Partial Fractions: Solving Numerator Issues

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SUMMARY

The discussion focuses on the requirements for numerators in partial fraction decomposition, specifically addressing the equation \(\frac{1}{(x-1)(x-2)^2}\equiv \frac{A}{x-1}+\frac{Bx+C}{(x-2)^2}\). It is established that the numerator must be a polynomial of lower degree than the denominator. In this case, since the second fraction's denominator is quadratic, its numerator must be linear. This conclusion is supported by the principle that during synthetic division, the remainder is always one degree less than the divisor.

PREREQUISITES
  • Understanding of polynomial degrees
  • Familiarity with partial fraction decomposition
  • Knowledge of synthetic division
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study polynomial long division techniques
  • Learn more about synthetic division and its applications
  • Explore advanced topics in partial fraction decomposition
  • Practice solving partial fractions with varying degrees of polynomials
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Students studying algebra, mathematics educators, and anyone looking to deepen their understanding of partial fraction decomposition in calculus or algebra courses.

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I'm having trouble understanding what the numerator needs to be in the partial fractions.
e.g.

[tex]\frac{1}{(x-1)(x-2)^2}\equiv \frac{A}{x-1}+\frac{Bx+C}{(x-2)^2}[/tex]

Notice how the first numerator has a constant A, while the second is linear Bx+C.
Actually... just now I think I may understand it. Does it have to do with the fact that during synthetic division, the remainder is always 1 degree less than the divisor? The second fraction's denominator is a quadratic, so its numerator should be linear?
 
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"Does it have to do with the fact that during synthetic division, the remainder is always 1 degree less than the divisor? The second fraction's denominator is a quadratic, so its numerator should be linear?"

Essentially - the numerator should always be the "most general" polynomial of lower degree than the denominator
 

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