 | John Bayley (fellow of Emmanuel College, Cambridge.) - 1830 - 428 sider
...square from the given quantity ; to twice this quotient annex the quotient next found, for a divisor ; multiply the whole of this divisor by the second term of the quotient ; subtract this product from the remaining terms of the proposed quantity ; and repeat this... | |
 | Thomas Sherwin - 1841 - 314 sider
...second term of the root, also at the right of the divisor. Multiply the divisor, with the term annexed, by the second term of the root, and subtract the product from the dividend. 4. The remainder will form a new dividend, which is to be divided by twice the whole root... | |
 | William Scott - 1844 - 568 sider
...term of the root. Multiply the algebraic sum of double the first and once the second term of the root by the second term of the root, and subtract the product from the first remainder ; divide the first term of the second remainder by double the first term of the root; the quotient... | |
 | Ormsby MacKnight Mitchel - 1845 - 308 sider
...first term of the root, by the second term, plus the square of the second term. Multiply the complete divisor by the second term of the root, and subtract the product from the remainder, and bring down the next remainder. Divide the first term of this remainder by three times... | |
 | Joseph Ray - 1848 - 250 sider
...found, and annex the result both to the root and the divisor. Multiply the divisor thus increased, by the second term of the root, and subtract the product from the remainder. 3d. Double the terms of the root already found, for a partial divvsor, and divide the first... | |
 | James Elliot - 1850 - 116 sider
...must be annexed with its proper sign, as also to the divisor. 4. Multiply the divisor, thus enlarged, by the second term of the root, and subtract the product from the last dividend. 5. To the remainder bring down two more terms of the given quantity, and use that as... | |
 | Joseph Ray - 1852 - 408 sider
...already found, and annex the result to both the root and the divisor. Multiply the divisor thus increased by the second term of the root, and subtract the product from the remainder. 3rd. Double the terms of the root already found for a partial divisor] then divide the first... | |
 | G. Ainsworth - 1854 - 216 sider
...quotient, and also annexed to the new divisor as a second term. Multiply the new divisor by the last found term of the root, and subtract the product from the first remainder. To the new remainder bring down the two following terms of the compound quantity for a new dividend,... | |
 | Thomas Sherwin - 1855 - 262 sider
...second term of the root, also at the right of the divisor. Multiply the divisor, with the term annexed, by the second term of the root, and subtract the product from the dividend. 4. The remainder will form a second dividend, which is to be divided by twice the whole root... | |
 | Theodore Strong - 1859 - 570 sider
...root to the incomplete divisor, and the result will be the complete divisor. Multiply the complete divisor by the second term of the root, and subtract the product from the first remainder, cutid the result will constitute the second remainder. 3. Take twice the root now fmmd for the incomplete... | |
| |
Multiply the divisor thus increased, by the second term of the root, and subtract the product from the remainder. 
