EXAMPLES. Involve the whole root, thus obtained, to the given power, and subtract from the given quantity. Divide the first term of the last remainder by the former divisor, annex the result to the root, and proceed as before. 1. Required the 3d root of 8a3+36a3b+54ab2+27b3-48a2c-144abc-108b2c+96ac 8a3+36a2b+54ab2+27b3-48a2c-144abc-1086°c +96ac +144bc2-64c3 (2a+3b-4c 12a2) S6a2b 8a3 8a3+56a2b+54ab2+27b3±(2a+3b)3 12a2) -48a2c 8a3+36a2b+54ab2+27b3-48a c-144abc-1086 c+96ac+144bc2-64c 2. Required the 3d root of 27x3-54x2y+36xy2—8y3. Ans. 3x-2y. 3. What is the 3d root of x2-6x5+15xa—20x3 +15x2 +6x+1? Ans. x2-2x+1. 4. Required the 5th root of 32x5—80x80x3-40x2 +10x-1. Result, 2x-1. 5. Required the 4th root of 16a2-96a3x+216a2x2216αx3+81x4 Ans. 2a-3x. 6. Find the 5th root of 10+10x9 +35x+40x7-30x6 -68x+30x+40x3-S5x2+10x-1. Result, x2+2x-1. 39. Note.-The roots of compound quantities may sometimes be obtained, with great facility, by a method of trial. To effect which, it may be observed, that the power of a binomial consists of as many terms as there are units in the exponent, increased by one. Hence, if the number of terms of a quantity, whose root is to be extracted, exceeds this sum, we may conclude, that the root consists of more terms than two. When the root is judged to be a binomial, take the roots of the extreme terms, and connect the results by the signor; but when the root appears to contain more terms than two, take also the roots of one or more of the other given terms, and connect the roots thus obtained, by the signs + or, as may be judged proper. Involve this supposed root to the given power, and if the result agrees with the quantity given, the root is manifestly correct. This expedient, however, is not likely to abridge the labour of å student in the early periods of his course. SECTION III. ALGEBRAIC FRACTIONS. 40. Algebraic fractions depend upon the same principles as those in common arithmetic, and are managed by similar rules, proper regard being paid to the signs and algebraic modes of operation. EXAMPLES. 1. Required the greatest common divisor of x3-—b2x, and x2+2bx+b2.* First to find a common divisor. 3a2-2α-1) 4a3—2a2— 3a+ 1 * In examples of this kind, we are frequently obliged to adopt expedients which are not usual in common arithmetic. When every term of the divisor has a factor, which does not contain a letter or number common to every term of the dividend, we divide by it, and use the resulting quotient as a divisor. When the divisor has been thus reduced, it often happens, that the division is still impracticable; in which case, we multiply the dividend by such a number as will make its first term a multiple of the first of the divisor. D |