(3) Divide +ax2+bx+c by x—r. x—r) x3+ax2+bx+c (x2+(r+a)x+(r2+ar+b) In the preceding and similar examples, the remainder differs only from the dividend in having r instead of x. That this is always the case when the divisor is x minus some quantity, will be shown hereafter. (Art. 238, Pr. I.) (4) Divide x3-ax2+bx—c by x—r. (5) Divide x3-(a+b+c)x2+(ab+be+ca)x-abc by x-a. (6) Divide (a+2)x2+(2a+b)x-2b by x-2. (7) Divide 11ab-19abc+10a3-15a2c+3ab2+15bc2-5b c by 5a2+3ab -5bc. (8) Divide x3-(a+b+d)x2+(ad+bd+c)x−cd by x2—(a+b)x+c. (9) Divide x+pxTM¬1+qx2¬2+rxTM¬3+, &c....+tx+u by x-a. m-3 *N.B. The signs of the products to subtract are actually changed in this example before they are written; a method which is sometimes practised. Their first terms need not be written, since they are cancelled by the first terms of the corresponding dividends. (10) Divide at x++a3 \x3-a3b [x2+a1b3 \x—a2b6 by -a3b -a1b -2a1b2 +2a3b1| +a2b2 +a3b2 -ab5 -ab3 a2b3 +a2b3 a2 \x2+a3x—a2b3. — ab / 2° + a3 When there are negative exponents of the letter of arrangement, they come after the term containing ro, i. e., the term in which x does not appear, those which have the greatest absolute value being placed last. (11) Divide -3—x2+10x+}—13x1—3x2+3x¬3 by x2-2x−2+}x¬1 +3232-2. ANSWERS. (4) x2+(r-a)x+(r-ar+b), and remainder is rar+br-c. (5) x2-(b+c)x+bc. (6) x2-ax+b. (7) 2a+b-3c. (8) x-d. (9) xm-1+ax2+a23 \xm¬3+a3 x+, &c.....+am-1 +ap +a2p +aq +r +am-2p 21. In those cases in which the division does not terminate, and the quotient may be continued to an unlimited number of terms, the quotient is termed an infinite series, and then 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. 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 : (x+a)(x+b)=x2+(a+b)x+ab.....(1) (1) Resolve ar+bx2-cx2 into its component factors. Here ar+bx-cx2x2(a+b-c). (2) Transform the expression n3+2n2+n into factors. Here n3+2n2+n=n(n2+2n+1) =n(n+1)(n+1) by (6) (3) Decompose the expression r2-x-72 into two factors. By inspecting formula (3), we have −1=-9+8, and -72=-9×8; hence x2-x-72=(x−9)(x+8). (4) Decompose 5a2bc+10ab2c+15abc into two factors. (5) Transform 3m*n®—6m3n3p+3m2n1p2 into factors. (6) Transform 363c-3bc3 into factors. (7) Decompose x2+8x+ 15 into two factors. (8) Decompose x3-2x2-15x into three factors. (9) Decompose x-x-30 into factors. (10) Transform a2—b2+2bc-c2 into two factors. (11) Transform a2r-r3 into factors. (4) 5abc(a+2b+3c). (5) 3m2n1(mn-p)3. (6) 3bc(b+c)(b-c). (7) (x+3)(x+5). ANSWERS. (8) x(x+3)(x-5). (9) (x+5)(x-6). (10) (a+b-c)(a−b+c). (11) x(a+x) (a-x). 23. By the usual process of division we might obtain the quotient of a"-b" divided by ab, when any particular number is substituted for n; but we shall 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. a—b) a"-b" (a-1+ a"-an-b Rem. an-1b-b": Rem. under another form, b(a-1 —bn−1). b(an—1 —b"—1) a-b Now it appears from this result, that a"-b" will be exactly divisible by a—b, if a11—ba-1 be divisible by a-b; that is, if 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 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 since a3—b3 is divisible, a1-b1 must be divisible, and so on ad infinitum. 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, (5) 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) a1—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). (10) (a3-b3)(a—b)=a2+ab+b2. (11) (a3+b3)÷(a+b)=a2—ab+b2. (13) (ab)(a—b)=a+a3b+a2b2+ab3+6. DIVISION BY DETACHED COEFFICIENTS. 24. Arrange the terms of the divisor and dividend according to the successive 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 omitted letters in the divisor, and the result will be 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 But a1÷a=a3, and the literal parts of the successive terms, are, therefore a3, a3, a1, ao, or a3, a, a, 1; hence, 2a3+4a2+8a+16= quotient. (2) Divide 8a3 —4a1x—2a3x2+a2x3 by 4a2—x2. 4+0—1) 8—4—2+1 (2—1 8+0-2 -4+0+1 -4―0+1 Now, a5a2=a3; hence a3 and a'r are the literal parts of the terms in the quotient, for there are only two coefficients in the quotient; therefore 2a3-ax quotient required. (3) Divide 1-3ax3 — 8a2x2+18a3x-8a1 by x2+2ax—2a2. (4) Divide 3y3+3xy-4x2y-4x3 by x+y. (5) Divide 10a-27 a3x+34a2x2-18ax2-8x1 by 2a2-3ax+4x2. (6) Divide a+4a5-8a-25a3+35a2+21a-28 by a2+5a+4. 25. (1) Divide the divisor and dividend by the coefficient of the first term in * The rule here given for Synthetic Division is due to the late W. G. Horner, Esq., of Bath, whose researches in science have issued in several elegant and useful processes, especially in the higher branches of algebra, and in the evolution of the roots of equation of all dimensions. In the common method of division, the several terms in the divisor are multiplied by the first term in the quotient, and the product subtracted from the dividend; but subtraction is performed by changing all the signs of the quantities to be subtracted, and then adding the several terms in the lower line to the similar terms in the higher. If, therefore, the signs of the terms in the divisor were changed, we should have to add the product of the divisor and quotient instead of subtracting it. By this process, then, the second dividend would be identically the same as by the usual method. We may omit altogether the products of the first term in the divisor by the successive terms in the quotient, because in the usual method the first term in each successive dividend is cancelled by these products. Omitting, therefore, these products, the coefficient of the first term in any dividend |