Case 4 — g(x) Has a Repeated Irreducible Quadratic Factor

The rule. This case combines the ideas of Cases 2 and 3. If an irreducible quadratic $(ax^2+bx+c)$ occurs $n$ times in $g(x)$ (i.e. $(ax^2+bx+c)^n$ divides $g(x)$ but the next power doesn't), it contributes a whole chain of $n$ partial fractions, each with a linear numerator, stepping up through every power of the quadratic from $1$ to $n$:

where all the $A_i$'s and $B_i$'s are real constants to be determined (with at least one of $A_n, B_n$ nonzero). This is the most general of the four rules — Case 3 is just the special case $n=1$, in the same way Case 1 was the special case $n=1$ of Case 2.

The method. Clear denominators as usual to get a single polynomial identity, then equate coefficients of matching powers of $x$ on both sides. Because there are $2n$ unknown constants here, you'll generally need $2n$ independent equations — one for each power of $x$ from the highest down to the constant term — and then solve that system. There isn't a real root to substitute (the quadratic is irreducible), so equating coefficients is the standard route, though substituting a couple of convenient real values of $x$ (like $x=0$) can still simplify the arithmetic alongside it.

Worked example. Resolve $\dfrac{x^3+2x^2+5x+3}{(x^2+2)^2}$ into partial fractions.

First confirm $x^2+2$ is irreducible: it has the form $x^2+0x+2$, discriminant $=0-4(1)(2)=-8<0$ — irreducible, and it appears squared in the denominator, so $n=2$. The fraction is proper (numerator degree $3<$ denominator degree $4$). Write

Clear denominators by multiplying through by $(x^2+2)^2$:

Expand the right side: $(Ax+B)(x^2+2) = Ax^3 + 2Ax + Bx^2 + 2B$, so the whole right side is

Now equate coefficients with $x^3+2x^2+5x+3$ power by power:

• $x^3$: $A = 1$
• $x^2$: $B = 2$ …