...divisor, and annex the result both to the root and to the divisor. Multiply the divisor thus completed by the term of the root last obtained, and subtract the product from the former remainder. Divide the first term of the remainder as before, for the next term of the root ;...

...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...

...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...

...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...

...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,...

...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,...

...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...

...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,...

...found, and annex the result to the -root, and also to the divisor. Multiply the divisor as it now stands by the term of the root last obtained, and subtract the product from the dividend. If there are other terms remaining, continue the operation in the same manner as before....

...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...