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56. 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 operation, and acquire the habit of calculating promptly, we will treat this last example in a different manner, at the same time indicating the simplifications which should be introduced. These, consist in subtracting each partial product from the dividend as soon as this product is formed.

- 40y5 + 68xy2 + 25x2y3 + 21x3y2 - 18x2y - 56x5 || 5y2 - 6xy - 8x2 1st rem. 20xy2 - 39x2y3 + 21x3y2 8y3 + 4xy2 - 3x2y + 7x3

2d rem.

3d rem.

Final remainder

15x2y3 + 53x3y2 - 18x+y

35x3y2 - 42x2y - 56x5

0.

8y3 for the quo

First, by dividing - 40y5 by 5y2, we obtain tient. Multiplying 5y2 by - 8y3, we have - 40y5, or by changing the sign, + 40y5, which destroys the first term of the divi

dend.

In like manner, 6xy X 8y3 gives + 48xy, and for the subtraction - 48xy, which reduced with + 68xy, gives 20xy for a remainder. Again, 8x2 X 8y3 gives +, and changing the sign, -64x2y3, which reduced with 25x2y3, gives 39x2y3 Hence, the result of the first operation is 20xy - 39x2y3, 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, to separate this new dividend from the primitive by a line, and to operate upon this new dividend in the same manner as we operated upon the primitive, and so on.

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12. Divide yo - 3y4x2 + 3y2x4 - x6 by y3 - 3y2x + 3yx2

13. Divide 64a4b6

Ans. y3 + 3y2х + Зух2 + х3.

3.

25a2b8 by 8a2b3 + 5ab4. Ans. 8a263 5ab+. 14. Divide 6a3 + 23a2b+ 22ab2 + 563 by 3a2 + 4ab + b2. Ans. 2a + 56. 15. Divide 6ax + 6ax2y + 42a2x2 by ax + 5ах. Ans. x5 + xy + 7ах. 15a2 + 37a2bd – 29a2cf – 20b2d2 + 44bcdf - 8c2f 5bd + cf. Ans. 5a2 + 4bd 8cf.

16. Divide

by 342

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20. Divide 6x65x5y2-6x+y2 + 6x3y2 + 15x3y3 - 9x2y + 10x2y5 +15y5 by 3x3 + 2x2y2 + 3y2. Ans. 2x3 - 3x2y2 + 5y

Remarks on the Division of Polynomials.

57. 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 shall find that a division is impossible, when the first term of any one of the partial dividends is not divisible by the first term of the divisor.

We will add, as to polynomials, that it may often be discovered by mere inspection that they are not divisible. When the polynomials contain two or more letters, observe the two terms of the dividend and divisor, which are affected with the highest exponent of each of the letters. If these terms do not give an exact quotient, we may conclude that the total division is impossible.

Take, for example,

12a3

5a2b+ 7ab2

1163 || 4a2 + 8ab + 3b2

By considering only the letter a, the division would appear possible; but regarding the letter b, the division is impossible, since 1163 is not divisible by 362.

58. One polynomial A, cannot be divided by another B containing a letter which is not found in the dividend; for, it is impossible that a third quantity, multiplied by B which contains a certain letter, should give a product independent of that letter.

A monomial is never divisible by a polynomial, because every polynomial multiplied by either a monomial or a polynomial gives a product containing at least two terms which are not susceptible of reduction.

59. If the letter, with reference to which the dividend is arranged, is not found in the divisor, the divisor is said to be independent of that letter; and in that case, the exact division is impossible, unless the divisor will divide separately the co-efficient of each term of the dividend.

For example, if the dividend were

3ba4 + 9ba2 + 126,

arranged with reference to the letter a, and the divisor 3b, the divisor would be independent of the letter a; and it is evident that the exact division could not be performed unless the co-efficient of each term of the dividend were divisible by 36. The exponents of the leading letter in the quotient would be the same as in the dividend.

1. Divide 18a3x2

2. Divide 25a4b

36a2x3 12ax by 6x.

Ans. 3a3x 6a2x2 30a2b+ 40ab by 56.

Ans. 5a

2a.

6a2 + 8a. 60. Although 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.

From the third remark of Art. 45, it appears that the term of the dividend affected with the highest exponent of the leading letter, and the term affected with the lowest exponent of the same letter, may each be derived without reduction, from the multiplication of a term of the divisor by a term of the quotient. Therefore, 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.

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 selected. If the same letter is preserved, it is only 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.

61. Among the different examples of algebraic division, there is one remarkable for its applications. It is expressed thus:

The difference between the same powers of any two quantities is always divisible by the difference between the quantities.

Let the quantities be represented by a and b; and let m denote any positive whole number. Then,

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will express the difference between the same powers of a and b, and it is to be proved that am bm is exactly divisible by a — b.

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Dividing am by a the quotient is am-1, by the rule for the exponents. The product of a - b by am-1 being subtracted from

the dividend, the first remainder is am-16

put under the form b (am-1 - bm-1).

bm, which can be

Now, if the factor (am-1 - bm-1) of the remainder, be divisible by ab, it follows that the dividend am bm is also divisible by a-b: that is,

If the difference of the same powers of two quantities be divisible by the difference of the quantities, then, the difference of the powers of a degree greater by unity is also divisible by it. But by the rules for division, we have

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63

Hence, we know, from what has just been proved, that a3 is divisible by a - b, and from that result we conclude that b, and so on, until we reach any

at

64 is divisible by a

exponent at m.

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