4. To divide algebraic fractions, invert the terms of the divisor, and proceed as in multiplication. 3 x z + 3x z3 23 z3 by x-z. x- z)x3-3x2z + 3 x z3 — z3(x2 — 2 x z + z' quotient. 5. Divide a b3 by ab. ( -- blus x(x+b)(x—b) x (x2 9. Divide 96 + 2 a3. 10. Divide 10 a3 +11 ab-19 abc-15 a°c + 3 a b2 + 15bc5bc by 3 a b + 5 a-5b c. Quot. 2 a+b-3 c. + y' 11. Divide x + y2 + 1⁄4 by x + y + 1. Quot. x − y y2 SECTION V.-Involution. 1.—To involve or raise monomials to any proposed power. Involve the coefficient to the power required, for a new co efficient. Multiply the index of each letter by the index of the required power. Place each product over its respective letter, and prefix the coefficient found as above: the result will be the power required. All the powers of an affirmative quantity will be + ; of a negative quantity, the even powers, as the 2d, 4th, 6th, &c. will be; the odd powers, as the 3d, 5th, 7th, &c. will be-. To involve fractions, apply these rules to both numerator and denominator. Examples. 1. Find the fourth power of 2 a3. 2×4=8, new exponent. Hence 16 a3 the answer. 2. The fifth power of 3 y is 243 y1o. 3. The fourth power of 4. The sixth power of 2z 64 26 4 x is 256 x12. is 3 y 729 y12 2. To involve polynomials. Multiply the given quantity into itself as many times, want ing one, as there are units in the index of the required power and the last product will be the power required. The operation required by the preceding rules, however simple in its nature, becomes tedious when even a binomial is raised to a high power. In such cases it is usual to employ Sir Isaac Newton's Rule for involving a Binomial. 1. To find the terms without the coefficients.-The index of the first, or leading quantity, begins with that of the given power, and decreases continually by 1, in every term to the last; and in the following quantity the indices of the terms are 0, 1, 2, 3, 4, &c. 2. To find the unciæ or coefficient. The first is always 1, and the second is the index of the power and in general, if the coefficient of any term be multiplied by the index of the leading quantity, and the product be divided by the number of terms to that place, it will give the coefficient of the term next following.* Note. The whole number of terms will be one more than the index of the given power; and when both terms of the root are +, all the terms of the power will be +; but if the second term be all the odd terms will be +, and the even terms Examples. 1. Let a + be involved to the fifth power. The terms without the coefficients will be a3, aa x, a3 x3, aa x3, α x1, x3, and the coefficients will be And therefore the fifth power is The same theorem applied to fractional exponents, and with a slight modifica tion, serves for the extraction of roots in infinite series; as will be shown a little farther on. Here we have, for the sake of perspicuity, exhibited sepa ately, the manner of obtaining the several terms and their respective coefficients. But in practice the separation of the two operations is inconvenient. The best way to obtain the coefficients is to perform the division first, upon either the requisite coefficient or exponent (one or other of which may always be divided without a remainder), and to multiply the quotient into the other. Thus, the result may be obtained at once in a single line, nearly as rapidly as it can be written down. 2. (x + y)2 = x2 + 7 x 6 z + 21 x3 z2 + 35 x1 z3 + 35 x3 z1 + 21 x3 z3 + 7 x z° + z2. —z)3 3. (x-2)=x3 — 8 x7 z+28 x z2 56 x3 3 +70 x1 z1 56 x3 z5 + 28 x2 zó 8 x z2 + z3. For Trinomials and Quadrinomials.-Let two of the terms be regarded as one, and the remaining term or terms as the other; and proceed as above. Example. Involve x + y-z to the fourth power. - z as Let x be regarded as one term of the binomial, and y the other then will (x + y) = {x + (y — z)}* = x2 + 4 x3 ( y − z) + 6 x3 (y — z)2 + 4 x (y — z)3 + (y — z)*, where the powers of y-z being expanded by the same rule, and multiplied into their respective factors, we shall at length have x + 4x3 y — 4 x3 z + 6x3 y3 . 12 x y z + 6 xo z3 + 4 x y3-12 x y z + 12 x y z3 — 4 x z3 + y' — 4 y3 z + 6 y31⁄23 — 4 y z3 +, the fourth power required. Had (x + y) and z been taken for the two terms of the binomial, the result would have been the same. Note. The rule for the involution of multinomials is too complex for this place. 11 |