Case 3 — g(x) Has a Non-Repeated Irreducible Quadratic Factor

The rule. Sometimes $g(x)$ doesn't factor completely into linear pieces — it has a genuinely irreducible quadratic factor $(ax^2+bx+c)$, $a \neq 0$ (recall: irreducible means $b^2-4ac<0$, so it has no real roots and can't be split further). If this quadratic factor appears only once in $g(x)$, it contributes one partial fraction of the form

Notice the numerator here is linear ($Ax+B$, not just a constant) — that's the key difference from the linear-factor case, and it reflects the fact that the denominator is one degree higher.

Why the numerator must be linear and not just a constant. If we tried a constant numerator over a quadratic denominator, we generally wouldn't have enough free constants to match an arbitrary linear-or-lower remainder that can appear on top; allowing $Ax+B$ gives exactly two unknowns per quadratic factor, matching the two conditions needed (Case 1's single linear factor needs one constant $A$ per one condition; a quadratic factor needs two).

The method. Since $(ax^2+bx+c)$ has no real root, you can't use the "kill the other terms" substitution trick on it directly — but you can still substitute the real linear factors' roots to find their constants, and then equate coefficients (matching powers of $x$) to pin down $A$ and $B$ for the quadratic piece.

Worked example. Resolve $\dfrac{3x^2-2x+2}{x^3-1}$ into partial fractions.

First factor the denominator: $x^3-1 = (x-1)(x^2+x+1)$. Check that $x^2+x+1$ is irreducible: discriminant $= 1^2 - 4(1)(1) = -3 < 0$, so yes, it has no real roots and can't be factored further over the reals.

Since $(x-1)$ is a non-repeated linear factor and $(x^2+x+1)$ is a non-repeated irreducible quadratic, write

Clear denominators:

Put $x=1$ in (1) — the second term vanishes: $3-2+2 = A(1+1+1) \Rightarrow 3 = 3A \Rightarrow A=1$. …