 | John Davidson, Robert Scott (writing master) - Arithmetic - 1818 - 190 pages
...the divisor. The tum of these three parts will be the complete divisor, which multiply by the last figure of the root, subtract the product from the dividend, and to the remainder bring down the next part for a new dividend. Proceed in the same manner as before to find the divisor... | |
 | John Radford Young - Algebra - 1832 - 408 pages
...divisor's place, and the divisor will be completed. Multiply the complete divisor by the last term of the root, subtract the product from the dividend, and to the remainder connect the three next terms, and proceed as before. For (by Art. 37), the cube of a + b is a3 + 3a2¿... | |
 | Benjamin Peirce - Algebra - 1837 - 300 pages
...is also to be placed at the right of the divisor. Multiply the divisor, thus augmented, by the last figure of the root, subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. Double the root now found for a new divisor and continue... | |
 | Benjamin Peirce - Algebra - 1837 - 302 pages
...is also to be placed at the right of the divisor. Multiply the divisor, thus augmented, by the last figure of the root, subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. Double the root now found for a new divisor and continue... | |
 | John Radford Young - 1839 - 332 pages
...divisor's place, and the divisor will be completed. Multiply the complete divisor by the last term of the root, subtract the product from the dividend, and to the remainder connect the three next terms, and proceed as before. For (by Art. 37,) the cube of a+b is a»+ 3a2¿>... | |
 | John Husband (math. master, Berwick.) - 1841 - 126 pages
...right ; add together these two lines for the complete divisor; multiply the sum by the second figure in the root ; subtract the product from the dividend, and to the remainder annex the third period for a new dividend. Place the square of the second figure of the root under the complete... | |
 | George Roberts Perkins - Arithmetic - 1841 - 274 pages
...the result will be the TRUE DIVISOR. Multiply the true divisor by this second figure of the root, and subtract the product from the dividend, and to the remainder annex the next period,for a SECCUD DIVIDEND. . ft IV. To the last TRUE DIVISOR, add the Jastfgure of the root, for... | |
 | James Bates Thomson - Arithmetic - 1846 - 402 pages
...the root, also on the right of the partial divisor ; multiply the divisor thus completed by the last figure of the root ; subtract the product from. the dividend, and to the remainder bring down the next period for a new dividend, as before. IV. Double the root already found for a new... | |
 | James Bates Thomson - Arithmetic - 1846 - 354 pages
...the root, also on the right of the partial divisor ; multiply the divisor thus completed by the last figure of the root ; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend, as before. IV. Double the root already found for a new... | |
 | James Bates Thomson - Arithmetic - 1847 - 426 pages
...the right of the partial divisor ; multijily the divisor thus completed b;/ the figure last placed in the root ; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. IV. Double the root already found for a new partial... | |
| |
This sum will be the complete divisor. Multiply the complete divisor by the second figure of the root, subtract the product from the dividend, and to the remainder annex the next group for a new dividend. 
