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" Multiply each term of the multiplicand by each term of the multiplier, and add the partial products. "
The Inductive Algebra: Embracing a Complete Course for Schools and Academies - Page 46
by William James Milne - 1881 - 347 pages
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A Complete Course in Algebra for Academies and High Schools

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 the rule, we multiply За —...
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A Complete Course in Algebra

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 rule, we multiply Зa — 26...
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Shorter Course in Algebra

George Albert Wentworth - Algebra - 1886 - 284 pages
...-\-bп -|- сn -\- ар -\-bр + ср. That is, to find the product of two polynomials, 71. Multiply the multiplicand by each term of the multiplier and add the partial products ; or, multiply each term of one factor by each term of the other, and add the partial products. 72....
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The Normal Elementary Algebra: Containing the First Principles of the ...

Edward Brooks - Algebra - 1888 - 190 pages
...Adding the partial 2a2 — ab products, we have 2a2+3a6- 262. Therefore, etc. +4a6-26' 2a2 + 3a6-26« Rule. — Multiply each term of the multiplicand by...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 7. Multiply 3a - 26 by...
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Sheldons' Complete Algebra: Part I Being Sheldon's Elements of ..., Parts 1-2

Algebra - 1888 - 492 pages
...last partial product. The sum of these partial products gives the product required. Rule. — Multiply the multiplicand by each term of the multiplier and add the partial products. 83. 1. 4s. 6. 7. 8. 9. 10. EXERCISES. Multiply : a + b by a + b. 3ж + 2y by 2ж + 3y. 3ab + № by...
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College Algebra: For the Use of Academies, Colleges, and Scientific Schools ...

Edward Albert Bowser - Algebra - 1888 - 868 pages
...ad— 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...
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Shorter Course in Algebra

George Albert Wentworth - 1888 - 268 pages
...+ an + bn -\-en + ap + bp + cp. That is, to find the product of two polynomials, 71. Multiply ¿he multiplicand by each term of the multiplier and add the partial products ; or, multiply each term of one factor by each term of the other, and add the partial products. 72....
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Robinson's Progressive Practical Arithmetic: Containing the Theory of ...

Horatio Nelson Robinson - 1888 - 372 pages
...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, beginning with the lowest term in each, an I call the product of any two denominations the denomination...
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The Academic Algebra

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 —...
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College Algebra

Webster Wells - Algebra - 1890 - 604 pages
...Polynomials by Polynomials. By Art. 60, (1), = ac + be + ad + bd, bj 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 — 26 by 2a — 56. In accordance with the rule, we multiply 3a— 2b by 2 a, and then...
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