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8 + 6 + 0 + 3 + 7
5+3+0+9+9

3+3+06-2;

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and restoring the letters, we get 3a + 3a3 + 0a3 - 6a -2 = the difference of the given quantities.

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When a 10, we have 2a-8, 6a-4, and the difference becomes 3a1 + 3a3 + Oa2 — a (a — 4) — (a — 8) = 3a1+3a3 +0a2 — a2 + 3a + 8 = 3a1 + 2a3 + 9a2 + 3a +8= 32938, which equals the difference between the numbers 86037 and 53099, which are the values of the given quanti-, ties when a = : 10.

We have given the preceding examples for the purpose of showing the simplicity of the method of detached coefficients in addition and subtraction. We shall now give one or two simple examples to show the great importance of the method in multiplication.

For a first example, we shall multiply 3a3 + 2a2 + 5a + 6 by 9a1; first by the common method, and then by detaching the coefficients.

Since we must take the multiplicand as often as there are units in 9a+1, it is clear that if we multiply the multiplicand by 9a and then by 1, and add the products, we shall have multiplied the multiplicand by 9a + 1, as required.

Hence we have

3a3 + 2a2 + 5a + 6

9a + 1

27a+18a+ 45a2 + 54a

3a3 + 2a2 + 5a + 6

27a*+ 21a3 + 47a2 + 59a + 6

the sought product, as found by the common method, which consists in multiplying each term of the multiplicand (observing the rule of signs) by each term of the multiplier, and then adding the products by the rules of addition.

We shall now multiply by the method of detached coefficients.

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3 + 2 + 5+ 6

9+ 1

27+18+45 +54

3 + 2 + 5 + 6

27 + 21 + 47 + 59 + 6

the sum of the products of the coefficients; and if we supply the requisite powers of a, we readily obtain the same result that we found by the common method of multiplication. If a 10, the sum of the coefficients is easily changed to the following sum of coefficients, viz., 2 + 9 + .6 +2 +9 +6, which, by joining the proper powers of a to each term, becomes 2a + 9a* + 6a3 + 2a2 + 9a + 6 = 296296, which equals the product of 3256 by 91, as it ought to do; for when a 10, the given multiplicand equals 3256, and the multiplier equals 91.

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For the second example, we shall find the product of 8a5a+7 and 3a2 + 4. Detaching the coefficients, etc., we get

8+ 5+ 7 3+0+ 4

24 + 15 +21

0 + 0 + 0

32+20 +28

24 + 15 + 53 + 20 + 28

the sum of the coefficients; which, by supplying the appropriate powers of the letter, gives 24a+15a3 + 53a2 + 20a28 for the sought product.

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If a 10, it is easy to change the sum of the coefficients. into the sum of the coefficients 2+ 6 + 0 + 5 + 2 + 8, which, by supplying the proper powers of a, gives 2a + 6a1 +0a3 + 5a2 + 2a + 8 = 260528, which is the product of 857 and 304, which are the values of the multiplicand and multiplier when a 10.

The analogy of the above process, in the use of the naughts and the arrangement of the work, to the method of multiplying integral numbers, etc., in Arithmetic, is too evident to need any comment.

For the last example, we shall find the product of a3 + 3ab + 1363 and a2 + 262. Detaching the coefficients, etc., we get the following work and sum of coefficients :

1+ 3 + 0 + 13
1+0+2

1+3+0+ 13

0+0+0+0

2+ 6+0+26

1+ 3+2+19 + 0 + 26

the sum of the coefficients. When the requisite powers of the letters are supplied, we have a + 3a1b + 2a3b2 + 19ab0ab1 + 26b3 = að + 3a1b + 2a3b2 + 19a2b3 + 2665 = the required product, since the coefficient of ab* equals naught.

It may be noticed that the terms of the multiplicand are homogeneous in a and b, and the terms of the multiplier are homogeneous in the same letters, and that the quantities are arranged according to the descending powers of a, the exponents decreasing by a unit from each term to the next successive term on the right; and that the quantities are arranged according to the ascending powers of b, the exponents increasing by a unit from each term to the next successive term on the right; and since the sum of the exponents of the letters in each product is 5, and as a alone enters the first or left-hand term of the product, it is easy to see how to supply the powers of the letters that belong to each of the coefficients in the product.

