of 3, a, and b is 3ab; we may also take 3 and ab for the factors of 3ab, but ab is a composite factor. For a second example, we shall take 15a2bc; here the simple factors are 3, 5, a, a, b, b, b, c, since their product is 15a2bc. For another example, we shall take ac + be; here, since c enters as a factor into each term of the given quantity, it is easy to see that a + b and c are the prime factors required, and we have (a + b) × c = ac + bc. For another example, we shall take 3abc-15abcde + 45abc; here, since 3abc enters as a factor into each term of the given quantity, it is evident that 3a2bc and ba— 5de + 15 are factors of the given expression, or that we shall have (b2 — 5de + 15) × 3abc3a2bc-15a2bcde + 45abc; where it is to be noted that 3abe is composite, its factors being 3, a, a, b, and c. EXAMPLES. 1. Find the prime factors of 21abcd. 5. Resolve 25amn + 75anpq-10a3n pq into factors, and give all the prime factors. 6. Resolve 21a2x2 + 6ax-12aa into factors, and give all the prime factors. 7. Resolve 13a3 - 16963 into factors. 8. Resolve 42a3x5 - 28ax 5ax into factors. 9. Resolve - 92m3 √n — 69m3n + 15n into factors, and find the prime factors. 10. Has 97a-172963 any factors? (14.) We shall now proceed to show how to multiply one compound quantity by another. We shall denote the sum of the positive terms of the multiplicand by a, and the sum of the negative terms by b; we shall also represent the sum of the positive terms of the multiplier by c, and the sum of the negative terms by d; then the multiplicand will be expressed by a - b, and the multiplier by c-d. Then the product of a-b and c will be expressed by (a - b)c, which is clearly greater than the product of a — b by cd, since c is greater than c-d by d. Hence we must multiply ab by d, and subtract the product from (a - b)c, since ab is taken too often by the units in d, and of course we shall have (a — b) . (c — d) = (a — b)c — (a — b)d. But by the rule for the multiplication of a compound quantity by a monomial, we have (a — b)c = ac — bc, and (a - b)dad-bd; which, subtracted from ac-bc, gives (a - b). (cd) = ac-be- ad + bd, as required. From an examination of the product, it appears that each term of ab has been multiplied by each term of c-d, and that like signs have given +, and unlike signs have given for the corresponding products. Here, it may be proper to notice, that a monomial which is positive is generally written without any sign; also in a compound quantity, if the first or left-hand term is positive, it is generally written without any sign, the sign + being in these instances understood; but in other cases positive terms (generally) have the sign + written before or to the left of them. If any quantity or term is negative, it is always written with the sign before or to the left of it, unless the contrary is expressly stated. It will be convenient to premise the following observations, before giving the rule for the multiplication of compound quantities. 1. A compound quantity, whose terms involve different powers of the same letter, is said to be arranged according to the descending powers of the letter, when the term that involves the letter with the greatest positive exponent is written first or to the left, and the term which involves the next greatest positive exponent of the letter is written next to it on the right, and the term that contains the next greatest positive exponent of the letter is written next to the last term on the right, and so on; but if the terms of the compound quantity are written in a reverse or contrary order, it is said to be arranged according to the ascending powers of the letter. Thus, -7 +5x2 - 3x + 18, is arranged according to x1 the descending powers of ; but if we write the quantity in the reverse or contrary order, it becomes 18 - 3x + 5x2. 7x+x1; and it is arranged according to the ascending powers of x. In like manner, A+ Bata - Cx3a2 + Daa3, in which A, B, C, D are the coefficients, is arranged according to the descending powers of x; but if we write the quantity in a contrary order, it will be arranged according to the ascending powers of x. Again, since a enters the second term, and a2 enters the third term, and a3 enters the fourth term, counting from the left to the right, the quantity is arranged according to the descending powers of x, or according to the ascending powers of a, and so on in all analogous cases. 2. If the exponents of the letter, according to which any compound quantity has been arranged, are integers, and if they differ from each other by 1, in any two successive terms, we say the compound quantity is complete, or wants none of its terms; but if the difference of the exponents of the letter in two successive terms is greater than 1, we say that the compound quantity is incomplete, or wants some of its terms. 