{(x2-2x+3)+√({—2x)} {(x2—2 x + })—√({−2x)}, and {(x2-2x+1)+√—x2} {(x2—2x+1)—~—x2}, of which the latter are the simpler. As - is the same as xx-1, it follows that x2-x-1; and by making the change, the last two factors may be put in a still better form. EXAMPLES FOR EXERCISE. Required the square root of each of the following expressions: (1) 9x-6x+a3. (2) x1+2x2+3x2+2x+1. (3) 9x-12x+16x2-8x+4. (4) 16a2x2+8ax2+8ax+x2+2x+1. (5) 4x+12x+5x1-2x3+7x2-2x+1. (6) 4x-24x+60x1-80x3+60x2-24x+4. (7) 4x1—12x3+11x2—3x+4. (8) 49x+28x3-17x2—6x+24· (9) 4x1+6x3+224x2+15x+25. (10) 1+4x-2x2-4x3+25x1-24x2+16xo. (11) 64x+192x2+240x+160x3+60x2+12x+1. (12) 4x-16x16x+12x+32x2+24x2+8x+1. (13) x-7-2x+13·44x2-1·728x+0576. (14) 9a3+6ab+ 30 ac+6ad+b2+10 bc+2bd+25 c2 + 10cd+d2. (15) 4xy-12x3y3+17x1y2-12x3y+4xo. (16) am4a3m+n+6a2m+2—4a+3n+a1". (17) Required the fourth root of x-4x3y+6x3y—4xy3+y*. (18) Required the fourth root of 16x-96x3y+216x3y3— 216xy3+81y1. (19) Prove, by the extraction of the square root, that the expression x+4x3-3x2-8x+4 may be decomposed into the two factors (x2+(2+3)x−2}{x2+ (2-√3)x-2}. (See page 72.) (20) Prove, by the extraction of the square root, that the expression x-4x+7x-4x+1 may be decomposed into the two factors {x-(2-—1)x+1} {x2— (2+√−1)x+1}. 38. To extract the cube root of a compound quantity:The rule for the extraction of the cube root of a polynomial is founded upon an examination of the form of a complete cube, just as the rule for the square root was deduced from the ascertained composition of a complete square. Commencing, as before, with the form whose root is a binomial, a+b, which form we know to be a3+3a2b+3ab2+b3, we see that the first term a of the root is got at once from the first term a3 of the cube: it is the cube root of that first term which, being subtracted, leaves 3a2b+3ab2+b3. Attending, as before, only to the leading term of this, 3ab, we see that 3a2, a quantity now known, taken as a partial or incomplete divisor, suggests b, the other term of the root; and this b, multiplied by 3a+b, gives 3ab+b2; and this completes the divisor: that is, when connected to 3a2, the incomplete divisor of 3ab+3ab+b3, it gives the quantity 3a2+3ab+b2, which is the complete divisor corresponding to the quotient b. It thus appears that the first term of the root, found from the first term of the cube, suggests the first term of the true or complete divisor, which first term is called the trial divisor: that the trial divisor suggests the second term of the root, which second term, when found, enables us to complete the divisor and to finish the work; at the same time putting it to the test whether the proposed expression is really the cube of a binomial or not. In like manner, when the proposed cube is that of a trinomial, a+b+c, we know, from the expression for (a+b)3, in which, of course, anything may be put for a and b, that [(a+b)+c]3= (a+b)3+3(a+b)2c +3(a+b) c2+c3. Out of this form, a+b may be evolved, as already explained: after this, the trial divisor of the remainder, 3 (a+b)°c + 3 (a+b) c2+c3, will be 3(a+b); that is, as before, three times the square of the partial root already discovered. This trial divisor suggests c, and the divisor is completed by adding to it 3(a+b)c+c2, in imitation of what was done in the former case: the multiplication of the corresponding divisor and quotient terminates the operation, as before. And it is obvious that, by imitating this uniform process, the terms of the root may be found, one after another, when their number is four, five, or any whatever. Should a remainder exist, after what we have here regarded as the close of the operation, we conclude, of course, that the proposed polynomial is not a complete cube, and that an exact cube root of it does not exist. The rule for executing the steps above described, when expressed in words, is as follows: RULE. The terms of the polynomial being arranged as if for division, put the root of its first term in the quotient's place, and the cube of this root under the said first term: draw a line, as if for subtraction, and bring down under it the next three terms of the polynomial: these will form a dividend, and a place to the left is to be marked off for a divisor. Put three times the square of the root just found, in the divisor's place, and see how often this trial or incomplete divisor is contained in the first term of the dividend: the quotient is the next term of the root. To three times the product of this new term and the preceding part of the root, add the square of the new term: connect the result with the incomplete divisor, and the divisor will be completed. Multiply the complete divisor by the new term, subtract the product from the dividend, and connect the next three terms of the polynomial to the remainder. We shall thus have a second dividend, with which we are to proceed as before. NOTE. It is worthy of observation that, in the operation here described, as also in that for the square root, the leading term of every divisor is invariably the same; and that this leading term is all that is required for the determination of each of the terms of the root, one after another: but the complete divisors are, of course, necessary to enable us to obtain the successive dividends. (1) x6—6x5+0x440x3+0x2-96x-64[x2-2x-4 26 3x46x34x2] −6x5+0x4+40x3 (2) -6x512x48x3 3(¿2—2x)2—12(x2—2.x)+42\ -12x4+48x3+0x2—96x-6 ́ or, 3x4-12x3+0x2+24x+16/ -12x448x3+0x2-96x-64 In each of these examples, the polynomial is a complete cube: when such is not the case, there will be a remainder after all the terms have been brought down. The proposed polynomial will then be equal to the cube of the imperfect root, or quantity in the quotient, minus this remainder. EXAMPLES FOR PRACTICE. - Required the cube root of each of the following expressions:(1) x3-9x2+27x-27. (2) x-3x5+5x3-3x-1. (3) x+6x-40x3+96x-64. (4) x-6x+15x-20x3+15x2-6x+1. (5) x-12x+60x160x3+240x2-192x+64. (6) x6—3ax3+6a2x2-7a3x3+6a2x2-3a5x+ao. (7) a6—6a5x+15a3x2—20a3x3+15a2xa—6ax2+x¤. (8) 8x+48x3y+60x1y2—80x3y3—90x2y*+108xy5—27yo. (9) a+12am+2228a3 3n 6a2+1x". 