What is the Quotient of 4a2x-2? and how do you reason on the sign of the quotient? Of 6ac2÷2c? Of 9x3y÷3xy? Of a3÷a? What is the Quotient of -7a3a2? and how do you reason on the sign of the quotient? Of -12c2-4? Of a3bab? Of -b4b2? What is the Quotient of 8a2-a2? and how do you reason on the sign of the quotient? Of 10ab÷-2b? Of 4a2c-a? Of 5x2x2 ? Of What is the Quotient of -7a37a? and how do you reason on the sign of the quotient? Of 12ac3a? Of -9a3b÷3? -20a4÷4? What is the Quotient of 10a35a? and why is that quotient true? What is the Quotient of 12x2-3? and why is that quotient true? What is the Quotient of -9a3by÷3a3? and why is that quotient true? What is the Quotient of -20a4x3-5ax? and why is that quotient true? What is the Quotient of -100axy÷―xy? and why is that quotient true? When the Dividend or the Divisor is 0. (50.) The quotient of 0 divided by any quantity, is 0; but the quotient of any quantity divided by 0, is infinitely great. First, let a represent any quantity we please; then 0a gives the quotient 0, because this quotient multiplied into the divisor a, produces the dividend 0, (43). Secondly, a divisor may be taken so small as to be contained a great number of times in any given dividend, represented by a; if the divisor be still diminished, the quotient will be increased; and if the divisor be diminished without limit, the quotient will be increased without limit. If therefore the divisor were dimished to 0, the quotient would be unlimited, that is, infinitely great. The character ∞ is the sign of infinity; we have then 0÷a equal to 0; and a÷0 equal to ∞, infinity, The preceding principles, and the two following Rules, embrace the subject of Algebraic Division, when the Quotient is an integral. quantity. RULE V. (51.) To Divide a Monomial into a Polynomial. Find the quotient of the monomial divided into each term of the polynomial, separately, and connect these quotients in a polynomial, with the proper sign prefixed to each. (47), (48), (49). EXAMPLE. To divide 20a2x-15ax2+30axy2-5ax by 5ax. 5ax)20a2x-15ax2 +30axy2-5ax 4a -3x+6y2 -1 Dividing into the first term of the dividend, we find the quotient term 4a; dividing into the second term of the dividend, we find the quotient -3x; and so on, through the dividend. EXERCISE S. 1. Divide 3x3+6x2+3ax-15x by 3x. Ans. x2+2x+a-5. Ans. b-3ax+4a2y-1 Ans. 2a-b+3x-3. 4. Divide a3x+3a2x2-6ax3+3ax by a. Ans. a2x+3ax2-6x3+3x. 5. Divide 562-10b3+5b1y—15b5 by 5b. 6. Divide bx2+2x3-8cx2+7x5 by x2. 7. Divide 4a4-8a3-4a2b+8a by 2a. 8. Divide ay+a2y3 —a3y2 —ay by ay. 9. Divide -y+by2-5y3+3y* by -y. 10. Divide 363 —962+126—15 by + 3. 11. Divide -c5+3c4-6c3+c2 by -c2. 12. Divide 8y3—4ay+a2y2 —3y by y. 13. Divide 10+20a-15a2+20 by -5. 14. Divide a3b-a2b2+a3b2-a4b by ab. 15. Divide x1y1+x3y3 —x2y2 —xy by xy. Ans. b—2b2+b3y—3b4. Ans. b+2x-8cx2+7x3. Ans. 2a3-4a2-2ab+4. Ans. y3+ay2-a2y—1. Ans. 1-by+5y3 —3y3.. Ans. 63-362+46-5. Ans. c3-3c2+6c-1. · Ans. 8y2-4a+a2y-3 Ans. 2-4a+3a2-4. Ans. a2-ab+a2b-a3. Ans. x3y3+x2y2 —xy—1. RULE VI. (52.) To Divide a Polynomial into a Polynomial. 1. Arrange the divisor and dividend according to the ascending or the descending powers of the same letter, (22). 