Root of a Polynomial. 220. EXAMPLES. 1. Find the 4th root of 81 x8-216 x7 +336 x5 — 56 xa -224x364 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 x7 108 x64 X (3x2)3 | 81x8-216x+216 x6 —96 x3 +16x4=(3x2—2x)4 2d Rem. -216x6 | 108 x6 81x8-216x7+336x5–56x1–224x3+64x+16=(3x2—2x—2)4 2. Find the 3d root of a3+3a2b+3a2c+3 a b2 + 6abc+3ac2+b3+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+8b3-12 b2c+6 b c2-c3. Ans. a2b-c. 4. Find the 3d root of 343 x6-441 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 a2 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 64+ 1900 63 c + 20° b2 c2 + 160 b c3 +18c4. Ans. 3 a-5b-c. Root of a Polynomial. 12005 a8 b4 + 6. Find the 5th root of 16807 a10 b5 1715 a6 b3 24010 a4 b7 c— - 2 4 5 a 4 b2 686o a2 b6c+ f8a2 b+245 b5 c — 32+ 13720 a-2b9c2—35 a-2b4c289a-4b8 c2+2a-4b3 c +39 a-6 b7 c2+3939 a-8b11 c3 — § a−8 b6 c2 — 560 a-10 610 c3 + 34 a-12 69 c3 + 560 a-14 b13 c4a-16 b12c4+32a-20 615 c5. a−20 Ans. 7a2b-+ 3 a4 b3 c. 7. Find the 9th root of y27+27y25+324 y23+2268 y21 +10206 y19+ 30618 y17 + 61236 y15 + 78732 y13 + 59049 y1119683 y9. 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 x +16. Solution. In the following solution, the arrangement is similar to that in the example of art. 190. xo +4 x2 +20 x2 — 16 x +16 | x3 + 2 x2 —2x + 4. Ans. 4 x5+20 x2-16x+16|2x3 4x4 - 4 x4 +20 x2-16x+16|2x3 + 4x2 8x316x2-16x+16 | 2x3 + 4x2 − 4 x 0. 2. Find the square root of 25 a1 30 a3 b+49 a2 b2 ·24 a b3 + 16 ba. Ans. 5a2-3ab +4b2. 3. Find the square root of 4 x6 + 12 x5 +5 x1 - 2 x3 +7x2-2x+1. Ans. 2x33x2-x+1. Solution of Binomial Equations. SECTION V. Binomial Equations. 223. Definition. When an equation with one un2known 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 AxM0, and may be called a binomial equation. 224. Problem. To solve a binomial equation. Solution. Suppose the given equation to be 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. Examples of Binomial Equations. 226. EXAMPLES. 1. Solve the two equations xy7+2y7-4 y3-8x+16=0, x2y7-4 y7-4x y3+8 y3+32x-640. Solution. The elimination of y between these two equations, by the process of art. 155, gives being substituted in the first of the given equations, produces 4 as will be shown when we treat of the theory of equations Again, the value of x, x=- 2, being substituted in the first of the given equations, produces as will be shown in the theory of equations. |