# Partial Fraction Expansion: Irreducible Quadratic Factors

• PRACTICE (online exercises and printable worksheets)

Partial Fraction Expansion (PFE) renames a fraction of polynomials (i.e., a rational function), using smaller, simpler ‘pieces’. This is the third of three sections covering PFE:

• The first section introduces PFE, reviews all needed concepts, and presents a simple example (distinct linear factors).
• The second section gives the complete ‘how-to’ of PFE and a detailed example (repeated linear factor).
• This current section presents two detailed examples, both of which illustrate how to handle irreducible quadratic factors:

The steps indicated in the examples below follow the summary of the previous section.

## Example: PFE with an initial long division, factoring a difference of cubes, and distinct irreducible quadratic factor

Find the partial fraction expansion of   $\displaystyle \frac{2x^4 + 4x^2 - 5x + 2}{x^3 - 1} \,$.

• STEP 1 (check degree of numerator):
The degree of the numerator (4) is not less than the degree of the denominator (3).
Therefore, an initial long division is required, which yields: $$\frac{2x^4 + 4x^2 - 5x + 2}{x^3 - 1} = \color{red}{2x} + \color{green}{\frac{4x^2-3x+2}{x^3-1}}$$ In steps 2–4 below, PFE is applied to the remainder term, $\displaystyle\,\color{green}{\frac{4x^2-3x+2}{x^3-1}}\,$.
In step 5, we'll need to be sure not to forget the $\,\color{red}{2x}\,$!
• STEP 2 (factor denominator):
The denominator, $\,x^3 - 1\,$, must be factored into linear factors and irreducible quadratics.
The difference of cubes formula comes to the rescue: \begin{alignat}{2} A^3 - B^3 &= (A - B)(A^2 + AB + B^2) &\qquad&\text{(the difference of cubes formula)}\cr\cr x^3 - 1^3 &= (x - 1)(x^2 + x\cdot 1 + 1^2)&&\text{(apply the formula with \,A = x\, and \,B = 1\,)}\cr\cr x^3 - 1 &= (x-1)(x^2 + x + 1)&&\text{(simplify)} \end{alignat} The factor $\,x-1\,$ is a distinct linear factor.
The factor $\,x^2 + x + 1\,$ is a distinct irreducible quadratic factor, since its discriminant is negative: $$b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3 < 0$$
• STEP 3 (form of PFE, with unknown constants):
The distinct linear factor, $\,x-1\,$, gives rise to a single term in the PFE, with a single unknown: say, $\,\displaystyle\frac{A}{x-1}\,$.
The distinct irreducible quadratic factor, $\,x^2 + x +1\,$, gives rise to a single term in the PFE, with two unknowns: say, $\,\displaystyle\frac{Bx+C}{x^2+x+1}\,$.
Of course, names other than $\,A\,$, $\,B\,$, and $\,C\,$ could be used for the unknown constants.

The form of the PFE for $\,\displaystyle \frac{4x^2-3x + 2}{x^3-1}\,$ is therefore: $$\frac{4x^2 - 3x + 2}{(x-1)(x^2+x+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \qquad (\dagger)$$
• STEP 4 (clear fractions; solve for unknown constants):
Clear fractions in ($\,\dagger\,$) by multiplying both sides of the equation by $\,(x-1)(x^2+x+1)\,$.
This gives: $$\begin{gather} \frac{4x^2 - 3x + 2}{\cancel{(x-1)(x^2+x+1)}}\cdot \cancel{(x-1)(x^2+x+1)} = \frac{A}{\cancel{x-1}}\cdot \cancel{(x-1)}(x^2+x+1) + \frac{Bx+C}{\cancel{x^2+x+1}}\cdot (x-1)\cancel{(x^2+x+1)}\cr\cr 4x^2 - 3x + 2 = A(x^2 + x + 1) + (Bx + C)(x-1) \qquad (\ddagger) \end{gather}$$ It's always easy to solve for an unknown corresponding to a distinct linear factor.
Always get these easy unknowns first!
In this example, the distinct linear factor $\,x-1\,$ corresponds to the term $\,\displaystyle\frac{A}{x-1}\,$ with unknown $\,A\,$.
So, $\,A\,$ is easy to find: let $\,x = 1\,$ in ($\,\ddagger\,$), since this makes $\,x-1\,$ equal to zero: $$\begin{gather} 4\cdot 1^2 - 3\cdot 1 + 2 = A(1^2 + 1 + 1) + 0\cr 3 = A(3)\cr A = 1 \end{gather}$$ It now remains to solve for $\,B\,$ and $\,C\,$.
In what follows, it's important that when you look at $\,(Bx+C)(x-1)\,$, you actually ‘see’ (in your mind's eye) the multiplied-out version: $$(Bx+C)(x-1)\ \ =\ \ Bx^2 + Cx - Bx - C \ \ =\ \ Bx^2 + (C-B)x - C$$ (If necessary, multiply it out on paper.)
Thus, $\,B\,$ appears in the coefficients of both $\,x^2\,$ and $\,x\,$ terms.
Also, $\,C\,$ appears in the coefficient of an $\,x\,$ term, and in a constant term.

