then the quotient is a-d. In order to explain this, let us resume the division of am by a", or by am+dam Xad; by suppressing the factor am, m, which is common to the dividend and divisor, according to what has been demonstrated with regard to the division of letters (Art. 1 84), we have for the quotient- : therefore, ad This transformation is very useful in various analytical operations; in order to see more clearly the meaning of it, we may recollect that atd is the same as a Xa Xa, &c., continued to d factors; therefore, according to the acceptation and opposition of the signs, a-d must represent a Xa xa, &c., continued to d factors in the divisor. 1 Hence, according to the results (1), (2), and (3), the proposition is general, when m and n are any whole numbers whatever; thus, a3÷a3—3—5—a—2, or ; because the divisor multiplied by the quotient is equal to the dividend, a3× a—2—a 5—2—a 3 the dividend, and Χα α 1 a5 a2 a2 whole number whatever; hence the method of notation pointed out, (Art. 32), is evident. 87. If a compound quantity is to be divided by a compound quantity, it frequently occurs, that the division cannot be performed, in which case, the division can be only indicated, in representing the quotient by a fraction, in the manner that has been already described (Art. 8). 88. But if any of the terms of the dividend can be produced by multiplying the divisor by any simple quantity, that simple quantity will be the quotient of all those terms. Then the remaining terms of the dividend may be divided in the same manner, if they can be produced by multiplying the divisor by any other simple quantity; and by continuing the same method, until the whole dividend is exhausted; the sum of all those simple quantities will be the quotient of the whole compound quantity. The reason of this is, that as the whole dividend is made up of all its parts, the divisor is contained in the whole as often as it is contained in all its parts. Thus, (ab+b+ad+ed)÷(a+c) is equal to b+d: For bX(a+c) ab+cb; and d×(a+c)=ad+cd; but the sum of ab+cb and ad+cd is equal to ab+cb+ad+cd, which is equal to the dividend; therefore b+d is the quotient required. Also, (a2+3ab+2b2)÷(a+b) is equal to a+2b. For, it is evident in the first place, that the quotient will include the term a, since otherwise we should not obtain a2. Now, from the multiplication of the divisor a+b by a, arises a2+ab; which quantity being subtracted from the dividend, leaves a remainder 2ab+262; and this remainder must also be divided by a+b, where it is evident that the quotient of this division must contain the term 26: again, 26, multiplied by a+b, produces 2ab+2b2; consequently a +26 is the quotient required; which, multiplied by the divisor a+b, ought to produce the dividend a2+3ab+262. See the operation at length : a+b)a2+3ab+2b2(a+2b 2ab+2b3 2ab+262 * 89. SCHOLIUM. If the divisor be not exactly contained in the dividend; that is, if by continuing the operation as above, there be a remainder which cannot be produced by the multiplication of the divisor by any simple quantity whatever; then place this remainder over the divisor, in the form of a fraction, and annex it to the part of the quotient already determined; the result will be the complete quotient. But in those cases where the operation will not terminate without a remainder; it is commonly most convenient to express the quotient, as in (Art. 87). 90. Division being the converse of multiplication, it also admits of three cases. CASE I. When the divisor and dividend are both simple quantities. RULE. 91. Divide, at first, the coefficient of the dividend by that of the divisor; next, to the quotient annex those letters or factors of the dividend that are not found in the divisor; finally, prefix the proper sign to the result, and it will be the quotient required. Note. Those letters in the dividend, that are common to it with the divisor, are expunged, when they have the same exponent; but when the exponents are not the same, the exponent of the divisor is subtracted from the exponent of the dividend, and the remainder is the exponent of that letter in the quotient. 