Partial Fractions - Irreducible Quadratics

Partial fraction decomposition is a technique used to write a rational function as the sum of simpler rational expressions. A partial fraction has irreducible quadratic factors when one of the denominator factors is a quadratic with irrational or complex roots:

\[\frac{1}{x^3+x} \implies \frac{1}{x(x^2+1)} \implies \frac{1}{x}-\frac{x}{x^2+1}.\]

The process for irreducible quadratic factors is slightly different than the process for linear factors and repeated factors.

The partial fraction decomposition form for irreducible quadratics gives rational expressions with linear (not constant) numerators.

Partial Fraction Decomposition Form for Irreducible Quadratics:

• A denominator factor is irreducible if it has complex or irrational roots.

• For each linear non-repeated factor in the denominator, follow the process for linear factors.

• For each repeated factor in the denominator, follow the process for repeated factors.

• For each irreducible quadratic factor, write a rational expression with that factor as the denominator and a linear binomial as the numerator.

Once the form is down, you can follow the same process as with linear factors to solve for the coefficients.

Find the partial fraction decomposition of the following rational expression:

\[\frac{x^2+8}{x^3+8}.\]

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The denominator can be factored as a sum of cubes:

\[x^3+8=(x+2)\big(x^2-2x+4\big).\]

The quadratic factor has complex roots, so it cannot be factored any further. In the partial fraction decomposition, the \(x+2\) denominator will have a constant numerator, and the \(x^2-2x+4\) denominator will have a linear binomial numerator:

\[\frac{x^2+8}{x^3+8}=\frac{A}{x+2}+\frac{Bx+C}{x^2-2x+4}.\]

To solve for the coefficients, combine the fractions on the right hand side of the equation:

\[\frac{x^2+8}{x^3+8}=\frac{A(x^2-2x+4)+(Bx+C)(x+2)}{(x+2)(x^2-2x+4)}.\]

Since the denominators are equal, the numerators must also be equal:

\[x^2+8=A(x^2-2x+4)+(Bx+C)(x+2).\]

Setting \(x=-2\) in this equation gives \(A=1.\) Group the remaining terms by degree:

\[\begin{align} x^2+8 &= 1(x^2-2x+4)+(Bx+C)(x+2) \\ \\ &= x^2-2x+4+Bx^2+2Bx+Cx+2C \\ \\ &= \left(B+1\right)x^2+\left(2B+C-2\right)x+\left(2C+4\right). \end{align}\]

This gives the system of equations

\[\begin{align} 1 &= B+1 \\ 0 &= 2B+C-2 \\ 8 &= 2C+4. \end{align}\]

Solving this system gives \(B=0\) and \(C=2\). Then the partial fraction decomposition is

\[\frac{x^2+8}{x^3+8}=\frac{1}{x+2}+\frac{2}{x^2-2x+4}.\ _\square\]