Now, if we substitute for a, a', a", a'"" their values, the preceding expression becomes

(1+ 2x + 3x2 + 4x3)5 = 1 + 10x + 55 x2 + 220 x3

+ 690 x4 + 1772x5 + 3670 x6 + 7040 x7 + 10245 x8 + 18810x 9 + 17203 x10 + 16660 x11 + 13280 x12 + 8320 x 13 + 3840 x14 + 1024 x15.

2. Find the third power of a + bx + c x2.

Ans. a3 +3a2bx+3a2cx2+6abcx3

3. Find the 6th power of a + b + c.

Ans. a6 + 6a5 b + 6 a5 c + 15 a4 b2 + 30 a b c +

4

20 a3 b3 + 15 a4 c2 + 60 a3 b2 c + 15 a2 64 + 60 a3 b c 2 + 60 a2 b3 c + 6ab5 + 20 a3 c3 + 90 a2 b2 c2 + 30 a b c + b6 + 60 a2 b c3 + 60 a b3 c2 +6 65 c + 15 a2 c4 + 60 a b2 c3 + 15 b4 c2 + 30 a b c2 + 20 b3 c3 + 6ac5 +

15 62 c4 + 6 b c5 + c6.

Root of a Polynomial.

4. Find the 4th power of a3 - a2 x + ax2 - 13.

Ans. a12 20 a3 x3 + 31 a8 x4 4 all x + 10 a10 x2 - 40 α7 x5 + 44 a6 x 6 - 40 a5 x7 + 31 a4 x8 - 20 a3 x + 10 a 2 x 10 - 4 a x 11 + x 12.

SECTION IV.

Roots of Polynomials.

140. Problem. To find any root of a polynomial.

Solution. Let the root to be found be the nth, and let the polynomial be represented by P, the terms of P being arranged according to powers of either of its letters, as x; and let R be the required root arranged according to the powers of the same letter.

Let Q represent the term of R which contains the highest power of x, and R(1) the remaining terms of R; and we have

R = Q+R(1);

and, by the binomial theorem,

P = R" = (Q+R(1))"

= Qn + n Qn-1 R(1) + &c.

Now it is evident, from inspection, that the term of the second member of this equation which contains the highest power of x is Q"; and, therefore, QTM must be equal to the term of P, which contains the highest power of x; so that if the term of P which contains the highest power of 2 is represented by O, and the remaining terms of P by P(1), we have

Root of a Polynomial.

that is, 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 polyno

that is, the second term of the root.

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.

Root of a Polynomial.

Divide the dividend by the divisor, and the quo

tient is the second term of the root.

If now we denote the root already found by S, and the remaining terms of the root by T, we have

and

R=S+T

P = R" = (S + T)"

= Sn + n Sn-1 T + &c.

P-Sn = n Sn-1 T + &c.

and it is evident that the first term of

P-Sn

is the same with the first term of

n S-1T;

so that, if we divide the first term of

P-S

by the first term of

nSn-1,

the quotient must be the first term of Tor the next term of the root.

But since the first term of S is the first term of the root or Q, the first term of

the same with the divisor used for obtaining the second term of the root.

Root of a Polynomial.

Hence, when the first terms of the required root are found, 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, then, being once obtained, is to be used in each successive division, the successive dividends being the first terms of the successive remainders.

Solution. The operation is as follows, in which the root is written at the left of the given power, and the divisor at the left of each dividend or remainder; and only the first term of each remainder is brought down.

81 x8-216x7+336x5-56x4-224x3+16x+1|3x2-2x-2

81 x8

1st Rem.

216 x7 | 108 x6 = 4 × (3x2)3

4

81x8-216x7 + 216 x6 - 96x5 + 16x4 = (3 x2 - 2x)+

2d Rem.

- 216 x6

| 108 x6

81x8-216x7+336x5-56x4-224x3+16x+1=(3x2-2x-2)4

2. Find the 3d root of a3+3a2b+3a2c + 3ab2

+6abc+3 ac2 + b3 +362c + 3bc2 + c3.

Ans. a + b + c.