 | John Bonnycastle - Algebra - 1813 - 456 pages
...former rule. CASE IL IVhen títe divisor is a simple quantity, and thé dividend a compound one. RULE. Divide each term of the dividend by the divisor, as in the former case ; setting down such as will not divide in the simplest form they will admit of. EXAMPLES.... | |
 | John Bonnycastle - Algebra - 1818 - 326 pages
...by-^a CASE II. . When the divisor is a simple quantity, and the dividend a, compound one'. , RULE. Divide each term of the dividend by the divisor, as in the former case ; setting; down such as will not divide is the simplest form they will admit of. EXAMPLES.... | |
 | John Bonnycastle - Algebra - 1825 - 336 pages
...xy Ans. -- - - , an CASE II. When the divisor is a simple quantity, and the dividend a compound one. Divide each term of the dividend by the divisor, as in the former case ; setting down such as will not divide i» the simplest form they will admit of. EXAMPLES.... | |
 | George Lees - 1826 - 276 pages
...— • CASE II. 31. When the divisor is a simple quantity, and the dividend a compound one. RULE. Divide each term of the dividend by the divisor, as in the farmer case, setting down such as will not divide in their simvlest form. • EXAMPLES. (ab + bc -HI... | |
 | Thomas Sherwin - Algebra - 1841 - 314 pages
...— , the quotient must have thf sign — . Art. 42. Hence, we have the following RULE FOR DIVIDING A POLYNOMIAL, BY A MONOMIAL.. Divide each term of the dividend by the divisor, according to the principles given in Art. 4O for dividing one monomial by another, observing the rule... | |
 | William Scott - Algebra - 1844 - 568 pages
...signs, by the rule for the division of monomials. Whence, to divide a polynomial by a monomial, Rule. Divide each term of the dividend by the divisor, as in the case of monomials, and take the algebraic sum of the partial quotients for the quotient required. Examples of the division... | |
 | Scottish school-book assoc - 1845 - 444 pages
...Ans. l$x%y$z*. 34. CASE II. "When the dividend is a compound, and the divisor a simple quantity. RULE. Divide each term of the dividend by the divisor, as in the last case, and the sum of the separate quotients will be the answer. EXAMPLE. Divide 3a2b+4a3bc, by... | |
 | Joseph Ray - Algebra - 1848 - 250 pages
...the multiplicand by the multiplier; therefore, we have the following RULE, FOR DIVIDING A POLVNOMIAL BY A MONOMIAL. Divide each term of the dividend, by the divisor, according to the rule for the divvsion of monomials. EXAMPLES. 1. Divide 6x+l 2y by 3 ........... Ans.... | |
 | John Bonnycastle - 1848 - 334 pages
...Ans.y-*"-» = -^r¿When the divisor is a simple quantity, and the dividend a compound one. Rule. — Divide each term of the dividend by the divisor, as in the former case, setting down such as will not divide in the simplest form they will admit of. EXAMPLES.... | |
 | Joseph Ray - Algebra - 1848 - 250 pages
...each term of the multiplicand by the multiplier; therefore, we have the following BULB FOR DIVIDING A POLYNOMIAL BY A MONOMIAL. Divide each term of the dividend, by the divisor, according to th« rtde for the division of monomials. EXAMPLES. 1. Divide 6z+12y by 3 Ans. 2x+4y. 2.... | |
| |
To divide a polynomial by a monomial, divide each term of the dividend by the divisor and add the partial quotients. 
