... the first term of the quotient ; multiply the divisor by this term and subtract the product from the dividend. II. Then divide the first term of the remainder by the first term of the divisor... Manual of Algebra - Page 42by William Guy Peck - 1875 - 331 pagesFull view - About this book
| Jeremiah Day - Algebra - 1814 - 304 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 - 352 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 - 278 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... | |
| Bourdon (M., Louis Pierre Marie) - Algebra - 1831 - 446 pages
...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 term of the quotient ; multiply the divisor by this second term, and subtract... | |
| Peter Nicholson - Algebra - 1831 - 326 pages
...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.... | |
| Bourdon (M., Louis Pierre Marie) - Algebra - 1831 - 326 pages
...dividend 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... | |
| Charles Davies - Algebra - 1835 - 378 pages
...division polynomials we have the 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 from the dividend. II. Then divide the first term of the remainder by the fast term of the divisor, which gives... | |
| Ebenezer Bailey - Algebra - 1835 - 258 pages
...term, and subtract the product from the dividend. Divide the first term of the remainder by the Jirst term of the divisor, for the second term of the quotient. Multiply the whole divisor by this second term, and subtract the product from the remainder. Continue this series... | |
| Mathematics - 1836 - 530 pages
...divide the first term in the 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 di6 a 6s + б Ь3 vidend ; proceed to deal with the remainder, if any, in the same way. 43. By applying... | |
| Mathematics - 1836 - 488 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... | |
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