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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.

Root of a Polynomial.

220. EXAMPLES.

1. Find the 4th root of 81 x8-216 x7 +336 x5 — 56 xa - 224 x3 + 64 x + 16.

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.

81x8—216x7+336x5—56x1—224x3+64x+16|3x2-2x-2.

81 28

1st Rem.216x7 108 x64 × (3x2)3

81x8-216x+216 x6 — 96 x5 +16 x4 = (3 x2—2 x)4

2d Rem.

-216x6

| 108 x6

81x8-216x7+336x5–56x1–224x3+64x+16=(3x2-2x−2)4

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2. Find the 3d root of a3+3a2 b + 3 a2 c + 3 a b2 + 6abc+3 ac2+63 +3b2c+3bc2+c3.

Ans. a+b+c.

3. Find the 3d root of a3+6a2b−3 a2c + 12 a b212 a b c +3 ac2+8 63-12 b2 c + 6 b c2 — c3.

Ans. a+2b-c.

4. Find the 3d root of 343 26-441 x5 y +777 x4 y2 — 531 x3 y3+444 x2 y1 — 144 x y1 +64 y®.

Ans. 7x2-3 x y +4y2.

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Root of a Polynomial.

6. Find the 5th root of 16807 a10 b5

12005 a8b4+ 1715 a6 b3 — 24810 at b7 c — 245 at b2 — 686o a2 b c + 1a2 b+245 b5c — 3+13729 a-2b9c2—35 a-2b4c289a-4b8 c2+24 a−4 b3 c +39 a−6 b7 c2 + 3929 a−8 b11 c3 — § a−8 b6 c2 - 560 a-10 610 c3 + 34 a−12 69 c3 + 560 a-14 b13 c4-ff a-16 b12 c4 +32a-20 615 c5.

Ans. 7a2b-+}a-4b3 c.

7. Find the 9th root of y27 +27 y25 +324 y23+2268 y21 +10206 y19 + 30618 y17 +61236 y15 + 78732 y13 + 59049 y11+19683 yo.

Ans. y3+3y.

221. Corollary. When the preceding method is applied to the extraction of the square root, it admits of modifications similar to those of art. 189, and 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.

Square Root of a Polynomial.

Proceed in the same way, to find the other terms

of the root.

222. EXAMPLES.

1. Find the square root of 26 + 4 x5 + 20 x2 +16.

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Solution. In the following solution, the arrangement is similar to that in the example of art. 190.

x3+2

x6 +-4 x5 +20 x2-16x+16 | x3 + 2x2-2x+4. Ans.

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— 4 x4 +20 x2 — 16x+16 | 2x3 + 4x2
-4x48x3 + 4x2

8x316x216x+16 | 2 x3 + 4x2 - 4x
8x316x16x16

0.

2. Find the square root of 25 a4-30 a3 b + 49 a2 b2 24 a 63+ 16 b4. Ans. 5a2-3 ab+462.

3. Find the square root of 4x6 + 12 x5 +5x42x3 +7x22x+1. Ans. 2x+3x2 —x + 1.

4. Find the square root of a

2 a x3 + x1.

2 a3 x + 3 a2 x2 — Ans. a2-a x+x2.

ax

5. Find the square root of +6 x — 17 x2 - 28 x3

+49 4.

Ans. 2x-7 x2,

Solution of Binomial Equations.

SECTION V.

Binomial Equations.

223. Definition. When an equation with one unknown quantity is reduced to a series of monomials, and all its terms which contain the unknown quantity are multiplied by the same power of the unknown quantity, it may be represented by the general form

Ax"M=0,

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Hence, find the value of the power of the unknown quantity which is contained in the given equation, precisely as if this power were itself the unknown quantity, and the given equations were of the first degree. Extract that root of the result which is denoted by the index of the power.

225. Corollary. Equations containing two or more unknown quantities will often, by elimination, conduct to binomial equations.

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