preceded by the sign --, and the even ones by the sign + (38); and hence, these powers of b, multiplied into the powers of a, all of which have the sign +, will give alternately + and 2.3.4.5 In an example like the preceding, the series will not terminate, unless a numerical value is given to m. If we make m = 4, it terminates at the fifth term; for in that case am-4 ao. The coefficient of the fifth term then becomes unity, for and the fifth term is aob or b1. In like manner, if we make m= = 3, the series terminates at the fourth term ; and if m=5, it terminates at the sixth term. 3. Find the 6th power of a- X. (a-x)-a-ba3x+15a2x2-20a3x3+15a2x*—6ax+xo. 4. Find the 5th power of 2a2+3x. (2a2+3x)5 (2a2)5+5(2a2)*3x+10(2a2) (3x)2+10(2a2)2(3x)3 +5(2a2)(3x)*+(3x)5 32a10+240a3x+720ax2+1080a*x+810a2x2+243x. 5. (4ac2-5c3) (4ac2) — 4(4ac2) (5c3)+6(4ac2)*(5c3)2 · 4(4ac2)3(5c3) — 4(4ac2)(5c3)3 +(5c3)1 11 =256a1c3-1280a3c3+2400a2c1o-2000ac11+625c12. 6. (4a+3x) = (4a) + 6(4a)5(3x3) + 15(4a)1(3x3)2 +20(4a)3(3x3)3 + 15(4a)2 (3x3)1 +6(4a)(3.x·3)+(3.x3). =4096a+ 18432a5x+34560a1x+34560a3x+19440a2x12 +5832ax15+ 729x13. 18 7, (9a2-4b2)—6561a3—11664a*b*+7776a*b1-2304a2b +2566o. EVOLUTION. (44.) Evolution is the reverse of involution. It is the finding or extracting the root of a given quantity. The root of a quantity is that quantity which, raised to a power denoted by the index of the root, shall be equal to the given quantity. Thus, the square root of a quantity, raised to the second power is equal to that quantity. So also the 3d, 4th, 5th and 6th roots of a quantity, raised to the 3d, 4th, 5th and 6th powers, respectively, will be equal to that quan tity. The signs of the roots in algebra are the same as in arith 3— 4 5 — 6 metic. Va, Va, va, va, Va, &c. indicate the square root, the cube root, the 4th root, the 5th root, and the 6th root of a, respectively. The figure before the sign shows the degree of the root, and is called the index of the root. EXTRACTION OF THE ROOTS OF MONOMIALS. Extract the root of the coefficient, and divide the exponent of each letter by the index of the root. The reason for this rule is evident; for since a monomial is raised to any power by raising the coefficient to that power, and multiplying the exponent of each letter by the degree of the power, the root of that power will be obtained by reversing the process; that is, by extracting the root of the coefficient, and dividing the exponent of each letter by the index of the root. Thus, the cube root of 125a3b is 5ab2; for (5ab) 125a3b. (45.) When the exponents of the letters are not divisible. by the index of the root, the root cannot be extracted; but in that case algebraists have agreed to indicate the extraction by writing the index of the root beneath the exponent of the letter. Thus, if it were required to extract the square root of a3, we should indicate the operation by writing 2 beneath the exponent in a fractional form, and we should have va3a So also, Vaan. These are termed fractional exponents and may be considered expressions for roots which cannot be extracted. EXTRACTION OF THE SQUARE ROOT OF POLYNOMIALS. RULE. (46.) 1. Arrange the polynomial according to the powers of a particular letter. 2. Find the square root of the first term, which must always be a perfect square, and place this root in the quotient. Subtract the square of this root from the first term, and there will be no remainder. 3. Bring down the next two terms for a dividend, and double the root for a divisor; see how often the divisor is contained in the first term of the dividend, and add the quotient to the root already found, and also to the divisor. 4. Multiply the divisor so increased by the aforementioned quotient, and subtract the product from the dividend. To the remainder bring down the next two terms · for a new dividend, and double the two terms of the root already found for a new divisor, and proceed as before; continue the operation in the same manner until all the terms of the polynomial are brought down. For an example, find the square root of 16a+ 96a3x +216a2x2+216ax3 +81x1. The root. 16a1+96a3x+216a2x2+216ax2+81x4||4a2+12ax+9x2 16a1 1st div. 8a2+12ax 96a3x+216a2x2 96a3x+144a2x2 2d div. 8a2 + 24ax +9x2 172a2x2+216ax3+81x1 172a2x2+216ax3+81x1 In this example, we have 4a2, the square root of 16a1, for the first term of the root. Subtracting its square, which is equal to the first term, and bringing down the next two terms, we have, for a dividend, 96a3x +216a2x2; and doubling 4a2, we have 8a2 for a divisor, which divided into 96a3x, gives 12ax for a quotient; and this is placed both in the divisor and in the root. The divisor thus increased, becomes 8a2+12ax, which being multiplied by 12ax, the product subtracted, and the next two terms added to the remainder, produces the second dividend. The whole of the root found being doubled, gives 8a2+24ax for a second divisor. Dividing the first term of the dividend by the first term of this divisor, gives 9r2 for a quotient, which is also added to the divisor, and the sum multiplied by 9x2. To understand the reason for this rule, we have only to recur to the square of a binomial, which we have found to be composed of the square of the first term, twice the product of the first and second, and the square of the second. The first term of the root is, therefore, the square root of the first term of the square; and since the second term of the square is twice the product of the two terms of the root, we shall evidently obtain the second term of the root, by dividing this double product by double the root already found. Thus, since (a + b)2 = a2 + 2ab+b2, in extracting the root of the latter quantity, we take the square root of the first term, which is a, and having subtracted its square, there remains 2ab+b2, Doubling the root found, we have 2a for a divisor; and dividing the first term of this remainder by it, we have b for a quotient; b is then added to the divisor, so that the divisor, so increased, multiplied by b, may produce the dividend, in order to test the correctness of the work; thus, (2a+b)b = 2ab+b2. the root. a2+2ab+b2 || a + b a2 Divisor, 2a+b | 2ab+b2 2ab+b2 The same reasoning will apply when the root consists of three terms; for we have (a+b+c)2 = (a + b)2 + 2(a + b) c+c2. Now having found the root of (a+b)2, which is a + b, we have only to double this root and divide the next term, 2(a+b) c by it, to find the third term c, of the root. The ope ration may be thus represented. Root, 1 (a + b)2 + 2(a + b) c + c2 || a + b + c 2(a+b)+c | 2(a+b) c+c2 2(a + b) c + c2 |