Partial Fractions: Solving Simple Linear Equations

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SUMMARY

This discussion focuses on solving simple linear equations using partial fractions, specifically the integration of expressions like (x² - 8)/(x + 3) and (x² + 1)/(x - 1). The correct approach involves polynomial long division, where (x² - 8) is divided by (x + 3) to yield x - 3 + 1/(x + 3). Additionally, recognizing the numerator as a difference of squares aids in simplifying the fractions. The participants emphasize the importance of checking work by reversing the process for better understanding.

PREREQUISITES
  • Understanding polynomial long division
  • Familiarity with partial fraction decomposition
  • Knowledge of the difference of squares formula
  • Basic algebraic manipulation skills
NEXT STEPS
  • Practice polynomial long division with various expressions
  • Explore partial fraction decomposition techniques in calculus
  • Review the difference of squares and its applications in algebra
  • Learn how to verify solutions by working backwards in algebraic equations
USEFUL FOR

Students studying algebra, particularly those learning about integration techniques and partial fractions, as well as educators seeking to clarify these concepts for their students.

Help1212
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Homework Statement


Hello I'm trying to figure out how to integrate with the use of partial fractions but I'm a bit stuck on something that should probably be simple.. but I can't see it clearly

How do you split up x2-8/x+3 and make it equal to (x-3) + 1/x+3

or for instance x2 +1/x-1 split up into x+1 +2/x-1

Homework Statement


Homework Equations


The Attempt at a Solution


They seem simple.. but i think I'm missing something here because I only see one linear denominator so I don't know how to apply the sequence.. or I'm overlooking something.. a bitof hellp please?
 
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It is basically "long division". To divide [itex]x^2- 8[/itex] by x+ 3, note that x divides into [itex]x^2[/itex] x times. Taking x times x+3, we get [itex]x^2+ 3x[/itex] and subtract to get [itex]x^2- 8 - (x^2+ 3x)= -3x- 8[/itex]. That is, x+ 3 divides into [itex]x^2- 8[/itex] x times with remainder -3x- 8. x divides into -3x -3 times. Taking -3 times x+ 3 we get [itex]-3x- 9[/itex] and subtracting that from -3x- 8 gives a remainder of +1. That tells you that
[tex]\frac{x^2- 8}{x+ 3}= x- 3+ \frac{1}{x+ 3}[/tex].

Another way to see that, simpler but not "rote", is to note that
[tex]\frac{x^2- 8}{x+ 3}= \frac{x^2- 9+ 1}{x+ 3}= \frac{(x- 3)(x+ 3)+ 1}{x+ 3}[/tex]
[tex]= x- 3+ \frac{1}{x+ 3}[/tex]
again.
 
Help1212 said:

Homework Statement


Hello I'm trying to figure out how to integrate with the use of partial fractions but I'm a bit stuck on something that should probably be simple.. but I can't see it clearly

How do you split up x2-8/x+3 and make it equal to (x-3) + 1/x+3

or for instance x2 +1/x-1 split up into x+1 +2/x-1

Knowing what to look for it is easy enough even when you have left out the brackets.

One useful thing for you would be to work backwards and check it that way - may give some insights.

One thing to look for is express your numerator as a [difference of two squares (check the formula for that) that has your denominator as factor] + something.
 
Last edited:

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