| Jeremiah Day - Algebra - 1814 - 303 pages
...first term, for the first term of the required root: Subtract the power from the given quantity, and **divide the first term of the remainder, by the first term of the** root involved to the next inferiour power, and multiplied by the index of the given power ,-f the quotient... | |
| Jeremiah Day - Algebra - 1827 - 332 pages
...first term, for the first term of the required root : Subtract the power from the given quantity, and **divide the first term of the remainder, by the first term of the** root involved to the next inferior power, and multiplied by the index of the given power ;t the quotient... | |
| William Smyth - Algebra - 1830 - 264 pages
...whole divisor by the term of the quotient just found, and subtract the result from the dividend ; 3°. **divide the first term of the remainder by the first term of the divisor,** the result will be the second term of the quotient ; 4°. multiply the whole divisor by the second... | |
| Peter Nicholson - Algebra - 1831 - 309 pages
...next term of the root. W Raise this quotient to the given power, and subtract it from the dividend ; **then divide the first term of the remainder by the first term of the divisor,** and it will give the next term of the quotient, proceed in a similar manner till the whole is finished.... | |
| M. Bourdon (Louis Pierre Marie) - Algebra - 1831 - 304 pages
...by the first term on the left of the divisor, and the first term of the quotient is found ; multiply **the divisor by this term, and subtract the product from the dividend.** Divide the first term of this remainder by the first term of the divisor, and the second term of the... | |
| M. Bourdon (Louis Pierre Marie) - Algebra - 1831 - 389 pages
...by the first term on the left of the divisor, the result is the first term of the quotient; multiply **the divisor by this term, and subtract the product from the dividend.** Then divide the first term of the remainder by the first term of the divisor, which gives the second... | |
| M. Bourdon (Louis Pierre Marie) - Algebra - 1835 - 353 pages
...following RULE. on the left of the divisor, the result is the first term of the quotient ; multiply **the divisor by this term, and subtract the product...Then divide the first term of the remainder by the** fast term of the divisor, which gives the second term of the quotient ; multiply the divisor by this... | |
| Jeremiah Day - Mathematics - 1836 - 472 pages
...first term, for the first term of the required root. Subtract the power from the given quantity, and **divide the first term of the remainder by the first term of the** root, involved to the next inferior power, and multiplied by the index of the given power ; the quotient... | |
| Society for the Diffusion of Useful Knowledge (Great Britain) - Education - 1836
...dividend by the first in the divisor, and write the result as the first in the quotient ; multiply **the divisor by this term, and subtract the product from the dividend** ; proceed to deal with the remainder, if any, in the same way. 43. By applying the same reasoning as... | |
| Charles Davies, Bourdon (Louis Pierre Marie, M.) - Algebra - 1838 - 355 pages
...the first term on the left of the divisor, the result is the first term of the quotient ; multiply **the• divisor by this term, and subtract the product...of the remainder by the first term of the divisor,** which gives the second term of the quotient ; multiply the divisor by this second term, and subtract... | |
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