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<para>This module is from<link document="col10614">Elementary Algebra</link>by Denny Burzynski and Wade Ellis, Jr.</para><para>A detailed study of arithmetic operations with rational expressions is presented in this chapter, beginning with the definition of a rational expression and then proceeding immediately to a discussion of the domain. The process of reducing a rational expression and illustrations of multiplying, dividing, adding, and subtracting rational expressions are also included. Since the operations of addition and subtraction can cause the most difficulty, they are given particular attention. We have tried to make the written explanation of the examples clearer by using a "freeze frame" approach, which walks the student through the operation step by step.</para><para>The five-step method of solving applied problems is included in this chapter to show the problem-solving approach to number problems, work problems, and geometry problems. The chapter also illustrates simplification of complex rational expressions, using the combine-divide method and the LCD-multiply-divide method.</para><para>Objectives of this module: be able to divide a polynomial by a monomial, understand the process and be able to divide a polynomial by a polynomial.</para>


  • Dividing a Polynomial by a Monomial
  • The Process of Division
  • Review of Subtraction of Polynomials
  • Dividing a Polynomial by a Polynomial

Dividing a polynomial by a monomial

The following examples illustrate how to divide a polynomial by a monomial. The division process is quite simple and is based on addition of rational expressions.

a c + b c = a + b c

Turning this equation around we get

a + b c = a c + b c

Now we simply divide c into a , and c into b . This should suggest a rule.

Dividing a polynomial by a monomial

To divide a polynomial by a monomial, divide every term of the polynomial by the monomial.

Sample set a

3 x 2 + x 11 x . Divide every term of   3 x 2   +   x 11   by   x . 3 x 2 x + x x 11 x = 3 x + 1 11 x

8 a 3 + 4 a 2 16 a + 9 2 a 2 . Divide every term of   8 a 3   +   4 a 2 16 a   + 9   by   2 a 2 . 8 a 3 2 a 2 + 4 a 2 2 a 2 16 a 2 a 2 + 9 2 a 2 = 4 a + 2 8 a + 9 2 a 2

4 b 6 9 b 4 2 b + 5 4 b 2 . Divide every term of 4 b 6 9 b 4 2 b + 5 by 4 b 2 . 4 b 6 4 b 2 9 b 4 4 b 2 2 b 4 b 2 + 5 4 b 2 = b 4 + 9 4 b 2 + 1 2 b 5 4 b 2

Practice set a

Perform the following divisions.

2 x 2 + x 1 x

2 x + 1 1 x

3 x 3 + 4 x 2 + 10 x 4 x 2

3 x + 4 + 10 x 4 x 2

a 2 b + 3 a b 2 + 2 b a b

a + 3 b + 2 a

14 x 2 y 2 7 x y 7 x y

2 x y 1

10 m 3 n 2 + 15 m 2 n 3 20 m n 5 m

2 m 2 n 2 3 m n 3 + 4 n

The process of division

In Section [link] we studied the method of reducing rational expressions. For example, we observed how to reduce an expression such as

x 2 2 x 8 x 2 3 x 4

Our method was to factor both the numerator and denominator, then divide out common factors.

( x 4 ) ( x + 2 ) ( x 4 ) ( x + 1 )

( x 4 ) ( x + 2 ) ( x 4 ) ( x + 1 )

x + 2 x + 1

When the numerator and denominator have no factors in common, the division may still occur, but the process is a little more involved than merely factoring. The method of dividing one polynomial by another is much the same as that of dividing one number by another. First, we’ll review the steps in dividing numbers.

  1. 35 8 .  We are to divide 35 by 8.
  2. Long division showing eight dividing thirty five. This division is not performed completely.   We try 4, since 32 divided by 8 is 4.
  3. Long division showing eight dividing thirty five, with four at quotient's place. This division is not performed completely. Multiply 4 and 8.
  4. Long division showing eight dividing thirty five, with four at quotient's place. Thirty two is written under thirty five. This division is not performed completely Subtract 32 from 35.
  5. Long division showing eight dividing thirty five, with four at quotient's place. Thirty two is written under thirty five and three is written as the subtraction of thirty five and thirty two. Since the remainder 3 is less than the divisor 8, we are done with the 32 division.
  6. 4 3 8 .   The quotient is expressed as a mixed number.

Questions & Answers

what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
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im not good at math so would this help me
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f(x)= 2|x+5| find f(-6)
Prince Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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what is nano technology
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what is system testing?
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1
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