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Homework Help: What algebra trick is this?

  1. Aug 28, 2014 #1
    1. The problem statement, all variables and given/known data

    My book tried to simplify [itex] \frac{1}{4-u^2}[/itex] into [itex] \frac{1/4}{2-u} + \frac{1/4}{2+u} [/itex]. What algebra trick did it use?

    2. Relevant equations
    [itex] \frac{1}{4-u^2}[/itex]
    [itex] \frac{1/4}{2-u} + \frac{1/4}{2+u} [/itex]

    3. The attempt at a solution

    [tex] \frac{1}{4-u^2} \ = \ \frac{1}{(2-u)(2+u)} [/tex] so that's where the (2-u) terms come from but how do you manipulate that into the desired term?
  2. jcsd
  3. Aug 28, 2014 #2


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    it's adding fractions backwards. i.e. when you add two fractions you get a common denominator by multiplyingn their denominators. so backwards from this, if you have a fraction with a product denominator, you can try to write it as a sum of two fractions eachn one having one of the factors as a denominator. This is called "partial fractions"n decomposition.

    so just write 1/(2-u)(2+u) = a/(2-u) + b/(2+u), and solve for a and b after adding them.
  4. Aug 28, 2014 #3


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    For just a bit of expansion on the first response: the given fraction can be written as a sum of two others: the denominators are the factors of your fractions, the numerators are constants (constants in this case because each denominator is 1st degree). The first line below says the given fraction is the sum of two others: the remaining lines show how to eliminate denominators.
    \frac 1 {4-u^2} & = \frac A {2-u} + \frac B {2+u} \\
    \left(\frac 1 {4-u^2}\right)(4-u^2) & = \left(\frac A {2-u} + \frac {B} {2+u} \right) (2+u)(2-u) \\
    1 &= A(2+u) + B(2-u)

    Multiply the terms on the right side and solve the system of equations to find A and B.
  5. Aug 28, 2014 #4
    Yes that makes sense. I just needed to remind myself of partial fraction decomposition in order to get to that answer.
  6. Aug 28, 2014 #5


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    Or, set u to two different numbers to get two equations- most easily, set u= 2 to get 1= 4A and set u= -2 to get 1= 4B.
  7. Aug 28, 2014 #6
    This is called the Method of Partial Fractions, or Partial Fraction Decomposition.

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