 | William James Milne - Algebra - 1908 - 476 pages
...root, and subtmct the product from the first remainder. Find the next term of the root by dividing the first term of the 'remainder by the first term of the trial divisor. Extract the square root of : 10. 25 a'2 - 40 a + 16. 13. 4 ж4 - 52 ^ + 169. 11. 900 ж2 + 60 x + 1.... | |
 | Jacob William Albert Young - 1908 - 344 pages
...example, x) is the first term of the quotient. Multiply the entire divisor by this term and subtract. Divide the first term of the remainder by the first term of the divisor. The result (in this example, 2 у) is the second term of the quotient. Multiply the entire... | |
 | Jacob William Albert Young, Lambert Lincoln Jackson - Algebra - 1908 - 460 pages
...example, x) is the first term of the quotient. Multiply the entire divisor by this term and subtract. Divide the first term of the remainder by the first term of the divisor. The result (in this example, 2 y) is the second term of the quotient. Multiply the entire... | |
 | Jacob William Albert Young, Lambert Lincoln Jackson - Arithmetic - 1909 - 292 pages
...2b2 a- Multiply the entire divisor by a and 2 , T subtract, as in the case of numbers. Di_L O7)2 vide the first term of the remainder by the first term of the divisor again; the + ^ — quotient is 26. Multiply and subtract, as in the case of numbers. Test:... | |
 | Herbert Edwin Hawkes, William Arthur Luby, Frank Charles Touton - Algebra - 1910 - 374 pages
...term of the quotient. to write the terms of the remainder in the same order as those of the divisor. Divide the first term of the remainder by the first term of the divisor for the second term of the quotient and proceed as before until there is no remainder, or until... | |
 | William James Milne - Algebra - 1911 - 378 pages
...root, and subtract the product from the first remainder. Find the next term of the root by dividing the first term of the remainder by the first term of the trial divisor. 4a;2 4 ar + 3 x 4 ж2 + 6 x 4ж2 + 6ж-3 12 ж3 - 3 ж-' 12 ж3 + 9 ж2 -12Ж2 12 ж2 — 18 а; +... | |
 | Herbert Edwin Hawkes, William Arthur Luby, Frank Charles Touton - Algebra - 1911 - 288 pages
...subtract, being careful to write the terms of the remainder in the same order as those of the divisor. Divide the first term of the remainder by the first term of the divisor for the second term of the quotient, and proceed as before until there is no remainder, or... | |
 | Webster Wells, Walter Wilson Hart - Algebra - 1912 - 344 pages
...subtract the result from the given expression. 3. Double the root already found, for the trial divisor. Divide the first term of the remainder by the first term of the trial divisor. Add the quotient to the root and also to the trial divisor, obtaining the complete divisor. 4. Multiply... | |
 | Webster Wells, Walter Wilson Hart - Algebra - 1912 - 344 pages
...step 2. 5. Continue in this manner : (a) double the root already found for a new trial divisor ; (6) divide the first term of the remainder by the first term of this product for the new term of the root ; (c) add the new term of the root to the trial divisor,... | |
 | Webster Wells, Walter Wilson Hart - Algebra - 1912 - 504 pages
...step 2. 5. Continue in this manner : (a) double the root already found for a new trial divisor ; (ft) divide the first term of the remainder by the first term of this product for the new term of the root ; (с) add the new term of the root to the trial divisor,... | |
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Then divide the first term of the remainder by the first term of the divisor... 
