We first divide the term a2 of the dividend by the term a of the divisor, the partial quotient is a which we place under the divisor. We then multiply the divisor by a and subtract the product a2-ax from the dividend, and to the remainder bring down x2. We then divide the first term of the remainder, -ax by a, the quotient is -x. We then multiply the divisor by -x, and, subtracting as before, we find nothing remains. Hence, a-x is the exact quotient.

In this example we have divided that term of the dividend which is affected with the highest exponent of one of the letters, by that term of the divisor affected with the highest exponent of the same letter. Now, we avoid the trouble of looking out the term, by taking care, in the first place, to write the terms of the dividend and divisor in such a manner that the exponents of the same letter shall go on diminishing from left to right. This is what is called arranging the dividend and divisor with reference to a certain letter. By this preparation, the first term on the left of the dividend, and the first on the left of the divisor, are always the two which must be divided by each other in order to obtain a term of the quotient.

55. Hence, for the division of polynomials we have the following

RULE.

I. Arrange the dividend and divisor with reference to a certain letter, and then divide the first term on the left of the dividend by the first term 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, and if you obtain 0 for a remainder, the division is said to be exact.

FIRST EXAMPLE.

Let it be required to divide 51a2b2+10a1—48a3b—15b1+4ab3 by 4ab-5a2+362.

Dividend arranged.

+10a-8a3b- 6a2b2

Divisor.

-2a2+8ab-5b2

10a1-48a3b+51a2b2+4ab3-15b1||—5a2+4ab+3b2

REMARK. 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 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,

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 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.

Divide... 21x3y2+25x2y3+68xy1 — 40y5 — 56x5 — 18x1y by 5y2 — 8x2-6xy.

-40y+68xy+25x2y3+21x3y2-18x4y-56x5||5y2-6xy-8x2

57. REMARK.-In performing the division, it is not necessary to bring down all the terms of the dividend to form the first remain

der, 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 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+68xy1+25x2y3+21x3y2-18x1y-56x5||5y2-6xу—8x2

First, by dividing --40y5 by 5y2, we obtain -8y3 for the quotient. Multiplying 5y2 by -8y3, we have -40y5, or by changing the sign, +40y5, 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 20xy1 for a remainder. Again, -8x2× -8y3 gives +, and changing 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, 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.

THIRD EXAMPLE.

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

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

-35a3+15a2+55a-25

1st, rem.

2d. rem.

0.

8a-5

15. Divide 48x3-76ax2-64a2x+105a3 by 2x-3a.

Ans. 24x2-2ax-35a2.

16. Divide y6-3y1x2+3y2x1-x6 by y3-3y2x+3yx2-x3.

Ans. y3+3y2x+3yx2+x3.

17. Divide 64a4b6-25a2b® by 8a2b3+5ab1.

Ans. 8a2b3-5ab1.

18. Divide 6a3+23a2b+22ab2+5b3 by 3a2+4ab+b2.

19. Divide 6ax6+6ax2y6+42a2x2 by ax+5ax.

20. Divide by 3a2-5bd+cf.

Ans. 2a+5b.

-15a+37a2bd-29a2cf-20b2d2+44bcdf-8c2f2

Ans. x3+xy6+7ax.

Ans. 5a2+4bd-8cf.

Ans. x3+x2y+xy2+y3.

21. Divide x+x2y2+y1 by x2-xy+y2. Ans. x2+xy+y2.

23. Divide 3a4-8a2b2+3a2c2+564-362c2 by a2-b2.

Ans. 3a2-5b2+3c2.