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the factors 12, a3, b3, and c may be divided respectively by the factors 4, a, b, and c, or we may divide the two terms of the fraction by the numerator 4 a2 b c. Now the quantity 4 a2 bc, divided by itself, gives 1 for the quotient, and the quantity 12 a2 b3 cd, divided by the first, gives by the above rules 3 b2 d; the new fraction then is

1

3 b2 ď

40. It follows from the rules of multiplication, that when a compound quantity is multiplied by a simple quantity, this last becomes a factor common to all the terms of the former. We may make use of this observation to simplify fractions of which the numerator and denominator are polynomials, having factors that are common to all their terms.

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by examining the quantity 6 a 3 a2bc + 12 a2 c2, we see that the factor a2 is common to all the terms, since a4a2 X a3, and that, besides, 6, 3, and 12 are divisible by 3; so that,

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3a2be12a3c2 = 2aa × 3a2

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bc X 3a2 + 4c2 × 3a3. Also the denominator has for a common factor 3 a2; for the factors a2 and 3 enter into all the terms, and we have

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9ab15a2c + 24 a3 3b × 3a2 — 5c X 3a2 + 8 a× 3a2. Suppressing therefore the 3a2 as often in the numerator as in the denominator, the proposed fraction will become

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41. I pass now to the case where the numerator and denominator are both compound, and in which one cannot perceive at first whether the divisor is or is not a factor of the dividend.

As the divisor multiplied by the quotient must produce the dividend, it is necessary that this last should contain all the several products of each term of the divisor by each term of the quotient; and, if we could find the products arising from each particular term of the divisor, by dividing them by this term, which is known, we should obtain those of the quotient, after the same manner as in arithmetic we discover all the figures of the quotient by dividing successively by the divisor the numbers, which we regard as the several products of this divisor by the

different figures of the quotient. But in numbers the several products present themselves in order, beginning with the units at the last place on the left, on account of the subordination established between the units of each figure of the dividend according to the rank which they hold. But as this is not the case in algebra, we supply the want of such an arrangement by disposing all the terms of the dividend and divisor in the order of the exponents of the power of the same letter, beginning with the highest and proceeding from left to right, as may be seen with reference to the letter a in the quantities

5 a7 — 22 a® b + 12 a5 b2 6 ab3

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4 a2 b1 + 8 a2 b3, 5 a4 2a3b+4a2 b3,

of which one is the product and the other the multiplicand in the example of art. 32. This is called arranging the proposed quantities.

When they are thus disposed, it is evident, that whatever be the factor by which it is necessary to multiply the second to obtain the first, the term 5 a7, with which this begins, results from the multiplication of 5 aa, with which the other begins, by the term in the factor sought, in which a has the highest exponent, and which takes the first place in this factor when the terms of it are arranged with reference to the letter a. By dividing then the simple quantity 5 a7 by the simple quantity 5 a, the quotient a3 will be the first term of the factor sought. Now as the entire product ought by the rules of multiplication to contain the several particular products arising from the multiplication of the whole multiplicand by cach term of the multiplier, it follows that the quantity here taken for the dividend, ought to contain the products of all the terms of the divisor, 5 a 2 a3b + 4 a2 b3, by the first term of the quotient a3; and consequently, if we subtract from the dividend these products, which are 5 a7-2ab+4a3b3, the remainder - 20 a b + 8 as b2 — 6 a1 b3 — 4a3 b1 +8 a3 b5 will contain only those, which result from the multiplication of the divisor by the second, third, &c, terms of the quotient.

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The remainder then may be considered as a part of the dividend, and its first term, in which a has the highest exponent, cannot be obtained, otherwise than by the multiplication of the first term of the divisor by the second term of the quotient. But the first term of this part of the dividend having the sign it is necessary to assign that which is to be prefixed to the corres-pouding term of the quotient. This is easily done by the first

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rule art. 31, for the quantity 20 a b, being regarded as a part of the product, having a sign contrary to that of the multiplicand 5 a1, it follows that the multiplier must have the sign —. Division then being performed upon the simple quantities, — 20 a b and 5 a', gives -4ub for the second term of the quotient. If now we multiply this by all the terms of the divisor, and subtract the product from the partial dividend, the remainder

10a4b34 a3 b2 + 8 a2 bs will contain only the products

of the third &c. terms of the quotient.

Regarding this remainder as a new dividend, its first term 10 a b3 must be the product of the first term of the divisor by the third of the quotient, and consequently this last is obtained by dividing the simple quantities, 10 a 63 and 5 a4, the one by the other. The quotient 263 being multiplied by the whole of the divisor furnishes products, the subtraction of which, exhausting the remaining dividend, proves that the quotient has only three terms.

