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61. Product of homogeneous expressions. The product of any two homogeneous expressions must be homogeneous; for the different terms of the product are obtained by multiplying any term of the multiplicand by any term of the multiplier, and the number of dimensions in the product of any two monomials is clearly the sum of the number of dimensions in the separate quantities; hence if all the terms of the multiplicand are of the same degree, as also all the terms of the multiplier, it follows that all the terms of the product are of the same degree; and it also follows that the degree of the product is the sum of the degrees of the factors.

The fact that two expressions which are to be multiplied are homogeneous should in all cases be noticed; and if the product obtained is not homogeneous, it is clear that there is an error.

62. It is of importance to notice that, in the product of two algebraical expressions, the term which is of highest degree in a particular letter is the product of the terms in the factors which are of highest degree in that letter, and the term of lowest degree is the product of the terms which are of lowest degree in the factors: thus there is only one term of highest degree and one term of lowest degree.

63. Detached Coefficients. When two expressions are both arranged according to descending, or to ascending, powers of some letter, much of the labour of multiplication can be saved by writing down the coefficients only.

Thus, to multiply 3x2-x+2 by 3x2+2x-2, we write

3-1+2
3+2-2

9-3+6

6-2+4

-6+2-4

9+3-2+6-4

The highest power of x in the product is 4, and the rest follow in order. Hence the required product is

9x+3x3-2x2+6x-4.

When some of the powers are absent their places must be supplied by O's.

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Thus, to multiply - 2x2+x-3 by x2 + x3 − x −3,

we write

1+0−2+1-3

1+1+0-1-3

1+0−2+1−3

1+0-2+1-3

-1-0+2-1+3

-3-06-3+9

1+1-2-2-5-1+5+0+9

Hence the product is

x2 + x2 - 2x - 2x5 - 5x - x3 + 5x2 + 9.

This is generally called the method of detached coefficients.

64. We now return to the three important cases of multiplication considered in Art. 56, namely,

(a + b)2 = a2 + 2ab + b2.

(a - b)2= a2 - 2ab + b2..

(a + b) (a − b) = a2 — b2 ......................

(i),

(ii),

.(iii).

A general result expressed by means of symbols is called a formula.

Since the laws from which the above formulae were deduced were proved to be true for all algebraical quantities whatever, we may substitute for a and for b any other algebraical quantities, or algebraical expressions, and the results will still hold good.

We give some examples of results obtained by substi

tution.

Put - b in the place of b in (i); we then have

that is

{a + (− b)}2 = a2 + 2a (− b) + (− b)2,

(a - b)2 = a2 — 2ab+b2.

Thus (ii) is seen to be really included in (i).

Put √2 in the place of b in (iii); we then have

(a + √2) (a−√2) = a2 — (√2)2 = a2

2.

[We here, however, assume that all the fundamental laws are true for surds: this will be considered in a subsequent chapter.]

Put b+c in the place of b in (i); we then have

{a + (b+c)}2 = a2 + 2a (b+c) + (b+c)2;

2

:: (a+b+c)2 = a2+2ab+2ac + b2 + 2bc + c2..........(iv).

Now put c for c in (iv), and we have

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{a+b+(−c)}2 = a2 + 2ab+ 2a (− c) + b2 + 2b (− c) + (−c)2;

.. (a+b−c)2 = a2+2ab2ac + b2 — 2bc + c2.

Put b+c in the place of b in (iii); we then have

{a + (b + c)} {a − (b + c)} = a2 − (b + c)2 = a2 − (b2 + 2bc +c2); :: (a + b + c) (a − b − c) = a2 — b2 — 2bc — c2.

The following are additional examples of products which can be written down at once.

(a2 + 2b2) (a2 − 2b2) = (a2)2 − (2b2)2 = aa — 4b1.

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(a2 +√3b2) (a2 − √/3b2) = (a2)2 — (√3b2)2=aa — 3b1.

