 | James Bryce - Algebra - 1837 - 322 pages
...term, and subtract its cube from the given quantity; 3* take the quotient of the first term of the remainder by three times the square of the part of the root already found, and set it down as the next term of the root: then, to the treble of the square of the first term of... | |
 | Stephen Chase - Algebra - 1849 - 348 pages
...43200+1800+25 = 45025, 2d complete divisor. 225125~ NoTES. (1.) The terms of the power not being distinct, three times the square of the part of the root already found is only a trial er approximate (§ 174. N. 2) divisor, and the correctness of the next figure of the... | |
 | James William M'Gauley - 1854 - 284 pages
...the first term will be the first term of the root: subtract its cube from the given quantity. Divide three times the square of the part of the root already found, into the remainder : — the quotient will be the second term of the root. Subtract three times the... | |
 | Dana Pond Colburn - Arithmetic - 1855 - 396 pages
...period, and to the remainder bring down the next period, calling the result a dividend. Fourth. Find three times the square of the part of the root already found, and make it a trial divisor. Fifth. See how many times the trial divisor is contained in the dividend,... | |
 | Dana Pond Colburn - Arithmetic - 1856 - 392 pages
...period, and to the remainder bring down the next period, calling the result a dividend. Fourth. Find three times the square of the part of the root already found, and make it a trial divisor. Fifth. See how many times the trial divisor is contained in the dividend,... | |
 | Archibald Montgomerie - Algebra - 1857 - 116 pages
...answer. (2.) Subtract the cube of this term from the given quantity, and divide the first term of the remainder by three times the square of the part of the root found ; setting down the quotient for the next term of the answer. (3.) Subtract the cube of the binomial... | |
 | Dana Pond Colburn - 1858 - 288 pages
...left-hand period, and to the remainder bring down the next period, calling the result a dividend. 4th. Find three times the square of the part of the root already found, and make it a TRIAL DIVISOR. 5th. See how many times the trial divisor is contained in the dividend,... | |
 | Isaac Todhunter - Algebra - 1858 - 530 pages
...column, and add the last three lines in this column ; thus we obtain 12ж4 — 36cж' + 27cV, which is three times the square of the part of the root already found. Now divide the remainder in the third column by the expression just obtained, and we arrive at c' for... | |
 | James Bryce - 1861 - 376 pages
...divisors, the first Зa2, and the second, formed after two additions, За' + 6а6е + Зb2, each being three times the square of the part of the root already found ; the complete divisors being the quantities found after one addition to each of these ; the second... | |
 | Isaac Todhunter - Algebra - 1863 - 302 pages
...second column, and add the last three lines in this column ; thus we obtain 12ar'-36«*+27ic2, which is three times the square of the part of the root already found. Now divide the remainder in the third column by the expression just obtained, and we arrive at 4 for... | |
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Divide this remainder by three times the square of the part of the root already found, with two ciphers annexed, and write the quotient as the next figure of the root. 