There are choices for how to solve for $\,B\,$ and $\,C\,$, and some are much better than others!
A few different approaches are shown in the table below—there are many more.
It's your job to discover the easy ways!
Definitely try to avoid approaches that require solving a system of equations, because these are more error-prone.

 The equation that the unknowns ($\,A\,$, $\,B\,$ and $\,C\,$) must satisy is repeated here for your convenience: $$4x^2 - 3x + 2 = A(x^2 + x + 1) + (Bx + C)(x-1) \qquad (\ddagger)$$ Here's the same equation with the right-hand side in standard form, so it's easier to compare the coefficients of each term type. In practice, however, try to ‘see’ all this from ($\,\ddagger\,$), without multiplying it out. $$4x^2 - 3x + 2 = x^2(A + B) + x(A + C - B) + (A - C)$$ Recall that it has already been determined that $\,A = 1\,$. (good) Equate $\,x^2\,$ coefficients in ($\,\ddagger\,$) to solve for $\,B\,$. $$\begin{gather} 4 = A + B\cr 4 = 1 + B\cr \color{green}{B = 3} \end{gather}$$ Equate constant terms in ($\,\ddagger\,$) to solve for $\,C\,$: $$\begin{gather} 2 = A - C\cr 2 = 1 - C\cr \color{red}{C = -1} \end{gather}$$ (okay) Equate constant terms in ($\,\ddagger\,$) to solve for $\,C\,$: $$\begin{gather} 2 = A - C\cr 2 = 1 - C\cr \color{red}{C = -1} \end{gather}$$ Equate $\,x\,$ coefficients in ($\,\ddagger\,$) to solve for $\,B\,$. $$\begin{gather} -3 = A + C - B\cr -3 = 1 - 1 - B\cr -3 = -B\cr \color{green}{B = 3} \end{gather}$$ (good) Let $\,x = 0\,$ in ($\,\ddagger\,$): $$\begin{gather} 2 = A - C\cr 2 = 1 - C\cr \color{red}{C = -1} \end{gather}$$ Equate $\,x^2\,$ coefficients in ($\,\ddagger\,$) to solve for $\,B\,$. $$\begin{gather} 4 = A + B\cr 4 = 1 + B\cr \color{green}{B = 3} \end{gather}$$ (horrible) Equate $\,x\,$ coefficients in ($\,\ddagger\,$): $$\begin{gather} -3 = A + C - B\cr -3 = 1 + C - B\cr -4 = C - B \end{gather}$$ Let $\,x = 2\,$ (arbitrarily chosen) in ($\,\ddagger\,$): $$\begin{gather} 4\cdot 2^2 - 3\cdot 2 + 2 = A(2^2 + 2 + 1) + (B\cdot 2 + C)(2-1)\cr 12 = 7A + 2B + C\cr 12 = 7 + 2B + C\cr 5 = 2B + C \end{gather}$$ Solve the system of equations: $$\begin{gather} -4 = C - B\cr 5 = 2B + C \end{gather}$$ Using substitution, the first equation gives $\,C = -4 + B\,$. Substitution into the second equation gives $\,5 = 2B - 4 + B\,$. Solving for $\,B\,$ gives $\,9 = 3B\,$, so $\,\color{green}{B = 3}\,$. Then, the first equation gives $\,-4 = C - 3\,$, so $\,\color{red}{C = -1}\,$.

