CASE II. 55. When the divisor only is a monomial. 1. Divide a x + ay + az by a. we divide each term of the dividend a x + a y+az by the divisor a, and connect the partial quotients by their proper signs. Hence, RULE. Divide each term of the dividend by the divisor, and connect the several results by their proper signs. 4. Divide 12 a x3- 24 a x2+42 a xy by 3 ax. 5. Divide 4 a2 x+8 a b x → 4 b2 x by 4x. by 3 a2x2. Ans. 2 x24 a x + 5 a2 x3. 7. Divide 12 at yo — 16 a5 y5 + 20 a y1 — 28 a y3 by 4 at y3. 8. Divide 5 x3 10 x2 - 15 x by 5x. 9. Divide 273 (a + x)2 — 91 (a + x) by 91 (a + x). Ans. 3 (a + x)-1=3a + 3 x − 1. a2 10. Divide 20 abc-4ac + 8acd12 a2 c2 by 4 ac. a* 11. Divide 16 a2x2- 32 a2x2+48 at x by 16 a2x2. 12. Divide 72 x2 y2 - 36 x3 y3 — 54 x y z3 by 18 x2 y2. Ans. 42 x y + 3x z3. 13. Divide 18 a x3 y — 54 x2 y2+ 108 c x5 y3 by 9x2y. 14. Divide 40 a b2+8 a2 b2-96 a3 b3 x3 by 8 a2 b2. 15. Divide 39 x1 z5 65 a x3 z6 + 13 x3 z5 by CASE III. 56. When the divisor and dividend are both polynomials. 1. Divide 3-3x2y + 3 x y2 -y by x2-2xy + y2. OPERATION. x2-2xy + y2) x3-3x2 y + 3 x y2 — y3 (x — y x3 x2 The divisor and dividend are arranged in the order of the powers of x, beginning with the highest power. x3, the highest power of x in the dividend, must be the product of the highest power of x in the quotient and in the divisor; therefore, =x must be the highest power of x in the quotient. The divisor x2 2 x y + y2 multiplied by x must give several of the partial products which would be produced were the divisor multiplied by the whole quotient. When (x2- 2 x y + y23) x = = x23 — 2 x2 y + xy is subtracted from the dividend, the remainder must be the product of the divisor and the remaining terms of the quotient; therefore we treat the remainder as a new dividend, and so continue until the dividend is exhausted. Hence, for the division of polynomials we have the following RULE. Arrange the divisor and dividend in the order of the powers of one of the letters. Divide the first term of the dividend by the first term of the divisor; the result will be the first term of the quotient. Multiply the whole divisor by this quotient, and subtract the product from the dividend. Consider the remainder as a new dividend, and proceed as before until the dividend is exhausted. NOTE. If the dividend is not exactly divisible by the divisor, the remainder must be placed over the divisor in the form of a fraction and connected with the quotient by the proper sign. 2. Divide + 4y by x2-2xy + 2y3. NOTE. By multiplying the quotient and divisor together all the terms which appear in the process of dividing will be found in the partial products. 4. Divide a xay + bx-by+z by x-y. 2 x − y) a x — a y + b x − b y + z (a + b + x 2 y 5. Divide 2 by 2 by -3 b2 y z + 6 b3 y + by z — y z Ans. 3b2y-by+y. by 2b-z. 7. Divide a3 + a2 + a2 x + a x + 3 ac + 3c by a + 1. Ans. a2+ax +3 c. 8. Divide a+b-d-ax-bx+ dx by a + b — d. 9. Divide 2 a 13 a3 y + 11 ay 8 ay3 + 2y by Ans. a6 ay +2y. 2 a2ay + y2. 10. Divide a3-3a2b+3ab3b3 by a22ab+b2. 19. Divide + 4x2 y2 + 3y by x+2y. 22. Divide 1 a1 by 1+ a + a2 + a3. Ans. 1 -a. 23. Divide 103-20 x2 y 30 y3 by x+y. 24. Divide 7 a x2 + 21 a x3 + 14 a by x + 1. 25. Divide 27 as y 8 ay by 3y2-2ay. SECTION VIII. DEMONSTRATION OF THEOREMS. 57. FROM the principles already established we are prepared to demonstrate the following theorems. THEOREM I. The sum of two quantities plus their difference is twice the greater; and the sum of two quantities minus their difference is twice the less. Let a and b represent the two quantities, and a > b; their sum is a+b; their difference, a — b. PROOF. 1st. (a+b)+(a − b) = a + b + a−b=2a; 2d. (a+b) (a — b) = a + b — a + b = 2b. Therefore, when the sum and difference of two quantities are given to find the quantities, RULE. Subtract the difference from the sum, and divide the remainder by two, and we shall have the less; the less plus the difference will be the greater. In the following examples the sum and difference are given and the quantities required. |