Irreducible Quadratic Factors

A quadratic factor $ax^2+bx+c$ of the denominator is called irreducible over the reals when it has no real linear factors, i.e. $b^2-4ac<0$. Since it has no real root to substitute, it cannot be assigned a constant numerator the way a linear factor is; instead the correct form is a linear numerator over it, $\dfrac{Ax+B}{ax^2+bx+c}$, and if the quadratic factor is itself repeated $n$ times, the full chain $\displaystyle\sum_{i=1}^n\dfrac{A_ix+B_i}{(ax^2+bx+c)^i}$ is needed, exactly mirroring the repeated-linear-factor pattern of §7.2.

When a fraction's denominator mixes a linear factor with an irreducible quadratic factor, the linear factor's constant is still found fastest by substituting its root (which kills the quadratic-factor term, since that root is not a root of the quadratic); the quadratic factor's two constants are then found by comparing coefficients of matching powers of $x$ in the cleared identity. For example, resolving $\dfrac{x+3}{(x-1)(x^2+1)}=\dfrac A{x-1}+\dfrac{Bx+C}{x^2+1}$: clearing denominators gives $x+3=A(x^2+1)+(Bx+C)(x-1)$; substituting $x=1$ gives $A=2$ at once; comparing the coefficient of $x^2$ (zero on the left) gives $B=-A=-2$; and comparing the constant term gives $C=A-3=-1$. …