7. Expand powers of x. Ans. 1- 2x + x2 1 = 1 + 2x + 3x2 + 4x3 + 5x + etc. 1+ 3x + 3×3 + 23 into a series of ascending 1 3x3 1+ 3x + 3x2 + x3 21+ etc. We will now proceed to treat of some cases of incomplete division, which will enable us to change certain algebraic expressions into new forms, which will be useful to us hereafter. = the remainder, which is the same as the given dividend if we put m for x; and since the dividend must (from the nature of division) equal the product of the divisor and quotient together with the remainder, we shall have ax + b = a(x — m) + am +b; so that ax + b is changed to a(x — m) + am + b. Ex. 2. To divide ax2 + bx + c by x ·m x max2 + bx + c | ax + am + b the remainder, which can be immediately obtained from the given dividend by changing x into m. By the first example, we can change the quotient ax + am + b into a(x — m) + 2am + b; and we shall of course have ax + bx + c = a(x − m)2 + (2am + b) (x — m) + am3 +bmc for the transformation of ax2 + bx + c. Ex. 3. To divide ax + bx2 + cx + d by x max3 + bx2 + cx + d | ax2 + (am + b)x + am2 + bm + c ax3 amx2 (am + b)x2 + cx (am + b)x2 — (am2 + bm)x (am2 + bm +c)x+d (am2 + bm +c)x-am3 —bm3 — cm am3 +bm3 +cm+d = the remainder, which is deduced at once from the given dividend by changing a into m. If in the last example we change b into am + b, and e into am2 + bm + c, ax2 + bx + c will become ax2 + (am + b)x + am2 + bm + c = a(x — m)2 + (3am + b)(x a(x — m)2 + (3am + b) (x − m) + 3am2 + 2bme the transformation of our quotient. Hence, we deduce ax + bx + cx + d = a(x − m)3 + (3am +b)(x — m)2 + (3am2 + 2bm + c)(x — m) + am3 + bm2 + cm +d, which is the transformation of ax3 + bx2 + cx + d. Ex. 4.-To transform Ax" + Bæn−1 +- Cx2-2 + + So +T, (A), so that shall be changed into am, supposing n to stand for a positive whole number, and the coefficients A, B, C,. . . . T to be independent of x. The transformation may clearly be effected as in the preceding examples; but we shall give another method, which is in some respects more simple. y, Thus, since x x − m + m, if we represent x -m by we have y+m, so that we may evidently put y + m for a in (A), which will change it to A(y+m)" + B(y + m)n−1 + C(y + m)n−2 + . . . . + S(y + m) + T, (B). .... Because (y+m)" = (y + m) × (y + m)"−1 = y(y + m)n−1 +m(y + m)"−1, we get A(y + m)" = y[A(y + m)"−1] + Am (y + m)n−1, (1); and adding B(y + m)"-1 to both sides of (1), we get A(y+m)" + B(y + m)"−1 = y[A(y + m)"~1] + (Am + B) (y + m)"-1, (2); .. if we change A in (1) into Am + B, and n into n 1, we get (Am + B) (y + m)n-1 = y[(Am + B) (y + m)n−2] + (Am2 + Bm)(y + m)"−2; consequently, if we add C(y + m)-2 to both sides of (2), we get A(y + m)” + B(y + m)n−1 + C(y + m)”−2 = y[A(y + m)”−1 + (Am + B)(y + m)"−2] + (Am2 + Bm + C) (y + m)"−2, (3). If we put Am2 + Bm + C for A in (1), and change n into n-2, it becomes (Am2 + Bm + C) (y + m)"−2 = y[(Am2 + Bm + C) (y + m)”−3] + (Am3 + Bm2 + Cm) (y + m)"−3; consequently, if we add D(y + m)2-3 to both sides of (3), we shall get A(y + m)” + B(y + m)”−1 + C(y + m)”~2 + D(y + m) n-3 = y[A(y + m)"−1 + (Am + B) (y + m)n−2 + (Am2 + Bm + C) (y + m)”−3] + (Am3 + Bm2 + Cm + D) (y +m)”—3, (4). It is easy to see, from the law of the preceding transformations, that if we continue the above process until it has been repeated n times, we shall have A(y + m)” + B(y + m)n−1 + C(y + m)n-2 + .... + S(y + m) + T = y[A(y + m)n−1 + (Am + B)(y + m)n-2 + (Am2 + Bm + C)(y + m)n-3 + (Am3 + Bm2 + Cm + D) (y +- m)" + .... + (Am”¬2 + Bm”¬3 + Cm”- + .. . . + R) (y + m) + Am"-1 + Bm"-2 + Cm2-3 + +S] + Am" + Bm"-1 + Cm"-3 + .... + Sm + T, (C). The quantity Am" + Bm"-1 + Cm"-2 + . . . . + Sm + T, which is independent of y in (C), is evidently the remainder which the division of (A) by - m would give, and it is deduced immediately from (A) by changing a into m. . . . . Hence, in order that (A) may be exactly divisible by xm, it will be necessary that m should be such as to reduce the aforesaid remainder to 0; or, which is the same, m must be a root of the equation Am" + Bm"-1 + Cm2-2 + ... Sm+T + Sm + T = 0; which is in conformity to the remarks made under Ex. 4 of the third case of Multiplication. .... The quantity A(y + m)n−1 + (Am + B) (y + m)”−2 + (Am2 + Bm + C) (y+m)n-3+ etc.....