6. Reversely, if we have to find the product of quantities which do not involve different powers of the same letter or letters, we may supply the powers of a letter or letters according to the nature of the case, so as to reduce the process to the method of multiplying quantities that involve different powers of the same letter or letters.

For we may regard the given terms as being the coefficients of different powers of a letter or letters, which have each been put equal to unity.

Thus, if we have to multiply a + b by c +d, we may multiply ax+bby cx+d, and putting a 1 in the product,

we shall of course have the product of a + b and c + d. Multiplying ax + b by cx+d, we have (evidently) the following process and result:

ax + b
cx + d

adx + bd

acx2+(cb + ad)x+bd

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= the product; which, by putting a 1, gives ac + cb + ad+bd for the product of a + b and c + d.

In a similar way, if we have to multiply 3a +56 + 7d by 4a6b+d, we may multiply 3ax2+5bx+7d by 4ax2 + 6b+d; and we get

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4αx2 + 6bx + d

12ax+20abx+28ada

18abx3 + 30b2x2 + 42bdx

3adx2+5bdx + 7d2

12a2x2+38abx3 + (31ad+30b2)x2 + 47bdx + 7d2

the product; and putting ax = 1, we have (3a + 5b+7d) × (4a + 6b + d) = 12a2 + 38ab +31ad +3062+47bd +7d2, as required.

7. It is easy to see that the number of products arising from the multiplication of one compound quantity by another will equal the product of the number of terms of the one quantity by the number of terms of the other quantity; since the products are formed from the multiplication of each term of the one quantity by each term of the other.

Consequently, if the products are all unlike, there will be as many terms in the sum of the products, or in the product of the two compound quantities, as there are units in the product of the number of terms of the one by the number of terms in the other. But if the products are not all unlike, two or more of the products will be reduced to one by addition, so that there will not be so many terms in the product of the two quantities as before.

Thus we have found that (a + b) (c + d) = ac + cb + ad +

bd, which contains four terms; this is in conformity to what has been stated above, since there are two terms in a + b and two terms in c + d, ... we have 2 × 2 = 4 = the _number of terms in the product of the two quantities.

Also, there are three terms in 3a +56 + 7d, and three terms in 4a + 6b + d, . · . 3 × 3 = 9 = the number of products arising from the multiplication of the one compound quantity by the other; but we have found that (3a +56 + 7d) x (4a + 6b + d) = 12a2 + 38ab+ 31ad +30b2 + 47bd + 7d2, which contains only six terms; since 20ab and 18ab have been reduced to 38ab, 28ad and 3ad to 31ad, 42bd and 5bd to 47bd; these reductions clearly result from some of the terms of one of the compound quantities being similar to some of the terms of the other.

But there are terms of the partial products which can not be united with other products. For, when we multiply the terms of the two quantities that contain the greatest positive exponents of the same letter by each other, the product evidently will be unlike any other terms of the partial products; and of course the product will appear in the final product without reduction; also, if there are terms of the quantities which do not involve any letters (or that contain their lowest powers, it is clear that their product must appear in the final product without reduction.

Thus, if we have (3+ 7+ 8x + 9) × (6x2 + 11x+8), we have the products 3 × 6x2 = 18x5 and 9 × 8 = 72, which must evidently appear in the complete product without reduction.

8. By 15, Sec. I., when we wish to show that all the terms of any compound quantity are to be taken together as a single quantity or letter, we draw a right line (called a vinculum or bar) over them, or we write them in a parenthesis. Hence we may evidently express the product of a compound and single quantity, or of compound quantities, in a way very similar to that of letters.

Thus, to show that ab + c is to be multiplied by m, we write either of the following expressions, viz., ab + c × m, a−b+c.m, (a − b + c) × m, (a − b + c) . m ; or, more simply, (a - b + c)m.

In a similar way, to indicate the product of a-b+c and

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