the multiplier ; and, then, placing like quantities under each other, the sum of all the terms will be the product required. Examples. 3 a2 + ab + b 6 Ն a Scholium. We may substitute for a and 6 (Ex. 1.) any determinate numbers ; so that the above example will furnish the following (31.) THBOREM. The square of the sum of any two numlers is equal to the sum of their squares, together with twice their product. From Example 2 we have another (32.) THEOREM. The product of the sum of two numbers multiplied by their difference, is equal to the difference of the squares of those numbers. And from this may be derived a third (33.) THBOREM. The difference of two square numbers is always a product, and divisible both by the sum, and by the difference of the roots of those two squares, consequently, the difference of two squares can never be a prime number. Note. All pombers, such as 2, 3, 5, 7, &c. which cannot be represented by factors, are called simple or prime numbers; whereas, others, as 4, 6, 8, 9, &c. which may be represented by the factors % 2, 2 X 3, 2 X 4, or 2X3 X 2, and 3 X 3, are called composite numbers. Miscellaneous Examples. 1. Multiply 10ac by 2a. Ans. 20a%c. & Multiply 3a2 - 26 by 3b. Ans. 9ab-664. 8. Multiply 3a + 26 by 32 - 21. Ans. Ga? _463. 4. Multiply r* - xy + yo by x + y. Ans. 20 tys. 3. Multiply a + ab + ab + l by e-b. Ans, at 24 6. Multiply a2 + ab + baby a- ab + b2. 7. Multiply 3.r– 2ry + 5 by I? + 2ry-6. · 8. Multiply 3az 2ar + 5x2 by 3a 4ax - 72 9. Multiply 3.1' + 2x2y3 + 3yo by 2.20 - 3røya + 3y. 10. Multiply az + ab + b2 by a - 26. DIVISION. + abc (34.) The same circumstances are to be taken into consideration, in the division of algebraic quantities, as in their multiplication, and consequently the four following rules must be attended to. 1. That if the dividend and divisor be like, then the sign of the quotient will be t; if on the contrary unlike, then the sign of the quotient will be 2. - That the co-efficient of the dividend is to be divided by the co-efficient of the divisor, to obtain the co-efficient of the quotient. 3. That all the letters common to both the dividend and divisor, must be rejected in the quotient. 4. That if the same letter be found in both the dividend and divisor, with different indices, then the index of that letter in the divisor, must be subtracted from its index in the dividend, to obtain its index in the quotient. (35.) From these rules we derive the four following expressions : 1st. + abc divided by + ac, or = +0 + ac 6 abc 2d. + 6 abc.... by 2 a, or 3bc -2a 10 XYZ 3d. 10 xy% +5 y, or 2.rz. + 5 y 20 a* re y. 4th. 20 a`zRy' ... 4 azy, or = + 5 ary Note.-Of Division, also, there are three cases, as in Multiplication. Case I. When the dividend and divisor are both simple terms. Rule. Divide the co-efficient of the numerator by the co-efficient of the denominator, expunge those letters which are common to both terms of the fraction, and add the odd ones, any, to the numeral quotient (Art. 34 and 35.) Examples. 1 2 18 ar? + 15 m2 1,2 = Ox Ans. 3al? Ans. 4 axy 3 ar 5 a 3d. . Divide - 28 ro yo by 5th. Divide 30 a?.x?y2 by ör? Case II. When the dividend is a compound quantity, and the divisor a simple one. Rule. Divide each term of the dividend separately by the imple divisor, and the resulting quantities will be the quotient required (35) 3a Examples 1st. Divide 42a2 + 3ab + 120° by 3a 42 a? + 3ab + 12 a? Here = 140 + 0 + 4a. Ans. 2nd. Divide 90ax3 18ax + 4a?? 2 ax by 2ar. 3rd. Divide 423 2x2 + 2 x by 2r. 4th. Divide 24ary - 3azy + 6 x'y® by - 3xy. 5th. Divide 14ab3 + 7a9 21a2b2 + 42 a3l by 7ab. Case III. When the dividend and divisor are both compouna quantities. Rule I. Arrange both dividend and divisor according to the powers of the same letter, beginning with the highest. 2. Then find how often the first term of the divisor is contained in the first term of the dividend, and place the result in the quotient. 3. Multiply each term of the divisor by this quantity, and place the product under the corresponding (i.e. like) terms in the dividend, and then subtract the one from the other. 