicated by the expression m+ (Part I. Art. 75). Hence the product of the monomial factors 5a2b3cd2, 7ab2c5 is 35a3b5c6d2. For 5a2bcd2x7ab2c5=5×a2 × b3 × c × d2 × 7×a× b2 × c5=5×7 × a2 ×a× b3 xb2 xcxc5 x ď2. Now 5x7=35, a2 ×a=a3, b3×b2=b5, c>c5=cổ. Making these substitutions, 5a2b3cd27ab2c5=35a3b5c6d2. d. The quantities whose products have been found are additive. Either factor or both factors may, however, be subtractive. The formulæ of Article 4 c, which are, (±m)×(±m')=+mm', serve to determine the sign of the product in every case. Whence, to multiply a literal quantity by any particular number, write the numeral factor on the left of the literal quantity. To multiply together any unlike literal factors, write them one after another in any order (the alphabetical order of the letters is generally followed). To multiply together any like literal factors, write each of the letters found in the factors once only, and affect each letter with an exponent equal to the sum of the exponents with which it is affected in all the factors. If any of the quantities have numeral coefficients, write the product of the coefficients on the left of the product of the literal factors. The following is, consequently, the rule for the multiplication of one monomial quantity by another: 1st. Multiply together the two coefficients. 2d. Annex to this product all the letters which are found in both the multiplicand and multiplier, affecting each letter with an exponent equal to the sum of the two exponents with which the letter is affected in the two factors. 3d. If a letter is found in one only of the factors, write it in the product with the exponent with which it is affected in that factor. 4th. If the monomials are both additive or both subtractive, the product is additive; but if one of them is additive and the other subtractive, the product is subtractive. It is required to find the products of the following monomial quantities; 1st. ab and abc?.... 2d. 3d. -5abc and 7ade ?... 4th. abd and -4cdef?. .........Ans. a2b2c. ..Ans. —35a2bcde. ...Ans. -15a3x1y2. ....Ans. abcd'ef. ..Ans. 21a5b4c2d2. .Ans. -91a7b3cd2. ........Ans. 10a3b1c2. .......Ans. 168a1b3c1d. 10. The product of a polynomial multiplicand, a-b+c, by a monomial multiplier, m, is indicated by the expression (a-b+c)m. The product can be obtained by writing the multiplicand in a horizontal line writing it again term under term in a second horizontal line, repeating this process as often as the multiplier m contains unity, and reducing the like terms, thus, ma b + b + b c3d repetition of the multiplicand. + cmth mb + mcm repetitions of the multiplicand. The first column containing m repetitions of the additive quantity a, its sum is ma. The second containing m repetitions of the subtractive quantity b, its sum is -mb. The third containing m repetitions of the additive quantity c, its sum is mc. Whence the sum of m repetitions of the polynomial a-b+c is ma-mb+mc; in other words, the product of a-b+c by m is ma-mb+mc. It is manifest that by this process each term of the multiplicand is repeated as often as the multiplier contains unity; otherwise, that each term of the multiplicand is multiplied by the multiplier, and that the sum of the partial products is the product required. Making the multiplier a subtractive quantity (m). 1st. The sum of m terms each equal to a, that is ma, is to be taken subtractively, a condition expressed by writing the quantity thus, -ma. 2d. The sum of m terms each equal to -b (or-mb) is to be taken subtractively. Now -mb, taken subtractively, becomes +mb (Art. 4 b). 3d. The sum of m terms, each equal to c, taken subtractively is ―mc; therefore (a-b+c)×(−m)=—ma+mb—mc; now (a-b+c)× (+m)=ma—mb+mc; therefore (a-b+c)x(+m)=+ma+mb+mc. The formation of the product of a polynomial by a monomial is hence reduced to a series of operations with monomial factors, in which the rule for the multiplication of such quantities (Art. 9) is followed. Hence, to multiply a polynomial by a monomial quantity, Rule. Form the product of each term of the multiplicand by the multiplier, by the rule for the multiplication of monomial factors, and take the algebraic sum of the partial products for the result required. 