Therefore 1023 is the product of 22 and the term con taining the highest power of x in the quotient; and hence the term containing the highest power of x in the quotient is 10 divided by 2x2, or 5x. Multiplying the divisor by 52, we have the product 10x3-15 x2- 20 x; which, when subtracted from the dividend, leaves the remainder -6x2+9x+12. This remainder is the product of the divisor by the rest of the quotient; therefore, to obtain the next term of the quotient, we regard −6x2+9x+12 as a new dividend. Dividing the term containing the highest power of a by the term containing the highest power of x in the divisor, we obtain 3 as the second term of the quotient. Multiplying the divisor by 3, we have the product −6x2+9x+12; which, when subtracted from the second dividend, leaves no remainder. Hence 5x-3 is the required quotient. It is customary to arrange the work as follows: 10-21-11x+12|2x2-3x-4, Divisor. Note. The example might have been solved by arranging the dividend and divisor according to the ascending powers of x. 102. From Art. 101 we derive the following rule: Arrange the dividend and divisor in the ame order of powers of some common letter. Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by this term, and subtract the product from the dividend, arranging the remainder in the same order of powers as the dividend and divisor. Regard this remainder as a new dividend, and proceed as before; continuing until there is no remainder. 103. The operation of divison may be shortened in certain cases by the use of parentheses. Example. Divide (a2+ab)x2+(2 ac +be+ad)x+c(c+d) by ax + c (a2+ab) x2+(2 ac + bc + ad)x + c(c + d) [ax + c (a2+ab)x2+( ac + bc ( ас (ac )x (a+b)x+(c+d) +ad)x+c(c+d) +ad)x+c(c+d) 104. It is evident from Art. 91 that, if the dividend and divisor are homogeneous, the quotient will also be homogeneous, and its degree will be equal to the degree of the dividend minus the degree of the divisor. Also, if the dividend or divisor are symmetrical with respect to any letters, the quotient, if it involves the letters at all, will be symmetrical with respect to them. 105. Division by Detached Coefficients. In finding the quotient of two expressions which are arranged according to the same order of powers of some common letter, the operation may be abridged by writing. only the numerical coefficients and signs of the terms. If the term involving any power is wanting, it may be supplied with the coefficient 0. Example. Divide 6x+2x-9x1+5x2+18x-30 by 3x3x2-6. 1 6+2 9+0+5+18-30|3+1+0-6 2+0-3+5 Therefore the required quotient is 2x3-3x+5. EXAMPLES. 106. Divide the following: 1. 62+15+51x-18 by 2x-4x2+7x-2 2. x-6xx-6 by x+2x+3. 3. a*+a2b2 +25b1 by (a - b) (a − 5 b) +3 ab. 4. m3 48 17 m3 +52 m + 12 m2 by m-2+ m2. 5. 2a53a2b3 — 49b3 — 7 a3b2-9ab by 2a2— 5 ab — 7 b2. 6. x-6x+5x-1 by x+2x2 - x − 1. 7. 2x-6y 12z+xy-2xz+17 yz by 2x+4z-3y. 8. a2n — b2m +2b"c-c2 by a" + bTM — c" ; n, m, and r being positive integers. 9. x-1-6x-3x2 by -2x2 — x + x3 — 1. -- 10. 12a14ab + 10 ab2 ab3-8 ab+4b5 by 6a-4ab3 ab2+2b3. 11. x2+(a+b+c) x2 +(ab+be+ca)x + abc by x2+(b+c)x + bc. 12. (b+c)a2 + (b2 + 3 bc + c2) a + be (b+c) by a+b+c 13. (x+y)-5(x+y)+6 by (x + y) - 2. 14. x2+(a+b−c) x2+(ab - beca)x- abe by x2+(bc)x-bc. 15. (mn)-2(m — n)2 + 1 by (mn)2-2 (m − n) + 1. 16. 2+(ab+c) x2+(acab - be)x-abc by x+c. 17. x4(3-)x+(c−3b-2) x2+(2b+3c)x-2c by x+3x-2. 18. (a-3 ab) x2+(2 a2+4 ab+3b2)x-(2ab+5b2) by axb. VI. FORMULÆ. 107. A Formula is the statement in algebraic symbols of a general rule. 108. The following results are of great importance in abridging algebraic operations: In the first case we have (Art. 8), (a+b)2= a2+2ab+b2. (1) This formula is the symbolical statement of the following rule: The square of the sum of two numbers is equal to the square of the first, plus twice the product of the first by the second, plus the square of the second. In the second case, (a - b)2= a2 - 2ab+b2. (2) That is, the square of the difference of two numbers is equal to the square of the first, minus twice the product of the first by the second, plus the square of the second. In the third case, (a+b) (a - b) = a2 — b2. (3) That is, the product of the sum and difference of two numbers is equal to the difference of their squares. Note. In the present chapter we shall, for the sake of brevity, use the expression "difference of two numbers" to denote the remainder obtained by subtracting the second from the first. 109. In connection with the formulæ of the present chap ter, a rule for raising a rational and integral monomial (Art. 23) to any power whose exponent is a positive integer, will be found convenient. 1. Required the third power of 5a3b. By Art. 8, (5a3b)3 =5a3b × 5a3b × 5a3b = 125 ab3. ( — b)1 = ( — b) × ( − b ) × ( − b ) × ( − b) = b1 (Art. 88). 3. Required the third power of - 3 m2. (-3 m2)3 = (-3 m2) × (-3 m3) × (−3 m3) — — 27m3. == 4. Required the value of (a)", where m and n are any positive integers. (aTM)n = aTM × aTM × am × ... to n factors =α m+m+m+to n terms = amn 5. Required the value of (ab)", where n is any positive. integer. (ab)" = ab × ab × ab × ... to n factors = (axax... to n factors) (bxbx to n factors) = an br. We then have the following rule: ... Raise the absolute value of the numerical coefficient to the required power, and multiply the exponent of each letter by the exponent of the required power. Give to every power of a positive term, and to every even power of a negative term the positive sign, and to every odd power of a negative term the negative sign (Art. 88). |