Polynomial Theorem. the derivative of V with reference to q is and that of Vn is x+1, n Vn-1x+1. Let, now, the required power be the nth and let ... ... ... and let the derivatives of A, B, C, P, Q, &c., with reference to p be A', B', C', P', Q', &c., and with reference to q, A", B", C", P", Q", &c.; the derivatives of the preceding equation with reference to p and q are nV1x=A'+ B'x + C' x2... + P' x2 + Q' x++1+ &c. nVn¬1xa+1=A"+B" x+C" x2... +P" x2+Q" x++1+ &c. the first of which, multiplied by x, is nVn¬1x*+1=A'x + B' x2+C' x3... +P'x✨+1+Q'x?+2+ &c., which, compared with the other, gives, by art. 163, for the coefficient of x¢+1‚ P'Q", that is, the derivative of any coefficient with reference to any letter p, is the same with that of the succeeding coefficient with reference to the succeeding letter q. Thirdly. Every term of Q, in which q is the letter farthest advanced in the alphabet, such as Tp" qo, must give in Q" a similar term 0 Tp" q-1, or else, if 0 = 1, a term Tp", in which there is no letter so far advanced as q. Every such term, as it belongs also to P', must be the derivative of a similar term in P, that is, of a term in which p is either the farthest advanced letter, or the next to the farthest advanced letter. The terms of PQ" Polynomial Theorem. are, then, obtained from those of P by derivation, while those of Q are obtained from those of its derivative Q" by a process, which is the reverse of derivation, and which, by art. 172, consists in multiplying by q, that is, in increasing its exponent by unity, and dividing by its exponent thus increased. This process is identical with the last paragraph of the rule in this article, and the three preceding paragraphs refer merely to those cases in which the increased exponent of q is unity, so that the division by it is superfluous. 217. Corollary. If x is put equal to unity in the value of (a + bx + c x2+&c.)", we have the value of (a+b+c+ &c.)", so that any power of a polynomial, the terms of which contain no common letter, is readily found by multiplying the successive terms, after the first, respectively by x, x2, x3, x1, &c., obtaining the power of the polynomial thus formed, and putting 1. Find the 5th power of 1+2x+3x2+4 x3. Solution. Represent the successive coefficients 1, 2, 3, 4 by a, b, c, and d, so that a = 1, b = 2, c=3, d= 4; and the given polynomial becomes a + bx + c x2 + d x3, Polynomial Theorem. the fifth power of which, is found by the rule to be +20bcd3 5ad4|x12+5bd4x13+5cd4x14+d5x15. +10c2d3 +30bc2d2 +10 c3 d2 Now, if we substitute for a, b, c, d, their values, the preceding expression becomes (1 + 2 x + 3 x2 + 4 x3)5 = 1 + 10 x + 55 x2 + 220 x3 690x+1772 x5+3830 x6 +7040 27+ 11085814970 x9 +17203 x10 +16660 x11 + 13280 x128320 x13+ 3840 x14+1024x15. 2. Find the 3d power of a + bx + c x2. Ans. a3+3 a2 b x + 3 a2 c | x2 キ 3 a c2x2 + 3b c2 x5 + c3 x6. 3 b2 c 3. Find the 6th power of a+b+c. Ans. a6 +6 a5 b + 6 a5 c + 15 aa b2 + 30 aa b с + 20 a3 b3 + 15 a1 c2 +60 a3 b2 c + 15 a2 b1 +60 a3 b c2 + Root of a Polynomial. 60 a2 b3 c +6 a b5 +20 a3 c3 + 90 a2 b2 c2 +30 a b c + b6+60 a2 b c3 +60 a b3 c2 + 6 b5 c + 15 a2 c4 + 60 a b2 c3 +15 b4 c2 +30 a b c1 +20 b3 c3 + 6 a c5 + 15 b2 c++ 6 b c + c°. 4. Find the 4th power of a3 — a2 x + a x2 — x3. a8 Ans. a12—4 all.x10 a10 x2—20 ao x3 +31 a3 x4 —— a12. -40 a7 x5-44 a6 x6-40 a5 x7+31 a4 x8-20 a3 x2+ 10 a2 x10 4 a x11+x12. 5. Find the square of a+bx+c x2 + d x3 + ex1 +ƒ x5. Ans. a2+2abx+2ac1x2+2ad x3+2 ae|x4+2af|x5 +2bc2bd2be + c2+2cd +2bfx6+2cf|x+2df | x2 + 2e ƒ x9 +ƒ2 x10. + b2 +2ce + d2 + 2 de + e2 219. Problem. mial. SECTION IV. Roots of Polynomials. To find any root of a polyno Solution. If the root is arranged according to the powers of either of its letters as x, whether ascending or descending, it is evident from the rule of art. 216, that The term of the required root which contains the highest power of x, is found by extracting the root of the corresponding term of the given polynomial. If, now, R is the first portion of the root, and R' the rest of it, and if P is the given polynomial of which the nth root is to be extracted, we have P=(R+R')" = R" + n R"-1 R'+ &c. and if, in P—R" and n R^-1, only the first term is retained, the first term of the quotient is the first term of R'; and a new portion of the root is thus found, which, combined with those before found, gives a new value of R and of P-R", which, divided by the value of n R-1 already obtained, gives a new term of the root, and so on. Hence to obtain the second term of the root, raise the first term of the root to the power denoted by the exponent of the root, and subtract the result from the given polynomial, bringing down only the first term of the remainder for a dividend. Also raise the first term of the root to the power denoted by the exponent one less than that of the root, and multiply this power by the exponent of the root for a divisor. Divide the dividend by the divisor, and the quotient is the second term of the root. The next term is found, by raising the root already found to the power denoted by the exponent of the required root, subtracting this power from the given polynomial, and dividing the first term of the remainder by the divisor used for obtaining the second term. This divisor, indeed, being once obtained, is to be used in each successive division, the successive dividends being the first terms of the successive remainders. |