77. Hence any factor may tor to the denominator of a fraction, or from the denominator to the numerator, by changing the sign of its exponent. be transferred from the numera that is, the denominator of a fraction may be entirely removed, and an integral form be given to any fractional expression. This use of negative exponents must be understood simply as a convenient notation, and not as a method of actually destroying the denominator of a fraction. 78. To divide a Polynomial by a Monomial.—We have seen, Art. 60, that when a single term is multiplied into a polyno mial, the former enters into every term of the latter. Thus, therefore (a+b)m=am+bm; (am+bm)÷m=a+b. Hence, to divide a polynomial by a monomial, we have the following RULE. Divide each term of the dividend by the divisor, and connect the quotients by their proper signs. EXAMPLES. Ans. c+4x-3a. 1. Divide 3x3+6x2+3ax-15x by 3x. Ans. x2+2x+a−5. 2. Divide 3abc+12abx-9a2b by 3ab. 3. Divide 40a3b3+60a2b2-17ab by -ab. 4. Divide 15a2bc-10acx2+5ac2d2 by -5a2c. 5. Divide 20x5-35x-15x3+75x2 by -5x2. 6. Divide 6a2x1y°—12a3x3y® +15a*x3y3 by 3a2x2y2. 7. Divide +1_x2+2+x2+3 - x2+ by x2. 8. Divide 12a1y—16a3y3+20a°y*—28a1y3 by —4a*y3. 79. To divide one polynomial by another. Let it be required to divide 2ab+a2+b2 by a+b. The object of this operation is to find a third polynomial which, multiplied by the second, will reproduce the first. It is evident that the dividend is composed of all the partial products arising from the multiplication of each term of the divisor by each term of the quotient, these products being added together and reduced. Hence, if we can discover a term of the dividend which is derived without reduction from the multiplication of a term of the divisor by a term of the quotient, then dividing this term by the corresponding term of the divisor, we shall be sure to obtain a term of the quotient. But, from Art. 64, it appears that the term a2, which contains the highest exponent of the letter a, is derived without reduction from the multiplication of the two terms of the divisor and quotient which are affected with the highest exponent of the same letter. Dividing the term a2 by the term a of the divisor, we obtain a, which we are sure must be one term of the quotient sought. Multiplying each term of the divisor by a, and subtracting this product from the proposed dividend, the remainder may be regarded as the product of the divisor by the remaining terms of the quotient. We shall then obtain another term of the quotient by dividing that term of the remainder which is affected with the highest exponent of a by the term a of the divisor, and so on. Thus we perceive that at each step we are obliged to search for that term of the dividend which is affected with the highest exponent of one of the letters, and divide it by that term of the divisor which is affected with the highest exponent of the same letter. We may avoid the necessity of searching for this term by arranging the terms of the divisor and dividend in the order of the powers of one of the letters. The operation will then proceed as follows: The arranged dividend is a2+2ab+b2 | a+b, the divisor. For convenience of multiplication, the divisor is written on the right of the dividend, and the quotient under the divisor. 80. Hence, to divide one polynomial by another, we have the following RULE. 1. Arrange both polynomials in the order of the powers of the same letter. 2. Divide the first term of the dividend by the first term of the divisor, for the first term of the quotient. 3. Multiply the whole divisor by this term, and subtract the product from the dividend. 4. Divide the first term of the remainder by the first term of the divisor, for the second term of the quotient. 5. Multiply the whole divisor by this term, and subtract the product from the last remainder. 6. Continue the same operation until a remainder is found equal to zero, or one whose first term is not divisible by the first term of the divisor. When a remainder is found equal to zero, the division is said to be exact. When a remainder is found whose first term is not divisible by the first term of the divisor, the exact division is impossible. In such a case, the last remainder must be placed over the divisor in the form of a fraction, and annexed to the quotient. EXAMPLES. Ans. a+b. Ans. x2-2ax+a2. 1. Divide 2ab+b3+2ab2+a3 by a2+b2+ab. Ans. a1+a3x+αx3· + x2. 4. Divide a-16a3x3+64x6 by a2-4ax+4x2. 5. Divide a+6a2x2-4a3x+x+-4ax3 by a2-2ax+x2. 6. Divide 32x+y3 by 2x+y. Ans. a2-2ax+x2. 7. Divide x+x3y2+xy*—x*y—x2y3—y3 by x3—y3. 8. Divide x+x3+5x-4x2−3 by x2-2x-3. 9. Divide ao-16 by a3+2a2b+2ab2+b3. 10. Divide x+1-2x3 by x2+1-2x. 11. Divide x+y+x2y2 by x2+y2+xy. 12. Divide 12x-192 by 3x-6. 13. Divide 6x6 - 6y by 2x2-2y2. Ans. 4x3+8x2+16x+32. 14. Divide a +3a2b*—3a+b2—b6 by a3-3a2b+3ab2—b3. Ans. a3+3a2b+3ab2+b3. 15. Divide x-6x+9x2-4 by x2-1. 16. Divide a*+a3b—8a2b2+19ab3-15b* by a2+3ab—5b2. 17. Divide x+y3+3xy-1 by x+y-1. 18. Divide a2b2+2abc2—a2c2—b2c2 by ab+ac-bc. 19. Divide a3-b3 by a-b. 20. Divide a1-b1 by a-b. 81. Hitherto we have supposed the terms of the quotient to be obtained by dividing that term of the dividend which is affected with the highest exponent of a certain letter. But, from Art. 64, it appears that the term of the dividend affected with the lowest exponent of any letter is derived without reduction from the multiplication of a term of the divisor by a term of the quotient. Hence we may obtain a term of the quotient by dividing the term of the dividend affected with the lowest exponent of any letter by the term of the divisor containing the lowest exponent of the same letter; and we may even operate upon the highest and lowest exponents of a certain letter alternately in the same example. 82. an-bn is always divisible by a-b. From the examples of Art. 80 we perceive that a3-b3 is divisible by a-b; and a*b* is divisible by a-b. We shall find the same to hold true, whatever may be the value of the exponents of the two letters; that is, the difference of the same powers of any two quantities is always divisible by the difference of the quantities. Thus, let us divide a5-65 by a-b: as-b5 a-b, divisor. a-aba, partial quotient. a+b-b5. The first term of the quotient is a1, and the first remainder is a1b-65, which may be written a1b—b3, b (a1 —b1). Now if, after a division has been partially performed, the remainder is divisible by the divisor, it is obvious that the dividend is completely divisible by the divisor. But we have already found that a1-b1 is divisible by a-b; therefore a5-65 is also divisible by a-b; and, in the same manner, it may be proved that ao-66 is divisible by a-b, and so on. 83. To exhibit this reasoning in a more general form, let n represent any positive whole number whatever, and let us attempt to divide a"-b" by a-b. The operation will be as follows: an_bn a-b, divisor. an-ban-1 an-1, quotient. The first remainder is ban-1—ơn. Dividing an by a, we have, by the rule of exponents, a”-ı for the quotient. Multiplying a-b by this quantity, and subtracting the product from the dividend, we have for the first remainder ban-1-b", which may be written b(an—1—bn−1). Now, if this remainder is divisible by a-b, it is obvious that the dividend is divisible by a-b; that is, if the difference of the same powers of two quantities is divisible by the difference of the quantities, then will the difference of the powers of the next higher degree be divisible by that difference. Therefore, since a*b* is divisible by a-b, a3-b5 must be |