3m 3n (10) x2+3x3+6x2+10xo+12x5+12x2+10x3+6x2+3x+1. There is a method of arranging the process for the cube root of a polynomial which saves some trouble in the formation of the several divisors; but as this arrangement presents more complication to the eye, and occupies more space, and is, moreover, less easy of recollection than the common rule given above, we have not thought it worth while to exhibit it here. When the cube root of number, instead of the cube root of an algebraic expression, is to be extracted, then the arrangement to which we allude is much to be preferred. The principal advantage of it consists in the facility with which it enables us to determine the successive trial divisors which in arithmetic it is in general necessary that we should obtain, in order to the accurate determination of the several figures of the root. But, in algebra, the entire trial divisor is not needed for this purpose, since, as observed in the preceding note (page 75), the leading term alone of each divisor is sufficient to conduct us to the corresponding term of the root. It is chiefly on account of this distinction between the arithmetical and the algebraical operations, that we have here thought it to be unnecessary to depart from the arrangement in general use. The inquiring student will, however, find every information respecting the improvements here hinted at in the "Analysis and Solution of Cubic and Biquadratic Equations," by the author of the present work. But as respects the extraction of the square root, an arrangement of the work has been suggested which is very conveniently suitable to algebra, though not to arithmetic: we shall here explain it. As headings for the two columns of which the work will consist, write the word Root on the left, and Remainder on the right. In the root-column, insert the root of the leading term of the polynomial, and also the quotient of the second term of the polynomial by double this root, the first and second terms of the sought root will thus be obtained. Subtract the square of the last term of the root from the polynomial and the remainder, neglecting the first two terms of the polynomial, place in the remainder-column. Divide the leading term of the remainder by double the leading term of the root; the quotient will be the third term of the root. Subtract double the product of this third term and the second, as also the square of the third term from the before-found remainder; and, neglecting the first term of that remainder, put the difference, which is a second remainder, in the remainder column. As before, divide the leading term of this second remainder by double the leading term of the root: the quotient will be the fourth term of the root. Subtract double the product of this fourth term by the two preceding terms, as also the square of the fourth term from the last remainder; and, neglecting the first term of that remainder, put the difference, which is a third remainder, under the former, and so on. For example: let the polynomial be 4x-12x+5x+2x3+7x2+2x+1; then the operation by the foregoing precepts will be as follows: ROOT. 2x-3x2 2x3-3x2-x-1 REMAINDERS. -4x2+2x3+7x2+2x+1 0 Again: let the polynomial be 16a6-24x3+25x1—20x2+10x2—4x+1. Lastly: let the polynomial be x2-4x3+7xa—4x+1. (See page 72). In the second mode of working this example, instead of the leading term of the root in the last step, we have taken the second term to suggest the final term of the imperfect root, as if the first step of the work had been arranged thus: Root, -2x+x2; Remainder, -4x+3x2+1. It is obvious that, by conducting the operation for the square root in the manner here pointed out, time and space are both saved; and that straggling appearance, which the common method exhibits to the eye, avoided. We should recommend the learner to work all the examples at page 73, conformably to this more compact arrangement.* If, in the extraction of the square root, the leading term of the polynomial require a factor to make it a square, we may multiply all the terms by this factor; and if, after the extraction, we divide the root by the square root of the factor, and the remainder by the factor itself, the results will be correct. And, in like manner, in the extraction of the cube root, if the polynomial be multiplied by a factor, the root must be afterwards divided by the cube root of that factor, and the remainder by the factor itself. CHAPTER III. ALGEBRAIC FRACTIONS. 39. The treatment of algebraic fractions is exactly the same as that of common fractions in arithmetic: the rules of operation * The operation for the extraction of the square root of a polynomial has been lately applied, by the author of this work, to the analysis of equations. Every one engaged in such analysis would greatly prefer the arrangement of the steps given above; and on this account, as well as from more general considerations, we think that the plan here proposed should supersede that hitherto adopted. See "The General Principles of Analysis," Part I. |