# Intuition Behind Polynomial Numerators in Partial Fractions

by Justin Skycak on

Each decomposition produces a system of linear equations where the number of unknowns equals the number of equations.

Consider the following partial fractions decomposition:

\begin{align*} \dfrac{7x^2-28x+17}{(x^2+1)(x-5)} = \dfrac{Ax+B}{x^2+1} + \dfrac{C}{x-5} \end{align*}

You might wonder, why does the $\dfrac{Ax+B}{x^2+1}$ term require a linear numerator instead of just a constant? Yes, we have to do that whenever the denominator is an irreducible quadratic… but why?

Partial fractions formulas are typically not proven in calc textbooks since the proof is rather complicated.

However, there is a hand-wavy intuitive way that I like to think about it. Each decomposition produces a system of linear equations where the number of unknowns equals the number of equations.

\begin{align*} \dfrac{7x^2-28x+17}{(x^2+1)(x-5)} &= \dfrac{Ax+B}{x^2+1} + \dfrac{C}{x-5} \\[5pt] 7x^2-28x+17 &= (Ax+B)(x-5) + C(x^2+1) \\[5pt] 7x^2-28x+17 &= (Ax^2-5Ax+Bx-5B) + (Cx^2+C) \\[5pt] 7x^2-28x+17 &= (A+C)x^2+(-5A+B)x+(-5B+C) \end{align*}

Here’s the system of linear equations:

\begin{align*} \begin{cases} 7 = A+C \\[5pt] -28 = -5A+B \\[5pt] 17 = -5B + C \end{cases} \end{align*}

Having the number of unknowns equal the number of equations is generally a “good” thing when solving equations. It makes a solution more likely to exist.

For instance, if you construct two linear equations at random, say $A+B=3$ and $A+2B=4,$ then a solution exists. (The only way a solution wouldn’t exist is if one of the equations happens to be “redundant,” introducing no new information – but it turns out the partial fractions setup also circumvents that possibility.)

If you do that same experiment with just one variable, then you’re basically doomed to fail unless the equations mean the same thing. The system $A=3$ and $A=4$ has no solution.

Now, back to our partial fractions decomposition – if you used a constant numerator on the quadratic fraction, $\dfrac{A}{x^2+1}$ instead of $\dfrac{Ax+B}{x^2+1},$ then you’d end up with a system of 3 equations and 2 unknowns, which would likely end up being unsolvable.

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