CASE III. When both the factors are compound quantities. RULE. Multiply every term of the multiplier into every term of the multiplicand, respectively, and set down the products one after another with their proper signs, and their sum will be the whole product required. hena is to be multiplied by +b, or +b by -a, this is the me as taking -a as many times as there are units in +b; and ce the sum of any number of negative terms is negative, it is dent that (—a)X(+b) or (+b)×(—a)=—ab. 3. When a is to be multiplied by -b: here a―a=0; erefore (a—a) X —b is also o, because o multiplied by any intity produces o; and since the first term of the product, or <'—b), is, by case 2, ab, the last term, or (-a)×(—b), st be ab, in order to make the sum (-abab)=0; con. quently (—a)X(—b)=+ab. 5. Multiply 3+x2y+xy2+y3 by x-y. Ans. x1-y1 ́ 6. Multiply x2+xy+y2 by x2—xy+y2. 7. Multiply 32-2xy+5 by x2+2xy-3. 8. Multiply 2a2-3ax+4x2 by 5a2-6ax-2x2. 9. Multiply 3x3+2x2y2+3y3 by 2x3-3x2y2+5y3. . DIVISION. CASE I. When the divisor is a simple quantity. RULE. 1. Place the dividend above a small right line, and the divisor under it, in the manner of a vulgar fraction.. 2. Expunge those letters which are common to both, the dividend and divisor, and divide the co-efficients of all. the terms by any number that will divide them without a remainder, and the result will be the quotient required. Note*. Like signs make +, and unlike signs —, the same as in multiplication. * That like signs give +, and unlike signs, in the quotient, will appear thus: The divisor, multiplied by the quotient, must produce the dividend; therefore, EXAMPLES. 1. It is required to find the quotient of a÷a; 8bc÷2b; and abc÷bcd. 2. It is required to find the quotient of 12xy ÷ 6x2; and (ab+b2)+26. 3. Divide 18x2 by 9x. 10. Divide 13a+3ax-17x2 by 21a. 11. Divide 3a2-15+6a+3b by 3a. CASE II. Ans. 2b+3c. Ans. 5a-9. Ans. 2x2-3y-1. When the divisor and dividend are both compound quantities. RULE. 1. Range the terms of both the quantities according to the dimensions of some letter in them, so that 1. When both the terms are +, the quotient is +, because (+) X(+) produces + in the dividend. 2. When they are both, the quotient is also +, because (+)X(-) produces in the dividend. 3. When one of them is + and the other, the quotient is -, because (-)x(+) produces. in the dividend; and (-)X(-) produces in the dividend. C the first term may contain the highest power of that letter, the second term, the next highest power; and so on. 2. Divide the first term of the first term of the divisor, and place the tient. dividend by the result in the quo 3. Multiply the whole divisor by the term thus found, and subtract the result from the dividend. 4. To this remainder bring down as many terms of the dividend as are requisite for the next operation, and divide as before; and so on, as in common arithmetic. Note. If the divisor be not exactly contained in the dividend, the quantity which remains after the operation is finished must be placed over the divisor, like a vulgar fraction, and set down at the end of the quotient, as in common arithmetic. EXAMPLES. x+y)x2+2xy+y2(x+y x2+ xy xy+y2 xy+y2 * a+x)a3+5a2x+5ax2+x3 (a2+4ax+x2 a3+ a2x 4a2x+5αx2 ax2+x3 |