An Elementary Treatise on Algebra

Root of a Polynomial.

60 a2 b3 c +6 a b5 +20 a3 c3 +90 a2 b2 c2+30 a b c + bε +60 a2 b c3 + 60 a b3 c2 + 6 b5 c + 15 a2 c4 +60 a b2 c3 + 15 b4 c2 +30 a b ca + 20 b3 3 + 6 a c5 + 15 b2 c4 + 6bc5+ c6.

4. Find the 4th power of a3-a22 x + a x2 — x3.

Ans. a12-4 all x + 10 a1o x2—20 a9 x3 +31 a3 x4 — -40 a7x5-44 a6 x6-40 a5 x7 +31 a4 x8-20 a3 x2+ 10 a2 x10 4 a xll+x12.

5. Find the square of a + bx + c x2 + dx3 + ex2+ƒx3. Ans. a2+2abx+2ac | x2+2ad | x3+2 ac\x4+2af|x5

+2bc2bd

+ b2

+ c2

+2be +2cd

+2bf\x6 + 2 cf|x2 + 2 d ƒ | x® + 2 e ƒ x2 +ƒ2 x10.

+2ce + 2 de

+ d2

༠༩! ]° +

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 R1-1 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 secord 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.

81 28-216x7+336x5-56x4-224x3+64x+16|3x2-2x-2. 81 28

1st Rem. -216 r7 108x6 = 4 × (3 x2)3

81 x8 — 216 x7+216 x6 — 96 x3 +16 x1 = (3x2—2x)4

2d Rem.

-21626

| 108 x6

81x8-216x7+336x5–56x1–224x3+64x+16=(3x2—2x-2)1

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

Ans. a+b+c.

3. Find the 3d root of a3+6 a2 b — 3 a2 c + 12 a b2 — 12 a b c +3 a c28b3-12 b2c+6bc2-c3.

Ans. a2b-c.

4. Find the 3d root of 343 26-441 x5 y+777 x4 y2 — 531 x3 y3+444 x2 y4 — 144 x y4+64 yo.

Ans. 7x2-3xy+4y2.

5. Find the 4th root of 81 a4 — 540 a3 b

- 72 a3 c+

1350 a2 b2+ 360 a2 b c +24 a2 c2-1500 a b3 — 600 a b2 c

— 80 a b c2 — 32 a c3 +625 b4 + 1000 b3 c + 20° b2 c2 + · 180 b c 3 + 18 c1.

Ans. 3a-5b-c

Root of a Polynomial.

6. Find the 5th root of 16807 al0 65

1715 a6 63—24810 a4 b7 c 245 at b2 — 68 69 a2 b c + £§ a2 b+245 b5 c — 3+13720 a-2b9c2—35 a-2b4c289a-4b8 c2+2a-4 b3 c+3o a−6 b7 c2+3939 a−8 bl1 c3 560 a-10 610 c33 + 29 a−12 bε c3 +

a-8 b6 c2

a−20

569a-14 613 c4 - a-16 612 c4 +2a-20 615 c5. • 좋은.

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 y1119683 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 x6 + 4 x5 + 20 x2 +16.

. 16 x

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

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

ენ

4x520x216x+16|2x3

4x5+ 4x4

— 4 x4 +20 x2-16x+16|2 x3 + 4x2