3. When any of the terms of a compound quantity are wanting, we may suppose them to have vanished from the quantity on account of their coefficients becoming equal to naught or zero (0); since any quantity or term when multiplied by naught, must evidently give naught for the product. Hence we may give any incomplete compound quantity a complete form, by supplying the deficient terms with the powers of the letters that are wanting, by writing 0 (or naught) for their coefficients; observing that the terms supplied may evidently have either the sign or at will. - 4. Not regarding the coefficients of the terms in any compound quantity, if the sum of the exponents of the letters in any one term is the same that it is in any other term; then the compound quantity is termed a homogeneous quantity; otherwise, it is said to be a heterogeneous quantity. Indeed, any monomials which are such that the sum of the exponents of their letters (that are not regarded as coefficients) is the same in each, are called homogeneous; otherwise, they are heterogeneous. Thus, 7a6 and 13m3n are homogeneous, as are also abe, pq, rst; but ab m3n are heterogeneous, and so are rs, pq, When deficient terms are supplied in compound quantities whose terms are homogeneous, and composed of different letters, it will be proper that the terms supplied shall be homogeneous with the given terms of the quantity, and in its letters. Thus, the compound quantity a+b3 has deficient terms; which being supplied, it becomes a3 + 0a3b + Oab+b3, which is the complete form of the compound quantity. From the definition of homogeneous terms, it is clear that the multiplication of homogeneous terms by terms that are also homogeneous, will give products that are homogeneous. Thus, if we take the homogeneous terms - 3ab and 2cd, and multiply them by the homogeneous terms 4p3q' and 5vw, we shall get the homogeneous products 12a bp3q and 10cdow, where it will be noticed that the sum of the exponents of the letters in each term multiplied is 3, and the sum of the exponents of the letters in each multiplier is 5, and we have 3+58 the sum of the exponents of the letters in each product, as it evidently ought to do. = 5. When we have to use compound quantities involving different powers of the same letter, or letters, in calculation, it is often convenient to use the coefficients only, and, after the work has been done, to supply the powers of the letter, or letters, that ought to correspond to the coefficients in the result that has been found; this process is generally called the method of detached coefficients. Thus, to add 7a1 — 9a3b + 13a2b2 — 17ab3 + 6b1, 9a1 + 6a3b + 20ab2 + 4ab3 + 56 and 5a2 + 15ab26ab3; by detaching the coefficients, supplying the deficient terms, and putting the coefficients that belong to like terms under each other, and then adding, we get the sum of the coefficients; where 21 is the coefficient of a1 in the sum, - 3, that of ab, and so on: so that the sum, when the requisite letters are supplied, is 21a* — 3a3b + 48a2b-19ab3 + 116, as it ought to be. In a similar way, to subtract 5a2 - 7ab from 15a2 — 9ab + b2, by detaching the coefficient, etc., we get 159 +1 57+0 10-2+1 the difference of the coefficients; and the letters, when supplied, give 10a2 - 2ab+b2 for the required difference of the given quantities. For further illustration of the method of detached coefficients, we shall add the quantities 9a + 6a2 + 3a + 5, 4a3+5a+6, and Sa2 + 7a + 3; detaching the coefficients, supplying the deficient terms, and placing the coefficients of like terms under each other, and then adding the coefficients as in Arithmetic, we get 13 +14 + 15 + 14 the sums of the coefficients. Supplying the letters that correspond to the sums of the coefficients, we get 13a2 + 14a2 + 15a + 14 for the sum of the quantities, as required. If a = 10, the sum becomes a + 4a3 + 5a2 + 6a + 4 = 14564, which will be found to equal the sum of the numbers 9635, 4056, 873, which are the values of the given quantities when a 10. This example may serve to show the analogy that exists between the different powers of a, and the local values of figures in Arithmetic; it also shows the analogy between the naughts that are used to supply the deficient terms and the naughts that are used in Arithmetic to show the local values of figures. Again, we shall subtract 5a + 3a3 + 9a + 9 from 8a+ 6a3 + 3a + 7, by the method of detached coefficients. Detaching the coefficients, etc., we get |