248 Examples of the multiplication of polynomial by monomial quantities: 1st. (6a+3b-5f)×5b?. .....Ans. 30ab+15b2—25bf. Ans. 70a2bd-150ab2c-160a2bcf. 4th. (2a1-3ba3—5b2a2+b3a)×a3 ? Ans. 2a7-3ab—5a3b+a1b3. ......Ans. a1b—a3b2+a2b3—ab1. .Ans. am+n—a” b2 +a"c". 5th. (a3-abab2—b3)× ab?. 6th. (am-b+c") by a" ?.. 11. If the multiplicand is a monomial and the multiplier a polynomial quantity, the order of the factors may be changed, and this case reduced to the preceding. Otherwise, making m the multiplicand and a+b-d the multiplier, the product mx (a+b-d) can be obtained by taking a repetitions of m, then b repetitions of m, then -d repetitions of m, and forming the algebraic sum of all the terms, thus, m=1st m=2d m=3d ―m=a+b-d The column composed of m contains, 1st. a repetitions of m, taken additively, or am; 3d. d repetitions of m, taken subtractively, or -dm. Since the column contains m as often as the multiplier contains unity, it is evident that the algebraic sum of the column is equal to the product of the multiplicand and multiplier. The algebraic sum of the column is am+bm―dm ; Therefore mx (a+b−d)=am+bm—dm. It seems unnecessary to repeat the process for a subtractive multiplier, -(a+b-d), or the rule, which is the same as that deduced from the preceding case. 12. Let both multiplicand and multiplier be polynomial expressions, and let the former be denoted by a-b+c-d, and the latter by m—n+r—s. If the product is formed by addition, it is evident that the product of the multiplicand by the multiplier is composed of the product of each term of the multiplicand by the polynomial multiplier, and that the product of a term of the multiplicand by the polynomial multiplier is composed of the product of that term by each term of the polynomial multiplier, as in Article 11. Hence the product of two polynomials is composed of the product of every term of the multiplicand by every term of the multiplier. The partial products are found by the rule of Article 9. Otherwise, since a-b+c-d=a+c-b-d=a+c-(b+d), and m―n+r—s=m+r¬n—s=m+r−(n+s); and since a+c, or the sum of the additive terms of the multiplicand, can be represented by a single letter, p, and b+d, the sum of the subtractive terms, by another letter, q, a multiplicand which is composed of terms of which some are additive, others subtractive, may be represented by the expression p-q. Similarly, a multiplier composed of terms of which some are additive and others subtractive may be represented by the expression p'-q'. Now the product of p-q by p'-q is obtained by forming the sum of p-q repetitions of p-q, or by forming the sum of p' repetitions of p-q, then the sum of q' repetitions of p-q, and subtracting the second quantity from the first. Now p' repetitions of p-q=pp-pq (Art. 10), and repetitions of p-q=pq-qq'· Subtracting the second of these expressions from the first, (p−q) p'—(p−q) qʻ, or (p−q) (p′—q'), =p′p—p'q—pq'+qq'. The signs of the partial products obtained in this manner agree with the signs obtained from the rule for the multiplication of monomials having the same or opposite signs (Art. 4 c). This reasoning is applicable to polynomials composed of any number of terms. Whence, to find the product of two polynomial factors, Rule. Multiply every term of the multiplicand by every term of the multiplier, and make reduction of the similar terms. a. In forming the product of two polynomial factors, the partial products and their algebraic sum are independent of any particular arrangement of the terms of the multiplicand and multiplier; but the reduction of the like terms among the partial products is facilitated by a particular arrangement of the terms of the factors whenever the same letter raised to different powers is found in several terms of both the factors. This arrangement consists in placing in both factors the term which contains the highest power of the letter which is common to many terms on the left; the term which contains the next lower power of the same letter, second; the term which contains the next lower power, third, &c.; and lastly, on the right, the term or terms which do not contain this letter. Factors arranged in this manner are said to be arranged according to the descending powers of the same letter. If this order is reversed, the factors are said to be arranged according to the ascending powers of the same letter. Both arrangements are equally convenient, and sometimes both are combined in the same expression, as in a3+a+x+a3bx2-a2cx3-ac2x2+x3, which is arranged according to the descending powers of a and the ascending powers of x. Examples of the multiplication of polynomial quantities: 1st. 2d. 3d. (a+b)x(a+b)? (a+b)x(a—b)? (a—b)x(a—b)? 