We first arrange the dividend and divisor with reference to a (¿ 44), placing the divisor on the right of the dividend. Divide the first term of the dividend by the first term of the divisor. The result will be the first term of the quotient, which, for convenience, we place under the divisor. The product of the divisor by this term (6 a2 + 6 a), being subtracted from the dividend, leaves a new dividend, which may be treated in the same way as the original one; and so on to the end of the operation. Since all similar cases may be treated in the same way, we have, for the division of polynomials, the following rule: Arrange the dividend and the divisor with reference to the same letter. 1 Divide the first term of the dividend by the first term of the divisor, for the first term of the quotient. Multiply the divisor by this term of the quotient, and subtract the product from the dividend. Divide the first term of the remainder by the first term of the divisor, for the second term of the quotient. Multiply the divisor by this term, and subtract the product from the first remainder, and so on. Continue the operation until a remainder is found equal to 0, or one whose first term is not divisible by that of the divisor. NOTES.-1. When a remainder is found equal to 0, the division is exact. 2. When a remainder is found whose first term is not divisible by the first term of the divisor, the exact division is impossible. In that case, write the last remainder after the quotient found, placing the divisor under it, in the form of a fraction. (2) Let it be required to divide 51a2b2 + 10 a1 — 48 a3b — 15 b1 +4ab3 by 4ab — 5a2+3b2. NOTE. First arrange the dividend and divisor with reference to a. 10a-48 a3b+51a2b2+ 4ab3 - 156-5a2+4ab+3b2 +10a. 8a3b 6a2b2 -40a3b+57a2b2+ 4ab3-156* 25 a2b2 - 20 ab3 — 15b1 25 a2b2 - 20 ab3 — 15 b1 NOTE. Here the division is not exact, and the quotient is fractional. (4) 1+ a 1 NOTE. 1- a 1+2a+2a3 +2 a3 +, etc. +2a +2a-2a2 +2a2 +2a2-2a3 In this example the operation does not terminate: it may be continued to any extent. 5. a3 — 5a*x+10a3x23 — 10 a2x3+5 ax1 — x5 by a2 — 2ax+x2. Ans. a3-3 ax + 3 ax2 — x3. 6. 48x3- 76ax2 — 64 a2x + 105a3 by 2x-3a. -- Ans. 24x2- 2ax — 35a2. 7. y® — 3y*x2+3y2x1 — x® by y3 — 3y2x + 3yx2 — x3. Ans. y+3yx+3yx2+x3. 8. 64 a'b-25a2b by 8a2b3+5ab'. Ans. 8a2b3-5 ab1. 9. 6a+23 ab+22 ab2 + 5 b3 by 3a2+4ab+b2. 10. 6ax+6axy + 42 a2x2 by ax+5ax. Ans. 2a+56. Ans. x5 + xy + 7 ax. 11.15a37a2bd-29 a'cf20b3d+44bcdf-8c'f' by Ans. 3a5b2+3c2. 14. 3a-8a2b2 + 3 a2c2+5b* - 3 b2c2 by a2 - b2. 15. 6x6 — 5x3у2 — 6x*y* + 6 x3y2 + 15 x3y3 — 9 x2y* + 10x2y3 - +15y by 32+2x2y2+3y2. Ans. 2x3-3x2y2+5y3. 17. 3x+4x3y - 4x2 - 4x3y2+16xy-15 by 2xy + x2-3. 18. +32y by x+2y. Ans. 3x2-2xy +5. Ans. x-2xy + 4x3y3 — 8xy3+16y. 19. 3a-26 ab - 14 ab3 +37a2b2 by 262-5ab+3a2. 20. a*b* by a3 + a2b+ab2 + b3. Ans. a2-7 ab. Ans. a - b. 21. x3-3x2y+y3 by x+y. Ans. x2-4xy+4y2 — 3ys x + y 22. 1+2a by 1-a-a2. Ans. 1+3a +4a2+7 a3 +, etc. CHAPTER III. FACTORING, GREATEST COMMON DIVISOR, AND LEAST COMMON MULTIPLE. USEFUL FORMULAS. 53. A formula is an algebraic expression of a general rule or principle. Formulas serve to shorten algebraic operations, and are also of much use in the operation of factoring. When translated into common language, they give rise to practical rules. The verification of the following formulas affords additional exercises in multiplication and division. 54. Formula 1. To form the square of a+b, we have (a+b)2 = (a+b)(a+b) = a2 +2ab + b2 = a2 + b2+2ab ; that is, The square of the sum of any two quantities is equal to the sum of their squares, plus twice their product. (1) Find the square of 2a+36. We have from the rule, (2a+3b)2 = 4a2 +962 + 12 ab. (2) Find the square of 5 ab+3ac. Ans. 25a2b2+9 a2c2+30 a2bc. (3) Find the square of 5a2+8a2b. Ans. 25 a 64 a1b2 + 80 a*b. (4) Find the square of 6ax +9 a2x2. Ans. 36a2x2+ 81 a*x* + 108 a3x3. |