3. Find the square root of 18 correct to 4 decimal places. Ans. 4.2420+. 4. Find the square root of 19 correct to 6 decimal places. Ans. 4.358898+. 5. Find the square root of 52.463 correct to 7 decimal places. Ans. 7.2431346+. 6. Find the square root of 7 correct to 8 decimal places. Ans. 2.64575131+. 7. Find the value of 53 correct to 5 decimal places. Ans. 11.18034+. CUBE ROOT OF POLYNOMIALS. 236. We may deduce a rule for extracting the cube root of a polynomial in a manner similar to that pursued in square root, Ly analyzing the combination of terms in the binomial cube. If the binomial, a+b, be cubed, we have a3+3a2b+3ab2+b3. We will now consider how the process may be reversed, and the root extracted from the power. We observe 1st. That the first term of the root may be obtained by taking the cube root of the first term of the power. Va3 = a. Thus, 2d. The second term of the root may be found by dividing the second term of the power by three times the square of the first term of the root. Thus, 3a2b÷3a2 = b. 3d. The last three terms of the power may be factored, and written as follows: (3a2+3ab+b2)b or {3a2+(3a+b)b}b. Thus we see that if to the trial divisor, 3a2, we add a correction, 3ab+b2, or (3a+b)b, the result will be a complete divisor, which multiplied by, will give the last three terms of the power. Hence, the whole operation of extracting the root, a+b, from the cube, a3+3a2b+3ab2+b3, may be written as follows: Having found a, the first term of the root, we take its cube from the whole expression, and obtain 3a3b+3al'+b3. Dividing the first term of this remainder by 3a2, we obtain b, the second term of the root. To complete the divisor, we first write the quantity 3a+b; and multiplying this by b, we have 3ab+b3, which added to the trial divisor, gives 3a2+3ab+b2, the complete divisor. Multiplying this by b, and subtracting the product from the dividend, there is no remainder, and the work is complete. 237. To recapitulate, we may designate the quantities employed in the foregoing operation, as follows: 238. Next, suppose there are three terms in the root, as a+b +c. Assume s = a+b; then s+c = a+b+c; and we have (s+c)'s3+3s3c+3sc2+c3. If we proceed as in the last example, we shall obtain a+b, or that part of the root represented by s, and subtract its cube from the whole expression. There will then be left 3s3c+3sc2+c3, which may be factored and written (3s2+3sc+c2)c or {3s2+(3s+c)c}c. And we perceive that 3s will be the new trial divisor to obtain c, and that (3s+c)c will be the new correction. The value of 382, or 3(a+b), may be obtained by multiplication. It will be more convenient, however, to derive it by the addition of three quantities already used in the operation. Thus, Last complete divisor, Last correction, 3a'+3ab+b' } (1) Square of last term of the root, 382 = 3(a+b)' = 3a2+6ab+3b Let it now be required to find the cube root of the polynomial x+3x-3x*—11x3+6x2+12x-8. 3x2+3x-2, -6x2-6x+4|3x2 + 6x3 —3x2-6x+4, −6x1—12x3 +6x2 +12x-8 Having arranged the polynomial according to the exponents of x, we proceed as in the former example, and obtain x, the first term of the root, 3x°—3x*—11x3+6x+12x--8 the first remainder, 3x the trial divisor, and x the second term of the root. To complete the trial divisor according to formula (a), we write three times the first term of the root plus the second, or 3x2+x, for the first factor of the correction. Whence we have (3x+x)x, or 3x2+x3, for the correction; 3x+3x+x for the complete divisor; (3x+ 3x+x)x, or 3x+3x+x', for the product; and— 6x-12x2+ 6x+12x-8 for the new dividend. To form the new trial divisor according to formula (b), we have (3x*+3x2+x3) + (3x2+x2) +x2 = 3x*+6x3+3x2; whence, by division, we obtain -2 for the third term of the root. To complete the new trial divisor, we have for the first factor of the correction, 3(x+x)-2 = 3x2+3x-2. This may be obtained in the operation from the former factor 3x+x, by simply multiplying its second term by 3, and anncxing the -2. We now find the correcbefore, and the work is tion, complete divisor, and product as finished. It is evident that three or more terms of the root will sustain the same relation to the next succeeding term, that the first sustains to the second, or the first and second to the third.. 239. From the foregoing analysis we derive the following RULE. I. Arrange the polynomial according to the powers of some letter, and write the cube root of the first term in the quotient; subtract the cube of the root thus found from the polynomial, and arrange the remainder for a dividend. II. At the left of the dividend write three times the square of the root already found, for a trial divisor; divide the first term of the dividend by this divisor, and write the quotient for the next term of the root. III. To three times the first term of the root annex the last term, and write the result at the left, and one line below, the trial divisor; multiply this result by the last term of the root, for a correction of the trial divisor; add the correction, and the result will be the complete divisor. IV. Multiply the complete divisor by the last term of the root, subtract the product from the dividend, and arrange the remainder for a new dividend. V. Add together the last complete divisor, the last correction, and the square of the last term of the root, for a new trial divisor; and by division obtain another term of the root. VI. Take the first factor of the last correction with its last term multiplied by 3, and annex to it the last term of the root, for the first factor of the new correction; with which proceed as before, till the work is finished. EXAMPLES FOR PRACTICE. 1. What is the cube root of 27a+108a2+144a+64? Ans. 3a+4. 2. What is the cube root of x+6x°—40x3+96x-64 ? Ans. x+2x-4. 3. What is the cube root of 8x°—36x1+66x*—63x3+33x2-9x +1? 4. What is the cube root of a°+9ab+24a*l2 +9a3l3—24a3b* +Ial3—la? Ans. 2x-3x+1. Ans. a2+3ab-b3. Ans. a3-2a+5a—2. 5. What is the cube root of ao-6a+27a-74a+159a234a*+257a'—174a3+60a—8 ? 6. What is the cube root of x-3x+6x-10x+12x-12x +10x2-6x+3x-1? Ans. x-x2+x—1. ·12ab+36abc + 6a3l3 — 7. What is the cube root of 8a36a'i'c-a'l'+5+al c2+9a l'c-27al'c+27b'e? 9. What is the cube root of x+6x-64x°-96x+192x*+ 512x-768x-512? Ans. x2—4x+Ã ̈* Ans. x+2x-4x-8. CUBE ROOT OF NUMBERS. 240. To establish a rule for extracting the cube root of a number, we must first ascertain the relative number of places in a cube and its root. This relation is exhibited in the following examples: Thus we perceive that a number consisting of one place, may have from one to three places in its cube; and that in all cases the addition of one place to the root adds three places to the cube. Hence, If a number be pointed off into three-figure periods, commencing at units' place, the number of periods will indicate the number of places in the root. 241. To ascertain how the several figures of the root are related in local value to the periods of the power, we may decompose any number, as 5423, and form the cubes of its several parts, as follows: 50003 125 000 000 000 54003 157 461 000 000 54203 159 220 088 000 = 5423' 159 484 621 967. Hence, |