= (x3-3x2)3 - 3(x3 – 3x2)2(2x+1)+3(x3 − 3 x2) (2 x + 1)2 − (2x+1)3 =x9—9x8+27x7 — 27 x6 – 3(x¤ — 6x5+9x1) (2x+1)+3(x3 −3 x2) (4x2+4x+1) − (8 x3 + 12 x2 + 6x + 1) =x9-9x8 +27 x7 27 x6 6x7 + 33 x6 – 36 x5 – 27 x2 + 12 x5 — 24 x1 — 33 x3 — 9 x2 - 8 x3 - 12 x2 - 6 x − 1 =x99x8 + 21 x2 + 6x6 - 24 x5 - 51 x 41 x3- 21 x2-6x-1. CHAPTER XIII EVOLUTION 224. Evolution is the operation of finding a root of a quantity ; it is the inverse of involution. 225. It follows from the law of signs in involution that: 1. Any even root of a positive quantity may be either positive or negative. 2. Every odd root of a quantity has the same sign as the quantity. √9 = +3, or −3 (usually written ±3); for (+3)2 and (−3)2 equal 9. $27 =- · 3, for (− 3)3 — — 27. Vā1 = ± a, for (+ a)a = aa, and (− a)1 = a1. √32 = 2, etc. 226. Since even powers can never be negative, it is evidently impossible to express an even root of a negative quantity by the usual system of numbers. Such roots are called imaginary numbers, and all other numbers are, for distinction, called real numbers. Thus 1 is an imaginary number, which can be simplified no further. EVOLUTION OF MONOMIALS 227. The following examples are solved by the definition of 228. To extract the root of a power, divide the exponent by the index. A root of a product equals the product of the roots of the factors. To extract a root of a fraction, extract the roots of the numerator and denominator. 5 a2. 3 3 38. VV-a. 31. √a2+2ab+b2. 35. √−32(m+n)3». 39. √.49 p2. 229. A trinomial is a perfect square if one of its terms is equal to twice the product of the square roots of the two other terms. (§ 118.) In such a case the square root can be found by inspection. Ex. 1. Find the square root of x6 — 6 x3y2+9 y1. 1. 1-4a+4 a2. 2. a1+16 b2 - 8 a2b. 3. a+1-2 a2. 4. 16 a + a + 8 a3. 5. 149 y 14 y2. 6. x2y2-6xyz +9 z2. 8. 16 x1-8 x2y2 + y*. 9. a+b8-2 a2b1. 10. 16 a1 · 120 a2bc + 225 b2c2. 13. a2+b2+c2+2ab+2 bc + 2 ac. 230. In order to find a general method for extracting the square root of a polynomial, let us consider the relation of a+b to its square, a2 + 2 ab + b2. The first term a of the root is the square root of the first term a2. The second term of the root can be obtained by dividing the second term 2 ab by the double of a, the so-called trial divisor; 2 ab a+b is the root if the given expression is a perfect square. In most cases, however, it is not known whether the given expression is a perfect square, and we have then to consider that 2 ab + b2 = b(2 a+b), i.e. the sum of trial divisor 2 a, and b, multiplied by b must give the last two terms of the square. The work may be arranged as follows: a2+2ab+b2 | a + b a2 2 a+b2ab+b2 2 ab + b2 Ex. 1. Extract the square root of 16 x 24 x2y3 +9 yo. Explanation. Arrange the expression according to descending powers of x. The square root of 16 x is 4 x2, the first term of the root. Subtracting the square of 4 x2 from the trinomial gives the remainder - 24 x2y3 +9 y6. By doubling 4 x2, we obtain 8x2, the trial divisor. Dividing the first term of the remainder, -24 x2y8, by the trial divisor 8x2, we obtain the next term of the root - 3y3, which has to be added to the trial divisor. Multiply the complete divisor 8 x2 - 3 y3 by — 3 y3, and subtract the product from the remainder. As there is no remainder, 4 x23 y3 is the required square root. - |