Adding to the trial-divisor 3 ab, that is, three times the product of the first term of the root by the second, and... Elements of Algebra - Side 135av Silvestre François Lacroix - 1825 - 276 siderUten tilgangsbegrensning - Om denne boken
| Adrien Marie Legendre - 1819 - 208 sider
...square of its first term must necessarily form the first term, 16 a* c*, of the proposed quantity ; **double the product of the first term of the root by the second** must give the second term, 24 a* 6* c, of the proposed quantity ; and the square of the last term of... | |
| Adrien Marie Legendre - 1825 - 224 sider
...in which the square I3 of the binomial m -\- n, produces, when developed, the terms m3 -f- 2mn + n3. **Now, after we have arranged the proposed quantity,...the square of these two terms, represented here by** I3 ; subtracting this square from the proposed quantity, we have for a remainder a quantity, which... | |
| Ormsby MacKnight Mitchel - 1845 - 294 sider
...term of the root. To get the third term of the root, complete the divisor by adding to it three times **the product of the 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,... | |
| Charles William Hackley - 1846 - 503 sider
...root Gx 3 ; multiplying the whole of this quantity, 6r > —2z > , by — 2x* (which produces twice **the product of the first term of the root by the second,** together with the square of the second), and subtracting the product from the first remainder, we obtain... | |
| John Fair Stoddard, William Downs Henkle - 1859 - 528 sider
...polynomial the square of the first term of the root, the first term of the remainder will be twice **the product of the first term of the root by the second** term of the root ; hence, we shall obtain the second term of the root by dividing the first term of... | |
| Shelton Palmer Sanford - 1879 - 332 sider
...the second term of the root. To form the complete divisor, we add to 362 (the trial divisor) 3 times **the product of the first term of the root by the second,** and also the square of the second. We shall then have 3b'+3bx + x2 as the complete divisor. Multiplying... | |
| Webster Wells - 1885 - 349 sider
...term of the root, we obtain I, the second term. Adding to the trial-divisor Зa6, that is, three times **the product of the first term of the root by the second,** and 62, that is, the square of the last term of the root, completes the divisor, Зa2 + Зa6 + б2.... | |
| Webster Wells - 1885
...term of the root, we obtain b, the second term. Adding to the trial-divisor 3ab, that is, three times **the product of the first term of the root by the second,** and 62, that is, the square of the last term of the root, completes the divisor, Зa8 + Зab + 62.... | |
| Webster Wells - 1885 - 349 sider
...root, 12 x4, we obtain — 3y as the second term of the root. Adding to the trial-divisor three times **the product of the first term of the root by the second,** — 18 x2y, and the square of the second term, 9т/2, completes the divisor, 12 ж* — I8x2y + 9ys.... | |
| Edward Albert Bowser - 1888 - 540 sider
...Also, since 3o26 + 3a62 + bs = (3a2 + Sab + b*)b, we add to the trial divisor Sab + &2, ie, three times **the product of the first term of the root by the second,** plus the square of the second, and we have the complete divisor 3a2 + Sab + b2 ; multiply this complete... | |
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