 | Simon Newcomb - Algebra - 1882 - 302 pages
...before. We have therefore the following rule for multiplying one polynomial by another. 119. RULE. Multiply each term of the multiplicand by each term of the multiplier, and add the products with their proper algebraic signs. EXERCISES. 1. (m - n) (p - q). Solution, (m... | |
 | Webster Wells - 1885 - 368 pages
...d was ac — be — aci + 6d. We have then the following rule for the product of two polynomials : Multiply each term of the multiplicand by each term of the multiplier, and add the partial products. EXAMPLES. 1. Multiply Зa — 26 by 2a — 56. In accordance with the... | |
 | Webster Wells - Algebra - 1885 - 324 pages
...d was аc — be — ad + &d. We have then the following rule for the product of two polynomials : Multiply each term of the multiplicand by each term of the multiplier, and add the partial products. EXAMPLES. 1. Multiply За — 2& by 2 а — 5&. In accordance with... | |
 | Horatio Nelson Robinson - 1888 - 372 pages
...is equal to the sum of the indices of those factors; thus: 7'x8'=56"; 4" x 5"'=2(X"". RULE. I. Write the several terms of the multiplier under the corresponding...of the multiplicand by each term of the multiplier, beginning with the lowest term in each, an I call the product of any two denominations the denomination... | |
 | Horatio Nelson Robinson, Daniel W. Fish - Arithmetic - 1888 - 372 pages
...those factors; thus: 7'x8'=56"; 4" x5'"=20""'. RULE. I. Write the several terms of the multiplier undef the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand by each term Oj the multiplier, beginning with the lowest term in each, and call the product of any two denominations... | |
 | Edward Albert Bowser - Algebra - 1888 - 868 pages
...bc+bd (Art. 33). . . (4) Hence, to multiply one polynomial by another, we have the following RULE. Multiply each term of the multiplicand by each term of the multiplier; if the terms multiplied together have the same sign, prefix the sign + to the product, if unlike, prefix... | |
 | Edward Brooks - Algebra - 1888 - 190 pages
...partial 2a2 — ab products, we have 2a2+3a6- 262. Therefore, etc. +4a6-26' 2a2 + 3a6-26« Rule. — Multiply each term of the multiplicand by each term of the multiplier, and add the partial products. a — 6 a +6 a2-a6 +a6-62 a3 -62 (6.) an-6" a2-6' a8 -62 an+Ja"68-a26"+6"+3... | |
 | William Frothingham Bradbury, Grenville C. Emery - Algebra - 1889 - 428 pages
...by-\-bz. Hence, for the multiplication of a polynomial by a polynomial, we have the following Ru1e. • Multiply each term of the multiplicand by each term of the multiplier, and find the sum of the several products. 2. Multiply 2 x2 + 3 xy — if by 3 x — 2 y. 2x* + 3xy... | |
 | Webster Wells - Algebra - 1890 - 560 pages
...Polynomials. By Art. 60, (1), = ac + be + ad + bd, by Art. 60, (5). "We then have the following rule : Multiply each term of the multiplicand by each term of the multiplier, and add the partial products. 1. Multiply 3a - 2 b by 2a - 56. In accordance with the rule, we multiply... | |
 | William James Milne - Algebra - 1894 - 216 pages
...a + 6 a times a + 6 = a2 + a6 6 times a + b = a6 + 62 (a + 6) times (a + 6) = a2 + 2 a6 + 62 RULE. Multiply each term of the multiplicand by each term of the multiplier, and add the partial products. 2. 2 а6 - 3 с 4а6 + c 8a2f>2-12a6c 2 aЬc - 3 с2 8a262- 10 a6c -3с2... | |
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X 5'" = 20"'". Hence the RULE. I. Write the several terms of the multiplier under the corresponding terms of the multiplicand. II. Multiply each term of the multiplicand by each term of the multiplier... 
