9. Partial Fractions

a2. General Decompositions

c. Non-Repeated Linear Factors

Find the partial fraction decomposition for \(\dfrac{x^2-11x+12}{x^3-5x^2+6x}\).

Since the degree of the numerator is less than the degree of the denominator, there is no polynomial part. We first factor the denominator: \[ x^3-5x^2+6x=x(x-2)(x-3) \] Then we write the general partial fraction expansion by putting a constant over each linear factor: \[ \dfrac{x^2-11x+12}{x(x-2)(x-3)}=\dfrac{A}{x}+\dfrac{B}{x-2}+\dfrac{C}{x-3} \] We must find \(A\), \(B\) and \(C\) so this is true for all \(x\). To do this we multiply through by the denominator: \[ x^2-11x+12=A(x-2)(x-3)+Bx(x-3)+Cx(x-2) \]

The naive, brute force way to solve this is to expand the right hand side, collect terms, equate coefficients of each power of \(x\) and solve the \(3\) equations for \(3\) unknowns. Show me how.

However: Don't simplify! Don't expand! Since this equation must be true for all \(x\), an easier way to solve is to plug in several values for \(x\). The obvious values are \(x=0,2,3\) which are the zeros of the original denominator \(x^3-5x^2+6x=x(x-2)(x-3)\). These give:

• \(x=0\): \[\begin{aligned} (0)^2-11(0)+12&=A(-2)(-3)+B(0)+C(0) \\ &\Longrightarrow \qquad 12=6A \qquad \Longrightarrow \qquad A=2 \end{aligned}\]
• \(x=2\): \[\begin{aligned} (2)^2-11(2)+12&=A(0)+B(2)(2-3)+C(0) \\ &\Longrightarrow \qquad -6=-2B \qquad \Longrightarrow \qquad B=3 \end{aligned}\]
• \(x=3\): \[\begin{aligned} (3)^2-11(3)+12&=A(0)+B(0)+C(3)(3-2) \\ &\Longrightarrow \qquad -12=3C \qquad \Longrightarrow \qquad C=-4 \end{aligned}\]

This immediately gives the coefficients \(A\), \(B\) and \(C\). However, this trick only works easily for non-repeating linear factors. Plugging the coefficients back into the partial fraction expansion, we conclude \[ \dfrac{x^2-11x+12}{x(x-2)(x-3)}=\dfrac{2}{x}+\dfrac{3}{x-2}-\dfrac{4}{x-3} \]

We check this by adding up the right hand side: \[\begin{aligned} \dfrac{2}{x}+\dfrac{3}{x-2}-\dfrac{4}{x-3} &=\dfrac{2(x-2)(x-3)+3x(x-3)-4x(x-2)}{x(x-2)(x-3)} \\ &=\dfrac{x^2-11x+12}{x(x-2)(x-3)} \end{aligned}\]