Of course, all correct approaches will lead to the same results.
(The author prefers the first column.)

Thus, we have $$\frac{4x^2-3x+2}{x^3-1} = \frac{1}{x-1} + \frac{3x-1}{x^2 + x + 1}$$
• STEP 5 (summarize results; spot-check):
Combine the results from step 1 and step 4 to get: $$\frac{2x^4 + 4x^2 - 5x + 2}{x^3 - 1} = 2x + \frac{1}{x-1} + \frac{3x-1}{x^2 + x + 1}$$ Spot-check with $\,x = 0\,$ to gain confidence in the result: $$\begin{gather} \frac{2}{-1} \overset{?}{=} \frac{1}{-1} + \frac{-1}{1}\cr\cr -2 = -2\qquad \text{Hooray!} \end{gather}$$

## Example: PFE with a Repeated Irreducible Quadratic Factor

Find the partial fraction expansion of $\displaystyle\,\frac{3x^2 - 5}{x^4 + 6x^2 + 9}\,$.

(The following solution is much more compact than the prior example.)

• STEP 1 (check degree of numerator):
The degree of the numerator (2) is less than the degree of the denominator (4). Check!
• STEP 2 (factor denominator):
The denominator is a ‘fake quadratic’ (also called a ‘pseudo-quadratic’): $$x^4 + 6x^2 +9 \ \ =\ \ (x^2)^2 + 6(x^2) + 9 \ \ =\ \ (x^2 + 3)^2$$ The quadratic $\,x^2 + 3\,$ is irreducible.
Why? Use any of the following equivalent characterizations:
• the discriminant is negative: $\,b^2 - 4ac = 0^2 - 4(1)(3) = -12 < 0$
• there are no real number solutions to $\,x^2 + 3 = 0\,$
• the graph of $\,y = x^2 + 3\,$ never crosses the $x$-axis
Thus, $\,(x^2 + 3)^2\,$ is a repeated irreducible quadratic factor.
• STEP 3 (form of PFE, with unknown constants):
The repeated irreducible quadratic factor, $\,(x^2 + 3)^\color{red}{2}\,$, gives rise to two terms in the PFE, each with two unknowns: $$\frac{3x^2 - 5}{x^4 + 6x^2 + 9} \ \ =\ \ \frac{3x^2 - 5}{(x^2 + 3)^2} \ \ =\ \ \frac{Ax+B}{x^2 + 3} + \frac{Cx+D}{(x^2 + 3)^2} \qquad (\dagger)$$
• STEP 4 (clear fractions; solve for unknown constants):
Clear fractions in ($\,\dagger\,$) to get: $$3x^2 - 5 = (Ax + B)(x^2 + 3) + Cx + D \qquad (\ddagger)$$ Equate $\,x^3\,$ coefficients in ($\,\ddagger\,$) to get $\,A = 0\,$.
Equate $\,x^2\,$ coefficients in ($\,\ddagger\,$) to get $\,B = 3\,$.
Equate $\,x\,$ coefficients in ($\,\ddagger\,$) to get: $$\begin{gather} 0 = 3A + C\cr 0 = 3(0) + C\cr C = 0 \end{gather}$$ Equate constants in ($\,\ddagger\,$) to get: $$\begin{gather} -5 = 3B + D\cr -5 = 3(3) + D\cr D = -14 \end{gather}$$
• STEP 5 (summarize results; spot-check):
Thus, $$\frac{3x^2 - 5}{x^4 + 6x^2 + 9} \ \ =\ \ \frac{3x^2 - 5}{(x^2 + 3)^2} \ \ =\ \ \frac{3}{x^2 + 3} - \frac{14}{(x^2 + 3)^2}$$ Spot-check with $\,x = 0\,$ to gain confidence in the result: $$\begin{gather} \frac{-5}{9} \overset{?}{=} \frac{3}{3} - \frac{14}{9}\cr\cr -\frac 59 = -\frac 59\qquad \text{Hooray!} \end{gather}$$ Or, zip up to WolframAlpha for the ultimate confidence booster:

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Introduction to Conic Sections

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