86). = Xa1-1Xx2-1=6Xa° Xx=6x. See (Art. 3 Ex. 2. Divide-48a2b2c2 by 16abc. In the first place, 48-16=3= the coefficient of the quotient, next, a2b2c2÷abc=a2-1 × 62-1 × c2—1— abc; now, annexing abc to 3, we have 3abc, and, prefixing the sign ; because the signs of the dividend and divisor are unlike; the result is 3ab, which is the quotient required. Or, the operation may be performed thus, -48a2b3c2 16abc 48 a2 b2 ca Ex. 3. Divide X X X 16 α b C =-3xaxbXc=-3abc. 21x3y3za by —7x3y3z3. Xc7-5=-4Xaa Xb3 Xca=-4a3b3c2. In order that the division could be effected according to the above rule ; it is necessary, in the first place, that the divisor contains no letter that is not to be found in the dividend in the second place, that the exponent of the letters, in the divi sor, do not surpass at all that which they have in the dividend; / finally, that the coefficient of the divisor, divides exactly that of the dividend. When these conditions do not take place, then, after cancelling the letters, or factors, that are common to the dividend and divisor; the quotient is expressed in the manner of a fraction, as in (Art. 84). Ex. 5. Divide 48a3b5c2d by 64α3b3c1e. The quotient can be only indicated under a fractional form thus, 48a3b5c2d : But the coefficients 48 and 64 are both divisible by 16, suppressing this common factor, the coefficient of the numerator will become 3, and that of the denominator 4. The letter a having the same exponent 3 in both terms of the fraction, it follows, that a3 is a common factor to the dividend and divisor, and that we can also suppress it. The exponent of the letter b is greater in the divisor than in the dividend; it is necessary to divide b5 by b3, and the quotient will be b2, or b5 63 ·=b5—3—b2, which factor will remain in the numerator. C4 With respect to the letter c, the greater power of it is in the denominator; dividing c by c2, we have c2, or =2, therefore the factor c2 will remain in the denominator. Finally, the letters d and e remain in their respective places; because, in the present state, they cannot indicate any factor that is common to either of them. By these different operations, the quotient, in its most sim3b2d ple form, is 4c2e Note. The division of such quantities belongs, properly speaking, to the reduction of algebraic fractions. Ex. 6. Divide 36x y2 by 9xy. Ex. 9. Divide-4ax2y3 by-axy3. Ex. 10. Divide 16ab3cx by -4a3bdy. Ans. Ex. 11. Divide -18a3b3c2 by 12a5 b3x. Ans. Ex. 12. Divide 17xyzw2 by xyzw. Aos. 4xy. Ans.-5ay. Ans. 6cxy. Ans. +4xy. 4a2 b3 cx dy 2a3bx Ans. 17w. Ans. 2a2b2c3 Ex. 14. Divide - 9xyz2 by x1y1z^. Ex. 15. Divide 39a9 by 13a5. CASE II. Ans. 9 x2 y2 z2 Ans. 3a4 When the divisor is a simple quantity, and the dividend a compound one. RULE. 92. Divide each term of the dividend separately by the simple divisor, as in the preceding case; and the sum of the resulting quantities will be the quotient required. EXAMPLE 1. Divide 18a3+3a2b+6ab2 by 3a. 18a3 Here, =6a2, =ab, and -=262 ; 3α Ex. 2. Divide 20a3 x3-12a2x2+8a3 x2-2a4x2 by 2ax2. 3a2b 6ab2 3a 18a3+3a2b+6ab3 therefore, =6a2+ab+2b2. 3a :10ax, -12a2x22ax2-6α, 8a3x2 20a2x3-12a2x2+8a3x2-2a+x2 hence --=10ax-6a+4a2 2ax2 -a3. 2ax2+4a2, and —2a1x2 ÷2αx2——α3 ; Ex. 3. Divide 20a2x-15ax2+30axy2-5ax by 5ax. Here 20a2x5αx=4a, —15ax2-5ax=-3x, 30axy2 5ax-6y2, and -5ax5ax=−1; therefore, 20a2x-15ax2+30axy2—5ax = 4a-3x+6y2-1. 5ax Ex. 4. Divide 5a x-25a3 x2 +50a1x3-50a3x4 +25a2x5 -x5; therefore, a5-5ax + 10a3x2-10a3x3+5ax-x5 is the quotient required. Ex. 5. Divide 3a2x2- 3a2x1 by -3a2x2. Ex. 6. Divide 21a3x3-7a2x2-14ax by 7ax. Ans. x2-a2. Ans. 3a2x2ɑX 2. |