9. Partial Fractions
a2. General Decompositions
b. Removing a Polynomial Part
Be sure to review Long Division of Polynomials.
Find the partial fraction decomposition for \(\dfrac{x^4-3x^2+5x-100}{ x^2+9}\).
Since the degree of the numerator is larger than the degree of the denominator, we first need to remove the polynomial part. We use long division of polynomials: \[\begin{aligned} &\quad x^2 \qquad\;-12\;+\;\dfrac{5x+8}{x^2+9} \\ x^2+9&\overline{\left)\;x^4+0\,x^3-3x^2+5x-100\;\right.} \\ & \;\;\,\underline{x^4\qquad\;\;\;+9x^2\qquad \qquad \quad } \\ & \qquad\qquad\,-12x^2+5x-100 \\ & \qquad\qquad\;\;\underline{-\,12x^2\qquad\;-108\qquad } \\ & \qquad\qquad\qquad\qquad\;\;5x\;+\;8 \end{aligned}\] Equivalently, \[ \dfrac{x^4-3x^2+5x-100}{x^2+9}=x^2-12+\dfrac{5x+8}{x^2+9} \] Since the denominator is an irreducible quadratic and the remaining numerator is linear, this is the partial fraction decomposition.
To check, we compute: \[\begin{aligned} (x^2+9)(x^2-12)+(5x+8) &=x^4-3x^2-108+5x+8 \\ &=x^4-3x^2+5x-100 \end{aligned}\]
Find the partial fraction expansion for \(\dfrac{x^3-2x^2+5x-4}{x-3}\)
\(\dfrac{x^3-2x^2+5x-4}{x-3}=x^2+x+8+\dfrac{20}{x-3}\)
By long division: \[\begin{aligned} &\quad x^2\;+\;x\;+\;8\;+\;\dfrac{20}{x-3} \\ x-3&\overline{\left)\;x^3-2x^2+5x-4\;\right.} \\ &\;\;\underline{\,x^3-3x^2\qquad\qquad\quad} \\ &\qquad\quad\;x^2+5x-4 \\ &\qquad\quad\underline{\,x^2-3x\qquad\;} \\ &\qquad\qquad\quad\;8x-\;4 \\ &\qquad\qquad\quad\;\underline{8x-24} \\ &\qquad\qquad\qquad\quad\;\;20 \end{aligned}\] Equivalently, \[ \dfrac{x^3-2x^2+5x-4}{x-3}=x^2+x+8+\dfrac{20}{x-3} \] Since the denominator is linear and the remaining numerator is constant, this is the partial fraction decomposition.
To check, we compute: \[\begin{aligned} (x-3)(x^2+x+8)+(20) &=x^3+x^2+8x-3x^2-3x-24+20 \\ &=x^3-2x^2+5x-4 \end{aligned}\]