2. Divide the first term of the divisor into the first term of the dividend; multiply the whole divisor by the quotient term; subtract the product from the dividend; divide into the remainder, as before, and so on,-observing to connect the several quotient terms in a polynomial, with the proper sign prefixed to each. EXAMPLE. To divide 6a+x+3a3x2-4a2x2+x6 by 2ax+x2 2ax+x2) 6a2x2+3a3x3-4a2x2+x6 ( 3a3x−2ax3+x3 6a2x2+3a3x3 -4a2x2+x6 2ax5+x6 The divisor and dividend are here arranged according to the ae scending powers of a, or the ascending powers of x. The first term 2ax of the divisor, divided into the first term 6a1x2 of the dividend, gives the quotient term 3a3x. Multiplying the whole divisor by this quotient term, and subtracting the product from the dividend, the remainder is -4a2x2+x6. We next divide the first term of the divisor into the first term, -4a2x2, of the remainder, and find the quotient term -2ax3; &c. By this Rule the divisor is multiplied by each part of the quotient, and the successive products subtracted from the dividend. When the dividend is thus exhausted, the product of the divisor and quotient is equal to the dividend; and the quotient is thus proved to be correct. The Divisor, when a polynomial, is sometimes set on the left of the dividend, and the quotient under the divisor. This arrangement is thought to afford greater facility in multiplying the divisor by the quotient term. EXERCISES. 16. Divide a2-2ax+x2 by a—x. 3 20. Divide 2x3-19x2+26x-16 by x-8. 24. Divide 4y-9y2+6y-1 by 2y2+3y—1. 26. Divide 4a5-64a by 2a-4. 27. Divide 6y6-96y2 by 3y-6. Ans. a-x. Ans. x2-2ax+a2. Ans. 2x2-3x+2. Ans. 2x2+3xc+c2. Ans. 2y2-3y+1. Ans. 2x3 +4x2+8x+16. Ans. 2y+4y+8y3+16y2. 28. Divide a⭑+4x1 by a2-2ax+2x2. 30. Divide y1+4y222-32z4 by y+2z. Ans. y3-2y2z+8yz2-16z3. (53.) The indicated Product of two or more factors is divided by any quantity, when either of the factors is divided by that quantity. Thus to divide 3a2x(x2+y2) by 3x, we divide the factor 3a2x, and find the quotient a2 (x2+y2). In the following exercises, the first of the given factors may be divided by the given divisor. 31. Divide (a2+2ay+y2) (b3—cd2+3) by a+y. Ans. (a+y) (b3-cd2+3). 32. Divide (2x3-6ax2+6a2x-2a3) (c2+3cy-y2) by x-a. Ans. (2x2-4ax+2a2) (c2+3cy—y2). 33. Divide (a1+4y1) (3xy2 −5y3+3y4—4) by a2 -2ay+2y2. Ans. (a2+2ay+2y2) (3xy2 — 5y3+3y1 —4.) 34. Divide (8a-2a3x-13a2x2-3ax3) (y2+2xy) (y3 +3x2y2 --x3) by 4a2+5ax+x2. Ans. (2a2—3ax) (y2+2xy) (y3+3x2y2 —x3). 27 CHAPTER III. COMPOSITE QUANTITIES.-COMMON MEASURE.-COMMON MULTIPLE. Composite and Prime Quantities. (54.) A composite quantity is one which is the product of two factors, each differing from unity; and a prime quantity is one which is not the product of such factors. Thus 3a2 is a composite, while a+b is a prime quantity. Is ab a composite or a prime quantity? Is a+5 a composite or a prime quantity? a?? 7x2? 2a+3b? 5a-5c2 ? Decomposition of Quantities. (55.) Decomposing a composite quantity consists in resolving it into its factors. Any divisor of the quantity, and the corresponding quotient, are two factors into which it may be resolved, (46.) Thus if the binomial 3x+6ax be divided by 3x, the quotient will be 1+2a; the binomial may therefore be resolved into the factors 3x(1+2a), 3x into the binomial 1+2a. Resolve 2a+4ax-6a2x2 into component factors. A monomial factor may generally be found by mere inspection, and a composite polynomial be resolved into a monomial and a polynomial factor, as above. No general Rule can be given for resolving a polynomial into the polynomial factors of which it may be composed. But there are particular Binomials which have well known binomial divisors, by means of which such Binomials may be resolved into two factors, (55). The following divisions will be found, on trial, to terminate without remainders. |