| William James Milne - Algebra - 1894 - 214 pages
...a + 6 а times a + b = a2 + ab b times a + b = ab + b1 (a + 6) times (a + 6) = a2 + 2 ab + 62 BULE. Multiply each term of the multiplicand by each term of the multiplier, and add the partial products. 2. 2a6-3c 4 ab + с 8 a262- 12aftc 2 a6c - Я с2 8cW- 10 a6c -3с3 Multiply : 4.... | |
| William Frothingham Bradbury, Grenville C. Emery - Algebra - 1894 - 144 pages
...this equals Hence, for the multiplication of a polynomial by a polynomial, we have the following Bule. Multiply each term of the multiplicand by each term of the multiplier, and find the sum of the several products. 2. Multiply За2 — 2а6 + 462 by 2a — 3b. 3 a* — 2 а... | |
| George Albert Wentworth - Algebra - 1894 - 204 pages
...+ bm + 6n + bp + cm -f- en + cp. To find the product of two polynomials, therefore, Multiply every term of the multiplicand by each term of the multiplier, and add the partial products. 88. In multiplying polynomials, it is a convenient arrangement to write the multiplier... | |
| William Frothingham Bradbury, Grenville C. Emery - Algebra - 1894 - 166 pages
...Hence, for the multiplication of a polynomial by a polynomial, we have the following Rule. Midtiply each term of the multiplicand by each term of the multiplier, and find the sum of the several products. 2. Multiply Зa2 — 2 a ¿ + 4 ¿2 by 2 a — 3 ¿. Зa2—... | |
| George P. Lilley - Algebra - 1894 - 522 pages
...х - 25 z10 + 2 а? + 7 z6 - 8 х4 - 3 х8 - 15 ж8 + 10 хг - 5 а; -25 Explanation. Multiplying each term of the multiplicand by each term of the multiplier and connecting these results with their proper signs, we have x10 — x9 + 2 xe — x6 — 5 x* + 3 z9... | |
| Wallace Clarke Boyden - Algebra - 1894 - 188 pages
...2. 3a^ -2ж +1 3я?.+ 4 ж4 + 6ж3- 4ж2 + 3x То multiply a polynomial by a polynomial, multiply the multiplicand by each term of the multiplier, and add the products. How is the first term of the product obtained ? How is the last term obtained ? The polynomials being... | |
| George Washington Hull - Algebra - 1895 - 358 pages
...partial products, we 2a2 + ab - 662, Ans. have 2aJ + 06 - 66Z. From this example we derive the following RULE. Multiply each term of the multiplicand by each term of the multiplier, and add the partial products. m2 + 2mn + w2, ^4ns. m2 — 2mn + n2 m'1 5. 6. a4 + rt'6 + a262 x*-x« + z4 - o,36... | |
| Edward Brooks - Arithmetic - 1895 - 424 pages
...placing terms of the same order in the same column, and draw a line beneath. II. Begin at the right, and multiply each term of the multiplicand by each term of the multiplier, writing the first term of each product under the term of the multiplier used to obtain it. III. Add... | |
| William Freeland - Algebra - 1895 - 328 pages
...За Зx-iy2 6a + 3&3-4c2 -3г 5cPe POLYNOMIALS BY POLYNOMIALS. 59. RULE. — Multiply all the terms of the multiplicand by each term of the multiplier, and add the partial products. 1. 6ж — 3y 24 a? -12 я?у 2. а-b — b3e + c4d + аbс ab!c4d - bV + bc'd +... | |
| Emerson Elbridge White - Algebra - 1896 - 418 pages
...polynomial according to the ascending or descending poicers of the same letter. Multiply all the terms of the multiplicand by each term of the multiplier, and add the several partial products. NOTE. In multiplying, observe carefully the laws of the signs. EXERCISES.... | |
| |