# Why does Partial Fraction Decomposition Result in Multiples of the Decomposed Fraction?

Let's say I have the rational function $\dfrac{x^3}{x^2 + x - 6}$. I use polynomial long division to get $x^3 = (x - 1)(x^2 + x - 6) + 7x -6 \implies \dfrac{x^3}{(x - 2)(x + 3)} = \dfrac{x^3}{x^2 + x - 6} = x - 1 + \dfrac{7x - 6}{x^2 + x - 6} = x - 1 + \dfrac{7x - 6}{(x - 2)(x + 3)}$.

$\dfrac{A}{x - 2} + \dfrac{B}{x + 3} = \dfrac{A(x + 3) + B(x - 2)}{(x - 2)(x + 3)} = \dfrac{Ax + 3A + Bx - 2B}{(x - 2)(x + 3)}$

$\therefore \dfrac{7x - 6}{(x - 2)(x + 3)} = \dfrac{Ax + 3A + Bx - 2B}{(x - 2)(x + 3)} \implies 7x - 6 = Ax + 3A + Bx - 2B$

$\implies 7x - 6 = (A + B)x + 3A - 2B$

$\therefore 7x = (A + B)x$ and $-6 = 3A - 2B$.

After solving this system of two linear equations, we get $B = \dfrac{27}{5}$ and $A = \dfrac{8}{5}$.

$\dfrac{7x - 6}{(x - 2)(x + 3)} = \dfrac{A}{x - 2} + \dfrac{B}{x + 3}$

$\therefore \dfrac{7x - 6}{(x - 2)(x + 3)} = \dfrac{8}{5x - 10} + \dfrac{27}{5x + 15}$

But when we expand $\dfrac{8}{5x - 10} + \dfrac{27}{5x + 15}$, we get $\dfrac{5(7x - 6)}{(x - 2)(x + 3)} \not = \dfrac{7x - 6}{(x - 2)(x + 3)}$.

Unless I'm consistently making the same error, I've noticed this phenomenon to be the same for other partial fraction decomposition problems: The result of the partial fraction decomposition is a constant multiple of the fraction that we were aiming to get.

I am confused as to why this happens. In doing the operations of partial fraction decomposition (as seen above), isn't our goal to get $\dfrac{7x - 6}{(x - 2)(x + 3)} = \dfrac{A}{x - 2} + \dfrac{B}{x + 3}$? Shouldn't our calculations (as seen above) generate exactly the result $\dfrac{7x - 6}{(x - 2)(x + 3)}$ rather than some multiple of it? What is the reasoning behind this? What am I misunderstanding?

I would greatly appreciate it if people could please take the time to explain this.

You made an arithmetic mistake; you lost track of the factor of $5$ in your common denominator:
$$\dfrac{8}{5(x-2)} + \dfrac{27}{5(x+3)} = \dfrac{8(x+3) + 27(x-2)}{\color{red}{\mathbf{5}}(x - 2)(x + 3)} = \dfrac{5(7x - 6)}{\color{red}{\mathbf{5}}(x - 2)(x + 3)}$$
• $x - 1 + \dfrac{8}{5x - 10} + \dfrac{27}{5x + 15} = x - 1 + \dfrac{40x + 120 + 135x - 270}{(5x - 10)(5x + 15)} = x - 1 + \dfrac{175x - 150}{5(x - 2)(x + 3)} = x - 1 + \dfrac{25(7x - 6)}{5(x - 2)(x + 3)} = x - 1 + \dfrac{5(7x - 6)}{(x - 2)(x + 3)}$ Can you see where I made an error? – The Pointer Mar 5 '17 at 3:23
• @ThePointer: If you calculate it that way, by multiplying the denominators rather than finding a least common multiple, the error is $(5x-10)(5x+15)=5(x-2) \cdot 5(x+3) = 25 (x-2)(x+3)$ – Hurkyl Mar 5 '17 at 3:27