II. Then divide the first term of the remainder by the first term of the divisor, for 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 operation until a remainder is found equal to 0, or till the first term of the remainder is not exactly divisible by the first term of the divisor. In the first case, (that is, when the remainder is 0,) the division is said to be exact. In the second case the exact division cannot be performed, and the quotient is expressed by writing the entire part obtained, and after it the remainder with its proper sign, divided by the divisor. SECOND EXAMPLE. Divide 21x3y2 + 25x2y3 + 68xy1 — 40y5 — 56x518x1y by 5y28x26xy. -- 40y5 + 68xy* + 25x2y3 + 21x3y2 — 18x1y — 56x5||5y2 — 6xy — 8.x2 — 8y3 + 4xy2 — 3x2y + 7x3 — 40y5 + 48xy1 + 64x2y3 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 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 +68xy* +25x2y3 + 21x3y2 — 18x1y — 56x5||5y2 — 6xy -- 8x2 1st rem. 20xy-39x2y3+21x3y2 -8y3+4xy2-3x2y+7x3 35x3y242x4y — 56x5 0. - 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 cancels the first term of the dividend. In like manner, 6xy8y3 gives +48xy+, or, changing the sign, 48xy4, 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 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. THIRD EXAMPLE. 95a ·73a2 + 56a4 — 25 - 59a3 by - 3a1 35a3+15a2 + 55a 0. GENERAL EXAMPLES. 5. Divide 6ax2 + 9a2x + a2x2 by ax. Ans. 6x + 9a + ax. 6. Divide a2 + 2ax + x2 by a +x. 7. Divide a3 - 3a2y + 3ay2 — y3 by Ans. a +x. a―y. 6ab. - 4a+2a2cb + 1. 96 by 3x-6. Ans. 2x3 + 4x2 + 8x + 16. a55a4x + 10a3x2 – 10a2x3+5ax1 20 Ans. a3 12. Divide y6 — 3y1x2 + 3y2x4 — x6 by y3 — 3y2x + 3yx2 — x3. 14. Divide 6a3 + 23a2b + 22ab2+5b3 by 3a2+4ab + b2. Ans. 2a +5b. 15. Divide 6ax6 + 6ax2y6 + 42a2x2 by ax + 5ax. 16. Divide Ans. x5 + xy6 + 7ax. 15a+37a2bd-29a2cf-20b2d2+ 44bcdf8c2f" Ans. 5a24bd8cf. by 3a25bd + cf. 17. Divide x1 + x2y2 + y1 by x2 - xy + y2. REMARKS ON THE DIVISION OF POLYNOMIALS. 58. The exact division of one polynomial by another is impossible: 1st. When the first term of the arranged dividend or the first term of any of the remainders, is not exactly divisible by the first term of the arranged divisor. It may be added with respect to polynomials that we can often discover 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 contain the highest powers of each of the letters. If these terms do not give an exact quotient, we may conclude that the exact division is impossible. Take, for example, 12a35a2b+ 7ab2 . 1163 || 4a28ab + 3b2. By considering only the letter a, the division would appear possible; but regarding the letter b, the exact division is impossible, since 1163 is not divisible by 362. 2d. When the divisor contains a letter which is not in the dividend. For, it is impossible that a third quantity, multiplied by one which contains a certain letter, should give a product independent of that letter. 3d. A monomial is never divisible by a polynomial. For, every polynomial multiplied by either a monomial or a polynomial gives a product containing at least two terms which are not susceptible of reduction. 4th. If the letter, with reference to which the dividend is arranged, is not found in the divisor, the divisor is said to be inde pendent of that letter; and in that case, the exact division is impossible, unless the divisor will divide separately the co-efficients of the different powers of the leading letter. For example, if the dividend were 3ba9ba2 + 126, arranged with reference to the letter a, and the divisor 36, the divisor would be independent of the letter a; and it is evident that the exact division could not be performed unless the coefficients of the different powers of a were exactly divisible by 3b. The exponents of the different powers of the leading letter in the quotient would then be the same as in the dividend. EXAMPLES. 1. Divide 18a3x2 — 36a2x3 — 12ax by 6x. From the 3d remark of Art. 46, it appears that the term of the dividend containing the highest power of the leading letter and the term containing the lowest power of the same letter are both 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 containing the lowest power 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 arranging the remainder with reference to some other letter and then proceeding as before. If the same letter is preserved, it is only because there is no reason for changing it; and because the polynomials are already arranged with reference to it. OF FACTORING POLYNOMIALS. 59. When a polynomial is the product of two or more factors, it is often desirable to resolve it into its component factors. This may often be done by inspection and by the aid of the formulas of Art. 47. When one factor is a monomial, the resolution may be affected by writing the monomial for one factor, and the quotient arising |