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on the left of the divisor, the result is the first term of the quotient; multiply the divisor by this term, and subtract the product from the dividend.

II. Then divide the first term of the remainder by the first term of the divisor, which gives the second term of the quotient; multiply the divisor by this second term, and subtract the product from the result of the first operation. Continue the same process until you obtain 0 for

a result; in which case the division is said to be exact.

When the first term of the arranged dividend is not exactly divisible by that of the arranged divisor, the complete division is impossible, that is to say, there is not a polynomial which, multiplied by the divisor, will produce the dividend. And in general, we will find that a division is impossible, when the first term of one of the partial dividends is not divisible by the first term of the divisor.

56. Though there is some analogy between arithmetical and algebraical division, with respect to the manner in which the operations are disposed and performed, yet there is this essential difference between them, that in arithmetical division the figures of the quotient are obtained by trial, while in algebraical division the quotient obtained by dividing the first term of the partial dividend by the first term of the divisor is always one of the terms of the quotient sought.

Besides, nothing prevents our commencing the operation at the right instead of the left, since it might be performed upon the terms affected with the lowest exponent of the letter, with reference to which the arrangement has been made. In arithmetical division

the quotient can only be obtained by commencing on the left.

Lastly, so independent are the partial operations required by the process, that after having subtracted the product of the divisor by the first term found in the quotient, we could obtain another term of the quotient by dividing by each other the two terms of the new dividend and divisor, affected with the highest exponent of a different letter from the one first considered. If the same letter is preserved, it is because there is no reason for changing it, and because the two

polynomials are already arranged with reference to it; the first terms on the left of the dividend and divisor being sufficient to obtain a term of the quotient; whereas, if the letter is changed, it would be necessary to seek again for the highest exponent of this letter.

SECOND EXAMPLE.

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21x3y2+25x2y3+68xy*—40y3—56xo — 18x1y by

—40y3 +68xy* +25x3y3 +21x3y3—18x*y—56xo||5y3 — 6xy — 8x2

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57. REMARK.-In performing the division, it is not necessary to bring down all the terms of the dividend to form the first remainder, but they may be brought down in succession, as in the example.

As it is important that beginners should render themselves familiar with the algebraic operations, and acquire the habit of calculating promptly, we will treat of this last example in a different manner, at the same time indicating the simplifications which should be introduced.

As in arithmetic, they consist in subtracting each partial product from the dividend as soon as this product is formed.

—40y3+68xy1+25x2y3 +21x3y3—18x1y—56x° || 5y2 — 6xy — 8x2

1st. rem. 20xy—39x2y3 +21 x3y3

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-8y3+4xy3-3x2y+7x2

-35 x3y2-42x1y-56x

0.

First, by dividing -40y5 by 5y3, we obtain -8y3 for the quotient. Multiplying 5y' by -8y3, we have -40y, or by changing the sign, +40y, which destroys the first term of the dividend.

In like manner, —6xy× — 8y3 gives +48xy' and for the subtraction -48xy, which reduced with +68xy*, gives 20xy for a remainder. Again, -8x3×—8y3 gives +, and changing sign, —64x3y3, which reduced with 25x3y3, gives-39x'y3. Hence the result of the first operation is 20xy-39x'y' followed by those terms of the dividend which have not been reduced with the partial products already obtained. For the second part of the operation, it is only necessary to bring down the next term of the dividend, separating this new dividend from the primitive by a line, and operate upon this new dividend in the same manner as we operated upon the primitive, and

so on.

Divide

1st. rem.

2d. rem.

THIRD EXAMPLE.

95a-73a2+56a1-25-59a3 by -3a2+5-11a+7a3

56a* — 59a3 — 73a2+95a—25||7a3-3a2—11a+5

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12. Divide 1 by 1-x.

13. Divide 6x*-96 by 3x-6.

Ans. 1+x+x2+x3, &c.

Ans. 2x+4x+8x+16.

14. Divide a-5a'x+10a3x-10a2x2+5ax-x by a1-2ax+x3.

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15. Divide 48x3-76ax3-64a2x+105a3 by 2x-3a.

16. Divide y°-3y*x2-3y2x*—x by y3-3y2x+3yx2—x3.

58. It may happen that one, or both, of the proposed polynomials contains in two or more terms the same power of the letter with reference to which the arrangement is to be made.

In this case, how should the arrangement be made, and the division be effected?

Divide

by

11a2b-19abc+10a-15a2c+3ab2+15bc2-5b2c

5a2+3ab-5bc.

In the first place, the two terms 11a2b-15a'c, can be placed under the form (11b-15c) a', or 116 a2, by writing the power a2

-15c

once, and placing to the left of it, and in the same vertical column, the quantities by which this power is multiplied; this polynomial multiplier is then called the co-efficient of a2.

The second manner of connecting the terms involving the same power, is preferable to the first, for two reasons. 1st. Because where there are many terms in the dividend and divisor, it would be difficult to write all on the same horizontal line. 2d. As the co-efficient of each power ought to be arranged with reference to a second letter, we are obliged, if the first term is subtractive, to subject the term to a modification, which might lead to error, in employing the first manner. Take, for example, — 15b2a2+7bca2 — 8c3a' the modification consists in putting this expression under the form

-(1562-7bc+8c2)a2

whereas, by the second, it is written thus:

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manner we have the advantage of preserving to each term the sign

with which it was at first affected.

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This being understood the operation may be performed in the following manner.

10a3+11b|a2+ 3b3 |a -56c+15bc2 ||5a2+3ba—5bc

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First divide 10 a' by 5a', the quotient is 2a. Subtracting the product of the divisor by 2a, we obtain the first remainder. Dividing the part involving a' in this remainder by 5a2, the quotient is b-3c. Multiplying successively each term of the divisor by b-3c, and subtracting the product, we have 0 for the result. Hence, 2a+b-3c is the required quotient.

59. Among the different examples of algebraic division, there is one remarkable for its applications. It is so often met with in the resolution of questions, that algebraists have made a kind of theorem of it.

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These are results that may be obtained by the ordinary pro cess of division. Analogy would lead to the conclusion that whatever may be the exponents of the letters a and b, the division could be performed exactly; but analogy does not always lead to cer

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