+ (Am"-2 + Bm"-3 + + R) (y + m) + Am"-1 + Bm"-2+....+S, which is within the braces in the right member of (C), and which is multiplied by y, is evidently of the same form as (B), and may be transformed by (C) merely by putting n-1 for n, Am + B for B, Am2 + Bm + C for C, and so on; then the part of the transformation which is independent of y will clearly be the coefficient of y in the transformation of (B), as re quired; and it is easy to see that nAm"-1 + (n − 1)Bm"-"+ (n − 2)Cmn-8 + ....+ 2Rm + S is the coefficient of y. The part of the last transformation that has y for a factor (as in (C)) being transformed as before by (C), the part of it which is independent of y will clearly be the coefficient of y, and so on, until the coefficients of all the powers of y in the transformation of (B) are obtained. It will not, however, be necessary formally to obtain them; for we see from what has been done, that they can easily be found by the following very simple rule. RULE. Write the coefficients of (A) or (B) in a horizontal line, supplying any term that may be wanting in the given quantity (A) (to make the exponents of x decrease by a unit as we pass from any term to the next successive term) by 0, thus we have A+B+C+ .... +S+T, (D). Then multiply A by m, according to their signs, and add B to the product, and denote the result by B', and we have Am + B = B'; multiply B' by m, according to their signs, and add C to the product, and denote the result by C1, and we have Am2 + Bm + C = B'm + C = C'; proceed in the same way as before with C and D, and so on; and the result, whose number is denoted by the number of units in n, will be the quantity in the transformed expression of (B), which is independent of y. The coefficient of y is found in the same way, by using B', C', etc., instead of B, C, etc.; observing that the coefficient of y is the result whose number is n - 1.. If we put Am + B' = B", B"m + C' = С", С"m + D' = D" and so on; then using A, B", C", etc., as before, and taking the result whose number is n-2, we have the coefficient of y. Proceeding in like manner with the preceding results, we get the coefficient of y', and so on, for the coefficients of y, y, etc., until all the coefficients are found. By writing the coefficients, (D), of the given equation, in the form of a dividend, and m for the corresponding quotient, we may express the rule, and the manner of applying it to particular examples, in the following and so on, until we arrive at the coefficient of y" in the transformed equation, which (coefficient) is clearly A, the coefficient of a" in the expression to be transformed. Where it is to be observed that T is used to express the term of the transformed equation which is independent of y, and S" expresses the coefficient of y, and R" that of y2, and so on. Remark. Our rule is substantially the same as that given by W. G. Horner, Esq. (of Bath, England), for the purpose of solving equations; but it has been obtained by a much more simple and obvious method. Horner's method consists in dividing (A) the equation to be transformed by x-m, then the remainder is the term of the transformed equation which is independent of y. Then he divides the quotient by x-m, and the remainder is the coefficient of y. And dividing the last quotient by x-m, the remainder is the coefficient of y. Proceeding in the same way with the last quotient, we get the coefficient of y3, and so on for the coefficients of y', y3, etc. Horner abridges the required divisions very much by detaching the coefficients as in (D), and then dividing by 1-m, by Synthetic Division, or Rule III. We will now illustrate the rule by some particular examples, which we shall perform according to the scheme, (E). EXAMPLES. Ex. 1.-To transform 3x2+5x+7, supposing that x = 3 +2. Here we have n = 2, m = 2, A = 3, B = 5, C = 7; then by (E) we have the following process. |