4th. (a+b)x(c+d)?.. ..Ans. a2+2ab+b2. ......Ans. a22-2ab+b2. ....Ans, ac+be+ad+bd. Ans. 2112-Im-271+22m2+99m. 5th. (a+b-c) x (d—e)?.........Ans. ad+bd-cd-ae-be-ce. 6th. (77-2m-9) × (31—11m)? 7th. (1-x+x2—x3)× (1+x)?. 8th. ....Ans. 1-x1. (a2+2ab+b2)x(a2-2ab+b2)?.........Ans. a1-2a2b2+b1. 9th. (4a3-5a2b—8ab2+2b3)× (2a2—3ab-4b2) ? Ans. 8a5-22a1b—17a3b2+48a2b3+26ab1—8b5. 10th. (3a2-5bd+cf)×(−5a2+4bd-8cf)? Ans. —15a1+37a2bd-29a2cf—20b2d2+44bcdf—8c2ƒ2. 11th. (3a3-5a2b-4ab2)x(-2a2-3ab+b2)? Ans. -6a+ab+19a3b2-3a2b3—}ab1. 12th. (a3+3a2b+3ab2+b3)×(a3—3a2b+3aba—b3) ? Ans. a6-3ab2+3a2ba—b®. 13th. (5a-2a3b+4a2b2) × (a3-4a2b+2b3)? 14th. Ans. 5a7-22a6b+12a5b2—6a4b3—4a3b1+8a2b3. (3a+5b-3c)x(a-2b+9c)? Ans. 3a2ab+7ac—10b2+52bc—63 c2. 15th. (x2—px + q) × (x + a) ? Ans. x3-(p-a)x2+(q—ap)x+aq. 16th. (am—a”+a2)×(am—a)? Ans. a2m-am+n+am+2—am+1+an+1—a3. 12'. Remarks on the multiplication of polynomial quantities: 1st. If the polynomial factors are composed of homogeneous terms, the product is composed of homogeneous terms; and the degree of each term of the product is equal to the sum of the degrees of any two terms, one taken in the multiplicand, the other in the multiplier; for, by the rule for the multiplication of monomial factors, the sum of the exponents in the product of two factors is equal to the sum of the exponents of the two factors. This remark may be employed to detect errors of calculation with respect to the exponents. 2d. Since the product of two polynomials is obtained by multiplying every term of the multiplicand by every term of the multiplier, it follows that when no reduction of like terms can be made, the number of terms of the product is equal to the product of the number of terms of the multiplicand by the number of terms of the multiplier. 3d. When some of the partial products are like, the whole number of terms of the reduced product may be less than the product of the number of terms in the multiplicand by the number of terms in the multiplier; but among the partial products there are at least two which cannot be reduced with any other; these are, the term arising from the multiplication of that term of the multiplicand which is affected with the highest exponent of one of the letters by that term of the multiplier which is affected with the highest exponent of the same letter, and the term arising from the multiplication of the two terms affected by the lowest exponents of the same letter; for these partial products must contain this letter, the first with a greater and the second with a less exponent than any of the other partial products. Consequently, these terms cannot be similar to any of the other partial products, and therefore cannot be reduced with them. DIVISION. 13. In this operation the data are a product and one of its factors, and that which is required is to find the other factor. Let 56a6bc3 and 8a2b+ be a product and one of its factors, and let it be required to find the other factor; or, in other words, let it be required to divide 56a6b+c3 by 8ab4. 56a6b4c3 The division is indicated thus, 84, and the required factor or quotient ought to be a quantity such that its product by 8a2ba is 56a6b+c3. Since the product of a polynomial by a monomial is a polynomial, and that the dividend is a monomial, it is evident that the quotient can contain only one term. Now it follows, from the rule for the multiplication of monomial factors, that 56, the coefficient of the product or dividend, is the product of the coefficients of the divisor and quotient; dividing, therefore, 56 by 8, the coefficient of the quotient is 7. From the same rule it follows, that the exponent, 6, in the dividend is the sum of the exponents of the letter a in the divisor and quotient; therefore the difference 6–2, or 4, is the exponent of |