- 531 x3 y3 +444 x2 y2 + 144 x y3 + 64 yo.

Ans. 7x23xy+4y2.

5. Find the 4th root of 81 a 540 a3 b 72 a3 c + 360 a2 b c + 24 a2 c2 1500 a b3 80 a b c2 32 a c3 + 625 ba + 19oo b3 c

+1350 a2 b2 600 a b2 c

+1a2b+24565c+1372o a−2 bo

9

980 a 4 b8 c2 + 24a-4b3 c + 29 a-6 b7 c2 + a-8b6c2 569 a 10 6 10 c3 +

3920 a- -8 611 c3

32 a 20 b15 c5.

29a-12 b9c3+560a-14 613 c4

243

a-16 b 12 c4 +

b3

Ans. 7 a2 b — + } a− 4 b3 c.

7. Find the 9th root of y27+27 y25 +324 y23 +2268 y21 +10206 y 19+ 30618 y 17+61236 y 1578732 y 13 + 59049 y1119683 y9.

Ans. y3+3y.

141. Corollary. When the preceding method is applied to the extraction of the square root, it admits of some modifications which shorten the labor of calculation. We have, in this case,

n = 2, n-1 = 1;

P= (S+T)2 = S2 + 2 S T + T2,

P-S22ST+ T2;

Square Root of a Polynomial.

so that the first term of 7 must be equal to the first term of P — S2, divided by the first term of 2 S; and, calling this term 'T, the next remainder must be

P. (S+ 'T)2 = P · S2 2 S T T2

=P-S2

that is, this new remainder is equal to the preceding remainder PS2 diminished by (2S+T) 'T.

Hence we have the following rule.

To extract the square root of a given polynomial. Arrange its terms according to the powers of some letter, extract the square root of the first term for the first term of the root.

Double the part of the root thus found for a divisor, subtract the square of this part of the root from the given polynomial, and divide the first term of the remainder by the divisor; the quotient is the second term of the root.

Double the terms of the root already found for a new divisor; subtract from the preceding remainder the product of the last term of the root multiplied by the preceding divisor augmented by the last term of the root. Divide the first term of this new remainder by the first term of the corresponding divisor, and the quotient is the next term of the root.

Proceed in the same way, to find the other terms of the root.

EXAMPLES.

1. Find the square root of 26 + 4 x5 + 20 x2

+16.

. 16 x

Square Root of a Polynomial.

Solution. In the following solution, the arrangement is similar to that in the example of the preceding article. x+4x520x216x+16x3+2x2-2x+4. Ans.

x6

— 4 x1 +20 x2 — 16x+162 x3 + 4x2

4 x4 8x34x2

3. Find the square root of 4x6 + 12 x3 + 5x4 2x3

+7x2 2x + 1.

4. Find the square root of a

2 ax3 + x1.

Ans. a2ax + x2.

5. Find the square root of 2+6x-17 x2-28 x3 + 49 x4. Ans. +2x-7 x2.

142. Corollary. Since any number, expressed by two or more figures, may be considered as composed of a certain number of units, tens, hundreds, &c. added together, it may be divided into portions separated by the sign+, and thus become a polynomial; and to this polynomial the rules for extracting the roots of polynomials may be applied with some changes, as in the following example :

Find the fourth root of 79502005521.

Solution. Since we have

14

1,

104 = 10000,

1004 = 10000 0000,

1000 10000 0000 0000,

&c.

Roots of Numbers.

it is evident that the fourth power of an integer between 1 and 10 cannot consist of more than four figures; that the fourth power of an integer between 10 and 100 must consist of more than 4 figures, and less than 8; that the fourth power of an integer between 100 and 1000 must consist of more than 8, and less than 12 figures, and so on. If, therefore, the given number is separated into portions of 4 figures beginning with the place of units, we can easily see, of how many figures its root must consist. This separation is thus effected by inserting periods, which is called dividing the number into periods,

795.0200.5521 ;

whence it appears that the required root must consist of three integral places.

Moreover, the given number is more than

its root must, therefore, be between 500 and 600; and the first figure on the left of the root must be 5, or we may consider the first term of the root as 500, and the other terms are obtained by the application of the rule as follows;

795.0200.5521 500+30 + 1 = 531. Ans.

(500)4625.0000.0000

1st Rem. 170.0200.5521/5000000004X (500) 3 Divisor.

=

Roots of Numbers.

But as the only object is to obtain the successive figures of the root, it is unnecessary to use so many cyphers, and the process may be thus written, more briefly,

795.0200.5521531

54 = 625

1st Rem. = 1700+1500 = 4 X 53 = Divisor.

(53) 4 =789.0481

2d Rem. 5.97+595,+= 4 × (5,3)3.

This method may be applied to any other case, and gives the following rule.

To find any root of a given number.

Point off the number into periods beginning with the place of units, each period containing the number of places denoted by the exponent of the power.

Find the greatest integral power contained in the left hand period; and the root of this power is the left hand figure of the root.

Raise the figure thus found of the root to the power, whose exponent is less by one than that of the required root, and multiply this power by the exponent of the root; this product is the divisor.

Raise the part of the root already found to the power denoted by the exponent of the root, subtract this power from the left hand period of the given number, bringing down the first figure of the next period for the remainder.