To divide a polynomial by a monomial, divide each term of the polynomial by the monomial: (Sab — 12ac) -i- 4a = 36 — 3c. A College Algebra - Page 19by James Morford Taylor - 1889 - 317 pagesFull view - About this book
| Robert Judson Aley, David Andrew Rothrock - Algebra - 1904 - 344 pages
...-*-k = a + b + c, which shows that k is distributed as a divisor to every term of the dividend. RULE. Divide each term of the polynomial by the monomial and add the results. , „ „ -4 a2г/2. EXERCISES. 1. (12 aY - 8 tfmf — 4 a'aA/3) ч- 4 ay. 2. (9 аЬсж4 — 18 aW... | |
| Henry Burchard Fine - Algebra - 1904 - 616 pages
...one monomial by another, form a fraction by writing the dividend over the divisor, and simplify. 2. To divide a polynomial by a monomial, divide each term, of the dividend by the divisor, and add the quotients so obtained. 4a2c Thus, common factor 2 ab'2 and applying... | |
| Henry Burchard Fine - Algebra - 1904 - 612 pages
...dividing by a monomial. From the definition of division and § 320, 4, we derive the following rules. 2. To divide a polynomial by a monomial, divide each term of the dividend by the divisor, and add the quotients so obtained. _ я (,''¡í'r Thus, - 8 a»f>2c -=- б... | |
| Arthur William Potter - Algebra - 1904 - 182 pages
...а&2-2 ab». 2 а+3 6 -б2 is therefore the quotient of 4 a26+6 ai2 -2 a#> divided by 2 aft. RULE. To divide a polynomial by a monomial, divide each term of the dividend 'by the divisor as in the case of monomials. Divide : 1. 6 ay— 3a2z/2 + 9 ay3 by Say. 2.... | |
| Walter Randall Marsh - Algebra - 1905 - 412 pages
...factor a are given, the quotient will be b + c. Whence is derived the following Rule for the Division of a Polynomial by a Monomial : Divide each term of the polynomial by the monomial and add the quotients thus derived. Divide a3 - 2 aïb + 8 aW by a3. a3 EXERCISE XVIII Perform the indicated divisions... | |
| John Charles Stone, James Franklin Millis - Algebra - 1905 - 776 pages
...quotient must be obtained by dividing the terms of the dividend by the divisor. Hence the rule : 5 To divide a polynomial by a monomial, divide each term of the dividend by the divisor, and take the sum of the resulting quotients. EXAMPLE. Divide ж' + аж5—... | |
| Arthur Schultze - Algebra - 1905 - 468 pages
...the product ax + bx + ex. But x(a + b + c) = ax + bx + ex. Hence аяе + bх + сх = а + b + о. x To divide a polynomial by a monomial, divide each term of the dividend by the monomial and add the partial quotients thus formed. Eg -- ^ 2 4- 5 L — 3xyz2 EXERCISE... | |
| Arthur Schultze - Algebra - 1905 - 674 pages
...expression which multiplied by x gives the product ax + bx + ex. But x(a + b + c) = ax + bx + ex. Hence To divide a polynomial by a monomial, divide each term, of the dividend by the monomial and add the partial quotients thus formed. Eg - 6 *У* - 15 *№ + 3 ^ = 2... | |
| George Albert Wentworth - Algebra - 1906 - 440 pages
...m — 1 = m + 1, as the exponents of a in the first and second terms of the quotient respectively. To Divide a Polynomial by a Monomial, Divide each term of the dividend by the divisor, and connect the partial quotients by their proper signs. EXERCISE 24 Divide... | |
| Arthur Schultze - Algebra - 1906 - 584 pages
...еж. But x(a + b + c) = ax + bx + cx. TT ax + bx -+• cx , ' Hence — — - — — = a + b + c. x To divide a polynomial by a monomial, divide each term of the dividend by the monomial and add the partial quotients thus formed. -- 2 3. , + 5 L -3xyz2 EXERCISE... | |
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