4. To the remainder bring down as many terms of the dividend as will make its number of terms equal to the number of terms in the divisor ; and then proceed as before, till the terms are all brought down as in common arithmetic. Examples. 1. Divide a 3a b + 3ab? - moby -b al) as за?ь + Заl? - 69 (a? - 2ab + b* Ans. as aՑՆ a Illus. In this example, the dividend is arranged according to the powers of a, the first term of the divisor. Having done so, we proceed by the following steps, which sufficiently illustrate the rule. 1. a is contained in a3, az times; and this (az) is put in the quotient. 2. a - b is multiplied by az, and it gives a3 -u? b. 3. From as 3uzh in the dividend, we subtract the product of (a - b) az or 03, u2b, and the remainder is 2q2b. 4. The next term + 3 abz in the dividend is now brought down. 5. a is contained in - 2a0b, Zab times, which we put in the quotient. 6. Multiply and subtract as before, and the remainder is ab2. 7. Bring down the last term of the dividend - 63. 8. a is contained in ub2, + be times; put this in the quotient. 9. Multiply and subtract as before, and nothing remains. The quotient, therefore, is az - 2 ah + b2, and we call it the ansu er. 10. To prove the operation, multiply the quotient az - 9 ab + b2 by the divisor a - b, and the product will be the dividend ; for by the nature of Division it is evident, that if one number be divisible by another, the quutient multiplied by the divisor will produce the dividend. Whence the reason of the above rules (Art. 54) and operations is evident from what was proved in Multiplication. (See Art. 28 and 29.) Eramples. 2. Divide a + 5a'r +100°r+ 10ar'+bar+r. Ans. ad + 3 ar + 3 ars + r. 3. Divide 12 xs 13 r* · 34 x + 35 x2 by 4x? — 76. Ans. 3.ro + 2r- 5r. 4. Divide 26 ta' ta + 2r-I by x2 + I-l. Ans, x* 33 + x2 3+ . 5. Divide y6 - 3y*r+ 3yoz* + x by yo — 3yor + 3yr — 20. Ans. y' + 3yor + 3yr + 6. Divide as 5a'r + Joa*.. 10a?r? + 5art - 205 by a' 2ar + re. Ans, a За'r + Зах? r. Miscellaneous Examples. 1. Divide a' - x by a - r. Ans, a' ta'r et a’ra + ax + *. 2. Divide 4.13 76ar? + 35a2x + 105a3 by 2.0 3а. . Ans. 2x 350x - 35a?. 3. Divide x3 + yby x + y. Ans. x2 + y - yr. 4. Divide 2a4 32 by a 2. Ans. 2a + 4a? t sa + i6. 5. Divide as t a'r – a'r? - 7aPro + 6x by a? x?. Ans, a + a? x - 6r. 6. Divide x + 3r2y + 3ry* + ys by x2 + 2xy + y. Ans. 2 + y. 7. Divide as - 3a?c + 4aca 2c3 by a? — 2ac + ca. aca Ans. Q6+ a2 · 2act 02 8. Divide 6.x 96 by 3r 6. Ans. 2x3 + 4x2 + 8r + 16. 9. Divide 1 - 50 + 10x2 10.x2 + 5x Is by 1 -2x + ?. Ans. I - 3.0 + 3.14 - 13. C FRACTIONS. Algebraic Fractions are subject to the same rules as are the fractions of commun arithmetic, and their reduction requires only the application of the principles laid down in the preceding rules. PROBLEM I. Rule. Multiply the integer by the denominator of the fraction, and to the product annex the numerator with its proper sign ; under this same place the former denominator, and the result will give the improper fraction required. Erample 1. 2.7 Reduce 3a + to an improper fraction. Here the integral part x denom. of the fraction + the numerato? За Х ba? + 2027 = 150% + 2ax, 15a + 2x Hence is the fraction required. 5a a PROBLEM II. Rule. Observe which terms of the numerator are divisible by the denominator without a remainder, the quotient will give the integral part ; to this annex the remaining terms of the numerator, with the denominator under them, with their proper signs, and the res:ut will be the mixed quantity required. Example 1. to a whole or mixed quantity 02 Here =at the integral part ; and the fractional, a, is the mixed quantity required. 23 - 30 2. Reduce Ans. 3a + 50 sa 3. Reduce 2r 21 12a + 40 Зc 4. Reduce 40 10x? 278 5. Reduce a a Ans. 2x + 3.20 PROBLEM III. Rule. Multiply each numerator into every denominator 'ut its own for the new numerators, and multiply all the denominators for a common denominator. S& multiplied by 7 * 35x ; and 2x - 3 subtracted from 35x leave 53345 |