If the question had been such as to require a greater number of terms, they might evidently have been found like the preceding, and if, as we have supposed, the dividend has the divisor for a factor, the subtraction of the product of this divisor by the last term of the quotient ought always to exhaust the corresponding dividend.

42. To facilitate the practice of the above rules;

1. We dispose the dividend and divisor, as for the division of numbers, by arranging them with reference to some letter, that is, by writing the terms in the order of the exponents of this letter, beginning with the highest ;

2. We divide the first term of the dividend by the first term of the divisor, and write the result in the place of the quotient;

3. We multiply the whole divisor by the term of the quotient just found, subtract it from the dividend, and reduce similar terms.

4. We regard this remainder as a new dividend, the first term of which we divide by the first term of the divisor, and write the result as the second term of the quotient, and continue the operation till all the terms of the dividend are exhausted.

Recollecting that when a product has the same sign as the multiplicand, the multiplier has the sign+, and, that when a product has the contrary sign to that of the multiplicand, the multiplier has the sign (31), we infer that, when the term of the dividend and the first term of the divisor have the same sign, the

quotient ought to have the sign +, and, if they have contrary signs, the quotient ought to have the sign; this is the rule for the signs. The individual parts of the operation are performed by the rule for the division of simple quantities.

We divide the coefficient of the dividend by that of the divisor; this is the rule for the coefficients.

We write in the quotient the letters common to the dividend and divisor with an exponent equal to the difference of the exponents of these letters in the two terms, and the letters which belong only to the dividend; these are the rules for the letters and exponents. 43. To apply these rules to the quantities,

5 a7 — 22 a® b + 12 a1 b2 · 6 aa b3 — 4 a3 b1 + 8 a2 b3,

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5a2a3b4a2 b2,

which have been employed as an example above, we place them as we place the dividend and divisor in arithmetic.

Dividend. Divisor. 5a7-22ab+12a*b3—6a+b3—4a3b*+8a2b3 5a+--2a3b+4a2b2

-5a2ab4a3b2

3

3

3

Rem.-20b+8α3b3—6α1b3—4a3b1 +8a3b* +20ab-8a5b2+16a4b3

rem.

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+10a4b3—4a3b1+Sa3b5
-10a4b34a3b1-8a2b5

0.

The sign of the first term 5 a7 of the dividend being the same as that of 5 a', the first term of the divisor, the sign of the quotient must be +, but, as it is the first term, the sign is omitted.

By dividing 5 a7 by 5 aa, we have for the quotient a3, which we write under the divisor.

Multiplying successively the three terms of the divisor by the first term a3 of the quotient, and writing the products under the corresponding terms of the dividend, the signs being changed to denote their subtraction (20), we have the quantity

5a2 + 2 a b 4 a5 b2,

which with the dividend being reduced,

·6 a b3·

we obtain for a remainder 4 a3 b1 + 8 a2 b5.

— 20 a® b + 8 a3 b2 By continuing the division with this remainder, the first term - 20 ab, divided by 5 as, will give for a quotient 4 a b, this quotient having the sign, as the dividend and divisor have

different signs. Multiplying it by all the terms of the divisor and changing the signs, we obtain the quantity

20 a b 8 as b2 + 16 a b3,

which taken with the dividend and reduced, gives for a remainder +10 a1 b3 — 4 a3 b1 + 8 a2 b5.

Dividing the first term of this new dividend, 10 a b3, by the first term, 5 a*, of the divisor, and multiplying the whole divisor by the result 263, writing the products under the dividend, the signs being changed, and making the reduction, we find that nothing remains, which shows that + 263 is the last term of the quotient sought. The quotient therefore has for its expression a3 — 4 a2 b + 263.

44. It is proper to remark here, that in division, the multiplication of the different terms of the quotient by the divisor often produces terms that are not to be found in the dividend, and which it is necessary to divide by the first term of the divisor. These terms are such as destroy themselves, since the dividend has been formed by the multiplication of the two factors, the quotient and the divisor. See a remarkable example of these reductions;

Let as b3 be divided by a — - b.

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The first term as of the dividend, divided by the first term a of the divisor, gives for the quotient a2; multiplying this quotient by the divisor. and changing the signs of the products, we have

a3a2b; the first term a3 destroys the first term of the dividend, but there remains the term a2 b, which is not found at first in the dividend. As it contains the letter a, we can divide it by the first term of the divisor, and obtain + a b. Multiplying this quotient by the divisor, and changing the signs of the products, we have a2 bab2; the term ab cancels the one above it, but there remains the term + a b2, which is not in the

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