(1.) Divide a x+ax1+b x1+ax3+bx3+cx3+ax2+bx2+cx2+bx+cx+e by a x2+bx+c. Arrange the terms of the dividend in the following manner, in order to keep the operation within the breadth of the page. ax2+bx+c) ax2+ax2+ax3+a\x2+b\x+c (x2+x2+x+1. ax2+a\x*+a\x2+a\x2+b\x+c 이 이 (2) Divide x3+a x2+bx+c by x—r. (r+a) x2+bx (r+a) x2—(r2+ar) x (r2+ar+b)x+c (y2+ar+b) x—(r3+ar2+br) +ar2+br+c Remainder. In the preceding and similar examples, the remainder differs only from the dividend in having r instead of s. (4.) Divide z3—(a+b+c)x+(ab+bc+c a) x—a b c by x-e (5.) Divide x3—(a+2) x2+(2a+b)x−2b by x−2. (6.) Divide 11 a2b-19 a b c +10 a3 — 15 a2 c + 3 a b2 + 15 b c2 —5 b2 e by 5a2+3a b-5b c. (7.) Divide 3—(a+b+d) x2+(ad+b d+c) x—c d by x2—(a+b) x+c. ANSWERS. (3.) + (ra)x +(r2—ar+b), and remainder is rar2+br-c. (4.) x2—(a+b)x+ab. (6.) 2a+ (b-3c). (5.) x2-ax+b. 21. In those cases in which the division does not terminate, and the quotient may be continued to an unlimited number of terms; then the quotient s termed an infinite series, and the successive terms of the quotient are generally regulated by a law, which in most cases is readily discoverable. The quotient in this case is called an infinite series, and the law of formation of this series is that any term in the quotient is the product of the immediately preceding term by x. (2.) Divide 1 by 1+z. (4.) Divide 1 by x+1. (5.) Divide r—a by x-b. Ans. 1—x+ x2—x3 +x1 — .... Ans. 1+2x+2x2+2x3+2x++. Ans. x-x+x3-x--x--... Ans. 1-(a-b) x--(a—b) bx-2-(a—b) b2x-3— Ans. 1+2x+3x2+4x3+5x1+ ... (6.) Divide 1 by 1—2x+x2. 22. When a polynomial is the product of two or more factors, it is often requisite to resolve it into the factors of which it is composed, and merely to indicate the multiplication. This can frequently be done by inspection, and by the aid of the following formulas: (1.) Resolve aux2-cx2 into its component factors. Here a ab x2—c x2=x2 (a+b—c). (2.) Transform the expression n3+2 n2+n into factors Here n3+2n2+n = n (n2 +2n+1) = n (n+1) (n+1) by (6) = n (n+1)2. (3.) Decompose the expression x2-x-72 into two factors. By inspecting formula (3) we have —1——9+8, and -72=-9x8; hence x2-x-72 = (x−9) (x+8). (4.) Decompose 5a2 b c +10a b2c+15a b c2 into two factors. (5.) Transform 3m3 no—6m3 n3 p+3m2 n4 p2 into factors. (6.) Transform 3b3 c-3b c3 into factors. (7.) Decompose x2+8x+15 into two factors. (8.) Decompose 2a3—2x2-15x into three factors. (9.) Decompose x2-x-30 into factors. (10.) Transform a2—b2+2b c—c2 into two factors. (11.) Transform a2x-x3 into factors. 23. By the usual process of division we might obtain the quotient of a"-b" divided by a-b, when any particular number is substituted for n; but we sal here prove generally that a"-b" is always exactly divisible by a-b. and exhibit the quotient. It is required to divide a"-b" by a-b. Now it appears from this result, that a"-b" wili ne exactly divisible by a—b, if a"—— be divisible by a-b; that is, it the difference of the same powers of two quantities is divisible by their difference; then the difference of the powers of the next higher degree is also divisible by that difference. But a-b2 is exactly divisible by a-b, and we have a2-b2 = a+b... a b (2). And since a2—b2 is divisible by a-b, it appears from what has been just proved, that a3—b3 must be exactly divisible by a-b; and hence, by putting 3 for n in formula (1), we get a2-b2 a-b Again, aa—ba must be exactly divisible by a-b, since a3-b3 is divisible by a-b; hence, by writing 4 for n in formula (1), we have Hence, generally, a"-b" will always be exactly divisible by a−b, and give the quotient In a similar manner we find, when n is an odd number, a"-b" a+b By substituting particular numbers for n, in the formulas (5), (6), (7), we may deduce various algebraical formulas, several of which will be found in the following deductions from the rules of multiplication and division. USEFUL ALGEBRAIC FORMULAS. (1.) a2—b2 = (a+b) (a−b). (2.) a' b1 = (a2+b2) (a2—b2) = (a2+b2) (a+b) (a—b). (3.) a3-b3 = (a2+ab+b2) (a−b). (4.) a3+b3 = (a2—ab+b2) (a+b). (5.) a-b (a3+b3) (a3-b3) = (a3+b3) (a2+ab+b2) (a—b). (6.) ao—b3 = (a3+b3) (a3—b3) = (a3—b3) (a2—ab+b2) (a+b). (7.) a b = (a3+b3) (a3-b3) = (a2-b2) (a1+a2 b2+b). (8.) a b = (a+b) (a−b) («2+ab+b2) (a2—að j«?? (9.) (a2—b2) ÷ (a−b) = a +b. (10.) (a-b3)+(a−b) = a2+ab+b2. (11.) (a3+b3) ÷ (a+b) = a2—ab+b2. (12.) (a1—b1) ÷ (a+b) = a3—a2b+a b2—b3. (13.) (ab) + (a−b) = u1+a3b+a2 b2+a b3+b1. (14.) (a+b)+ (a+b) = a'—a3b+a2¿2—a b3+b. (15.) (ab) + (a2—b2)=a1+a2b2+b1. DIVISION BY DETACHED COEFFICIENTS. 24. Arrange the terms of the divisor and dividend according to the suc cessive powers of the letter or letters common to both; write down simply the coefficients with their respective signs, supplying the coefficients of the absent terms with zeros, and proceed as usual. Divide the highest power of the omitted letters in the dividend by that of the suppressed letters in the divisor, and the quotient will give the literal part of the first term in the quotient. The literal parts of the successive terms follow the same law of increase or decrease as those in the dividend. The coefficients prefixed to the literal parts will give the complete quotient, omitting those terms whose coefficients are zero. EXAMPLES. (1.) Divide 6a1—96 by 3a-6. 3—6) 6+0+ 0+0−96 (2+4+8+16 6-12 12 12-24 24 24-48 48-96 48-96 But a1÷a=a3, and the literal parts of the successive terms are therefore a", a2, a1, ao, or a3, a2, a, 1; hence, 2a3+4a2+6u+io=quotient. (2.) Divide 8a5-4a* x—2a3 x2+a2 àa3 by 4a2—x2. 8+0-2 −4+0+1 Now, a3÷a2=a3; hence a3 and a2x are the literal parts of the terms in the quotient, for there are only two coefficients in the quotient; therefore 2a3-a2 x=quotient required. (3.) Divide x1—3a x3—8a2 x2 + 18a3 x—8a1 by x2+2a x—2a3. (4.) Divide 3y+3x y2—4x2 y—4x3 by x+y. (5.) Divide 10a1—27a3 x+34a2x2—18a x3—8x1 by 2a2—3a x+4x2 (6.) Divide aR+4a3—8a*—25a3+35a2+21a−28 by a2+5a+4. (3.) x2-5a x+4a* ANSWERS. (5.) 5a2-6α x—2xo. |