If we now begin the contraction, it is good to know beforehand on what number of additional root-figures we may reckon. We may be pretty certain of having nearly as many as there are figures in the divisor when we begin to contract--one less, or at least two less. Thus, there being now eight figures in the divisor, we may conclude that the contraction will give us at least six more figures. To begin the contraction, let the dividend stand, cut off one figure from the divisor, two from the column before that, three from the one before that, and so on. Thus, our contraction begins with | | | | |0002 1|704 5445|28 7734837|6 47339778 | | | | The first column is rendered quite useless here. Conduct the process as before, using only the figures which are not cut off. But it will be better to go as far as the first figure cut off, carrying from the second figure cut off. We shall then have as follows: | | | 1|704 5445|28 7734837|6 47339778(6 | 5455|5 7767570|6 734354 5465|7 7800364|8 5475|9 | | At the next contraction the column 1|704 becomes |001704, and is quite useless. The next step, separately written (which is not, however, necessary in working), is | | 54|759 780036|48 734354(0 | | Here the dividend 734354 does not contain the divisor 780036, and we, therefore, write 0 as a root figure and make another contraction, or begin with | | |54759 78003|648 734354(9 | 78008|5 32277 78013|4 | At the next contraction the first column becomes |0054759, and is quite useless, so that the remainder of the process is the contracted division. | 7801|34)32277(4137 | 1072 292 58 3 and the root required is 21·36094137. I now write down the complete process for another equation, one root of which lies between 3 and 4: it is _x_³ - 10_x_ + 1 = 0