(a − b + c) (a + b − c) = {a − (b − c)} {a+ (b −c)} = a2 − (b − c)2.

(a2 + ab + b2) (a2 − ab+b2) = {(a2 + b2) +ab} {(a2+b2) — ab}

(x2 + x2+x+1)(x3 − x2 + x − 1) =

= (a2 + b2)2 - (ab)2 = a1+a2b2+ba.

{(x3 +x) + (x2 + 1)} {(x3 +x) − (x2+1)}

= (x2 + x)2 - (x2 + 1)2= x2 + 2x2 + x2 - (x2 + 2x2 + 1) = x2 + x1 − x2 -- 1.

65. Square of a multinomial expression. We have found in the preceding Article, and also by direct multiplication in Art. 57, the square of the sum of three algebraical quantities; and the square of the sum of more than three quantities can be obtained by the same methods. The square of any multinomial expression can however best be found in the following manner.

We have to find

(a+b+c+d+...) (a+b+c+d+ ...).

Now we know that the product of any two algebraical expressions is equal to the sum of the partial products obtained by multiplying every term of one expression by every term of the other. If we multiply the term a of the multiplicand by the term a of the multiplier, we obtain the term a of the product: we similarly obtain the terms b2, c2, &c. We can multiply any term, say b, of the multiplicand by any different term, say d, of the multiplier; and we thus obtain the term bd of the product. But we also obtain the term bd by multiplying the term d of the multiplicand by the term b of the multiplier, and the term bd can be obtained in no other way, so that every such term as bd, in which the letters are different, occurs twice in the product. The required product is therefore the sum of the squares of all the quantities a, b, c, &c. together with twice the product of every pair.

Thus, the square of the sum of any number of algebraical quantities is equal to the sum of their squares together with twice the product of every pair.

For example, to find (a+b+c)2.

The squares of the separate terms are a2, b2, c2.

The products of the different pairs of terms are ab, ac and bc.

Hence

Similarly,

(a+b+c)2=a2 + b2 + c2+2ab+2ac+2bc.

(a+2b - 3c)2=a2 +(2b)2 + ( − 3c)2 + 2a (2b) + 2a ( − 3c) +2 (2b) ( − 3c)

= a2+4b2+9c2+4ab-6ac - 12bc.

And

(a − b + c − d)2 = a2 + ( − b)2 + c2 + ( − d)2 + 2a ( − b) + 2ac + 2a ( − d)
+2(-b)c+2(-b) ( − d) + 2c ( − d) = a2+b2 + c2 + d2 – 2ab+2ac
-2ad-2bc+2bd-2cd.

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After a little practice the intermediate steps should be omitted and the final result written down at once. To ensure taking twice the product of every pair it is best to take twice the product of each term and of every term which follows it.

66. Continued Products. The continued product of several algebraical expressions is obtained by finding the product of any two of the expressions, and then multiplying this product by a third expression, and so on. For example, to find (x+a) (x+b)(x+c), we have

x + a

x+b

x2 + ax
+ bx+ab

x2+(a+b)x+ab

x + c

x2+(a+b) x2+abx

+cx2+(a + b) cx+abc

x2+(a+b+c) x2+(ab+ ac + bc) x+abc

In the above all the terms which contain the same powers of x are collected together: it is frequently necessary to arrange expressions in this way.

Again, to find (x2+a2)2 (x+α)2 (x − a)2.

The factors can be taken in any order; hence the required product = [(x − a) (x+a) (x2 + a2)]2= [(x2 − a2) (x2 + a2)]2 = (x1 — a1)2 = x8 – 2a1x2 + a3.

67. We have proved in Art. 55 that the product of any two multinomial expressions is the sum of all the partial products obtained by multiplying any term of one expression by any term of the other.

To find the continued product of three expressions we must therefore multiply each of the terms in the product of the first two expressions by each of the terms in the third; hence the continued product is the sum of all the partial products which can be obtained by multiplying together any term of the first, any term of the second, and any term of the third.

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