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69. I. The rule for dividing one simple expression by another will be obtained from an examination of the corresponding case in Multiplication.

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Hence we have the following rule for dividing one simple expression by another; remove from the dividend all the factors which occur in the divisor, and prefix the sign + if the two expressions have the same sign, and the sign- if they have different signs.

70. Thus it will be seen that the Rule of Signs holds in Division as well as in Multiplication.

71. It may happen that the factors of the divisor do not occur in the dividend; in this case we can only indicate the division by the notation which we have appropriated for it. Thus, if 5a is to be divided by 3c, the quotient can only be indicated by 5a÷2c, or by

Ба

20

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Again, it may happen that some of the factors of the divisor occur in the dividend, but not all of them; in this case the expression for the quotient can be simplified by a

15a2b denoted by 6bc

principle already used in Arithmetic. Suppose, for example, that 15ab is to be divided by 6bc; then the quotient is Here the dividend 15a2b=5a2 × 3b; and the divisor 6bc=2c × 3b; thus the factor 36 occurs in both dividend and divisor. Then, as in Arithmetic, we may remove this common factor, and denote the quotient

5a2

by ; thus

2c

15a2b

5a2

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72. One power of any number is divided by another power of the same number, by subtracting the index of the latter power from the index of the former.

For example, suppose we have to divide a5 by a".

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In like manner the rule may be shewn to be true in any other case.

Or we may shew the truth of the rule thus;

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73. If any power of a number occurs in the dividend and a higher power of the same number in the divisor, the quotient can be simplified by Arts. 71, and 72. Suppose, for example, that 4ab2 is to be divided by 3cb5; then the 4ab2 quotient is denoted by The factor 62 occurs in both 3cb5

dividend and divisor; this may be removed, and the quo

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4ab2 4a

thus

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74. II. The rule for dividing a compound expression by a simple expression will be obtained from an examination of the corresponding case in Multiplication.

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Hence we have the following rule for dividing a compound expression by a simple expression; divide each term of the dividend by the divisor, by the rule in the first case, and collect the results to form the complete quotient. 4a3-3abc + a2c

For example,

α

=

4a2 - 3bc+ac.

75. III. To divide one compound expression by another, we must proceed as in the operation called Long Division in Arithmetic. The following rule may be given. Arrange both dividend and divisor according to ascending powers of some common letter, or both according to descending powers of some common letter. Divide the first term of the dividend by the first term of the divisor, and put the result for the first term of the quotient; multiply the whole divisor by this term and subtract the product from the dividend. To the remainder join as many terms of the dividend, taken in order, as may be required, and repeat the whole operation. Continue_the process until all the terms of the dividend have been taken down.

The reason for this rule is the same as that for the rule of Long Division in Arithmetic, namely, that we may break the dividend up into parts and find how often the divisor is contained in each part, and then the aggregate of these results is the complete quotient.

76. We shall now give some examples of Division arranged in a convenient form.

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a2-2ab+3b2)3a* - 10a3b+22a2b2 — 22ab3 + 15b1(3a2 - 4ab+5b2

3a4-6a3b+9a2b2

-4a3b+13a2b2-22ab3
-4a3b+ 8a2b2 – 12ab3

5a2b3-10ab3+ 15b4
5a2b2-10ab3 + 15b4

Consider the last example. The dividend and divisor are both arranged according to descending powers of a. The first term in the dividend is 3a and the first term in the divisor is a2; dividing the former by the latter we obtain 3a2 for the first term of the quotient. We then multiply the whole divisor by 3a2, and place the result so that each term comes below the term of the dividend which contains the same power of a; we subtract, and obtain −4a3b+13a2b2; and we bring down the next term of the dividend, namely, -22ab3. We divide the first term, -4a3b, by the first term in the divisor, a2; thus we obtain -4ab for the next term in the quotient. We then multiply the whole divisor by -4ab and place the result in order under those terms of the dividend with which we are now occupied; we subtract, and obtain 5a2b2-10ab3; and we bring down the next term of the dividend, namely, 1564. We divide 5a2b2 by a2, and thus we obtain 562 for the next term in the quotient. We then multiply the whole divisor

by 562, and place the terms as before; we subtract, and there is no remainder. As all the terms in the dividend have been brought down, the operation is completed; and the quotient is 3a2-4ab+5b2.

It is of great importance to arrange both dividend and divisor according to the same order of some common letter; and to attend to this order in every part of the operation.

77. It may happen, as in Arithmetic, that the division cannot be exactly performed. Thus, for example, if we divide a2+2ab+2b2 by a+b, we shall obtain, as in the first example of the preceding Article, a+b in the quotient, and there will then be a remainder b2. This result is expressed in ways similar to those used in Arithmetic; thus we may say that

a2+2ab+2b2
a+b

=a+b+

b2

a + b ;

that is, there is a quotient a + b, and a fractional part

b2

a + b In general, let A and B denote two expressions, and suppose that when A is divided by B the quotient is q, and the remainder R; then this result is expressed algebraically in the following ways,

A=qB+R, or A-qB=R,

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The student will observe that each letter here may represent an expression, simple or compound; it is often convenient for distinctness and brevity thus to represent an expression by a single letter.

We shall however consider algebraical fractions in subsequent Chapters, and at present shall confine ourselves to examples of division in which the operation can be exactly performed.

78. We give some more examples.

Divide a-5a3 +7x3 + 2x2-6x-2 by 1+ 2x − 3x2 + xa.

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