all positive, and the index in the dividend either odd or even. x2 + Y1 = x6 — X3y + x^y2 — x3y3 + x2y* — xy3 +yo, x + y and so on; the divisor being x + y, the terms in the quotient alternately positive and negative, and the index in the dividend always odd. x© — Y° = x3 — x1y + x3y2 — x2y3 + xy* — y3, x + y and so on; the divisor being x + y, the terms in the quotient alternately positive and negative, and the index in the dividend always even. IV. The expressions 2+ y2, x2 + y2, x2 + yε ... (where the index is even, and the terms both positive), are never divisible by x + y or x - y. All these different cases may be more concisely stated as follows: (1) x” — yn is divisible by x-y if n be any whole number. (2) x+y" is divisible by x + y if n be any odd whole number. (3) xn ·yn is divisible by x + y if n be any even whole number. (4) x + yn is never divisible by x + y or x -y when n is an even whole number. NOTE. General proofs of these statements will be found in Art. 106. DIVISION BY DETACHED COEFFICIENTS. 63. In Art. 52 we considered certain cases of compound expressions in which the work of multiplication could be shortened by using the Method of Detached Coefficients. In the same cases the labor of division can be considerably abridged by using detached coefficients, and employing an arrangement of terms known as Horner's Method of Synthetic Division. The following examples illustrate the method: Ex. 1. Divide 3x5-8 x1-5 x3 +26 x2−28 x +24 by x3-2 x2-4x+8. 13 8 5+2628 +24 Dividend Divisor Inserting the literal factors in the quotient according to the law of their formation, which is readily seen, we have for a complete quotient, 3x2 − 2 x + 3. EXPLANATION. The column of figures to the left of the vertical line consists of the coefficients of the divisor, the sign of each after the first being changed, which enables us to replace the process of subtraction by that of addition at each successive stage of the work. Dividing the first term of the dividend by the first term of the divisor, we obtain 3, the first term of the quotient. Multiplying 2, 4, and — 8, the remaining terms of the divisor, by this first term of the quotient gives the second horizontal line. We then add the terms in the second column to the right of the vertical line and obtain - 2, which is the coefficient of the second term of the quotient. With this coefficient as a multiplier, and using 2, 4, and — 8 again as a multiplicand, we form the third horizontal line. Adding the terms in the third column gives 3, which is the third term of the quotient. With this coefficient as a multiplier and the same multiplicand as before, we form the fourth horizontal line. As only zeros now appear in the quotient, the division is exact. Ex. 2. Divide 2 a7 + 7 ab + 12 a5b2 + 10 a4b3 — 4 a3b4 by 2 a3 + 3a2b-b3 to four terms in the quotient. Inserting literal factors, a1+2 a3b + 3 a2b2 + abs is the complete quotient, and -5a3b4 + 3 a2b5ab6 is the remainder. EXPLANATION. The term ab2 in the divisor is missing, so we write 0 for the coefficient of this term in the column of figures on the left of the vertical line. We add the columns as in Ex. 1, but as the first term of the divisor is 2, we divide each sum by 2 before placing the result in the line of quotients. We then use these quotients as multipliers, the multiplicand being in each case 3, 0, and 1, and form the horizontal lines as in Ex. 1. Having obtained the required number of terms in the quotient, the remainder is found by adding the rest of the columns and setting down the results without dividing by 2. By continuing the first horizontal line (dividend), as shown in this example, we at once see what literal factors the remainder must contain. EXAMPLES V. d. Divide: 1. a1 - 4 a3 + 2 a2 + 4 a + 1 by a22a - 1. 2. a1 3. a5 10 a+b+ 16 a3b2 — 12 a2b3 + ab1 + 2 b5 by (a - b)2. 4. x8-2b4x4 + b8 by x3 + bx2 + b2x + b3. 5. x5 − 3 x2y3 + 8 xy± − 5 y5 by x2 - 4 xy + y2 to four terms in the quotient. CHAPTER VI. REMOVAL AND INSERTION OF BRACKETS. 64. We frequently find it necessary to enclose within brackets part of an expression already enclosed within brackets. For this purpose it is usual to employ brackets of different forms. The brackets in common use are ( ), {}, []. Sometimes a line called a vinculum is drawn over the symbols to be connected; thus a-b+c is used with the same meaning as a -(b+c), and hence 65. To remove brackets it is usually best to begin with the inside pair, and in dealing with each pair in succession we apply the rules already given in Arts. 25, 26. Ex. 1. Simplify, by removing brackets, the expression a - 2b - [4 a 6 b −{3 a − c + (5 a − 2 b − 3 a − c + 2 b)}]. a 2b 5a+2b+3a c+2b] 4a+6b+3a - c + 5a-2b-3 a + c - 2b =2a, by collecting like terms. Ex. 2. Simplify the expression -[-2x-3y-(2x-3y)+(3x-2y)}+2x]. The expression [-2x-3y-2x+3y+3x-2y}+2x] == =2x-3y+2x-3y-3x+2y+2x] =2x+3y-2 x + 3y + 3x-2y-2x = x+4y. 3. a [2a-3b-(4c-2 a)}]. {c − (a - b)}. [b + c]). 7. -(-(-(-x)))− (− (− y)). 8. [ab (c − a)}]-[b - {c — (a - b)}]. 9. 10. [−{−(b + c − a)}]+[−{− (c + a − b)}]. 5x-3y-(2x-(2y-x)}]. 11. -(-(-a))-(− (−(− x))). 12. 3a-[a + b − {a + b + c − (a+b+c+d)}]. 13.2a-[3x+(3c-(4y+3x+2a)}]. 14. 3x [5y- {6z (4x-7y)}]. 15. [5x-(11 y − 3 x)]–[5 y − (3 x − 6y)]. 20. [a-{a + (x − a)-(x − a) a}-2 a]. 66. A coefficient placed before any bracket indicates that every term of the expression within the bracket is to be multiplied by that coefficient. NOTE. The line between the numerator and denominator of a x-5 fraction is a kind of vinculum. Thus is equivalent to }(x − 5). 3 Again, an expression of the form √(x + y) is often written √x + y, the line above being regarded as a vinculum indicating the square root of the compound expression x + y taken as a whole. Thus whereas √25+ 144 = √169 = 13, √25+ √144 = 5 + 12 = 17. 67. Sometimes it is advisable to simplify in the course of the work. |