ART. 111. From the preceding examples, also from what was shown in Article 104, relative to any power of a product, we infer that Any root of a product is the product of the roots, to the same degree, of each of the factors of that product. For example, the third root of 27 a3 b6 is 3 a b2, which is the product of the third roots of 27, a3, and 66, the factors of 27 a3 b6. In a similar manner, if any numerical quantity is separated into factors that are exact powers of the required degree, which may always be done when the number itself is an exact power of that degree, we may extract separately the root of each factor, and afterwards multiply these roots together. Thus, 12969.144, the second root of which is 3.12=36. ART. 112. Since, in extracting the root of a monomial, we divide the exponent of each letter by the number expressing the degree of the root, it follows, that if any exponent is not divisible by that number, the division must be expressed, and this gives rise to fractional exponents For example, the third root of a is at, that of a2 is a3. The expression a3 represents either the third root of a2, or the second power of a; for (a) a1× 2 = a3. = denotes either the fifth root of a3, or the third Also, a a2 power of as The radical sign, as well as fractional exponents, may be used to indicate a root of any degree, provided we place over this sign a number expressing the degree of the root. Thus, which is the same as ✔, indicates the second root;, the third root; 3 2 the fourth root. Hence, we have the following equivalent expressions, viz. 3 4 6 √ m = m3 ; & m2 = m3 ; √ m3 = m2; & m2 = m3, &c. We may, therefore, use indifferently either the radical sign or a fractional exponent, remembering that the number over the radical sign is the denominator, and that the exponent of the quantity under the sign is the numerator, of the fractional exponent. SECTION XXXVII. SECOND ROOTS OF POLYNOMIALS. ART. 113. It is required to extract the second root of 25 x2+60 x y +36 y2. Operation. 25 x2+60 x y +36 y2 (5x+6y. Root. 25 x2 60 x y +36 y2 (10x+6 y 60 x y +36 y2 0. By comparing this quantity with a2+2ab+b2, the second power of a+b, and recollecting the process of extracting the second roots of numbers, we shall readily see the mode of proceeding. The first term 25 x2 corresponds to a2; we therefore extract the root of 25 x2, which is 5x, (Art. 110,) place it at the right, and subtract its second power from the given quantity. The remainder 60 x y +36 y2, which we regard as a dividend, corresponds to 2ab+b2, or (2 a+b) b. Di viding the first term 60 x y corresponding to 2 ab, by 10 x corresponding to 2 a, we have 6 y answering to b. We now place 6y, with its proper sign, in the root, also at the right of our divisor, and have 10x+6y answering to 2 a+b. Multiplying 10x+6 y by 6 y, we obtain 60 x y +36 y2 corresponding to (2 a+b) b. Subtracting this product from the dividend, we have no remainder. Consequently, 5x+6y is the required root. When a quantity consists of more than three dissimilar terms, its second root will consist of more than two terms. But the process of finding the second root of a polynomial is, in all cases, so similar to that of extracting the root of a number, that it hardly needs a separate explanation. The following example will serve as an illustration. What is the second root of 9 a1 — 24 a3 b + 22 a2 b2 — 8 a b3+b4? Operation. · 24 a3 b+22 a2 b2 — 8 a b3 + b1 ( 3 a2 — 4 a b + b2. -24 a3 b+22 a2 b2-8 a b3b4 (6 a2-4 ab 6a2b2-8 a b3b4 (6 a2 — 8 ab+b2 0. The process of finding the first two terms of the root is precisely the same as in the first example of this article. Having obtained the second dividend, 6 a2 b2 — 8 a b3 +b4, we double the first two terms of the root, and have for a second divisor 6 a2-8 ab. Performing the division, we obtain b2 for the third term of the root, which we annex, with its proper sign, both to the preceding part of the root and to the divisor. Our divisor then becomes 6 a2-8 ab+b2, which we multiply by b2, and subtract the product from the second dividend. As there is no remainder, the root required is 3a2-4ab +b2, or 4 ab-3 a2-62, for the second power of either will produce the given quantity. ART. 114. From the foregoing examples and explanations, we derive the following RULE FOR EXTRACTING THE SECOND ROOT OF A POLYNOMIAL. 1. Arrange the quantity according to the powers of some letter. 2. Find the root of the first term, and place it as the first term of the root sought; subtract the second power of this term from the given polynomial, and call the remainder the first dividend. 3. Double the term of the root already found for a divisor, by which divide the first term of the dividend, and place the quotient, with its proper sign, as the second term of the root, also at the right of the divisor. Multiply the divisor, with the term annexed, by the second term of the root, and subtract the product from the dividend. 4. The remainder will form a second dividend, which is to be divided by twice the whole root found, and the quotient is to be placed, as the next term of the root, also at the right of the divisor. Multiply the divisor, with the term last annexed, by the last term of the root, and subtract the product from the last dividend. 5. The remainder will form a new dividend, with which proceed as before; and thus continue, until all the terms of the root are found. Remark 1. As we at first arrange the given polynomial according to the powers of some letter,.so the same arrangement must be preserved in each dividend. Remark 2. In dividing, we merely divide the first term of the dividend by the first term of the divisor; and it is manifest, from the manner in which the divisors are obtained, as well as from inspection, that the successive divisors will have their first terms alike. Extract the second roots of the following quantities. 4. x4—4 x y3+ y1 — 4 x3 y + 6 x2 y2. 5. 9x30x3y+25 x2 y2-42x2-70 x y +49. 8. 9m6-12m5+34 m4-20 m3 + 25 m2. ART. 115. The following additional remarks may be found useful. 1. No binomial can be an exact second power; for the second power of a monomial is a monomial, and the second power of a binomial necessarily contains three Thus, x2 y2 cannot be an exact second power. It wants +2xy to make it the second power of xy, and it wants -2xy to make it the second terms. 2. In order that a trinomial may be a perfect second power, it must be such, that, when it is arranged according to the powers of a particular letter, the extreme terms shall both be positive, and shall both be exact second powers, and the mean term shall be twice the product of the second roots of those powers. When these conditions are fulfilled, the second root may be found in the following manner. Extract the second roots of the extreme terms, writing these roots after each other, and giving them both the sign |