Why Repeated Quadratics Need a Full Template
When a rational function has a repeated irreducible quadratic factor in its denominator, its partial-fraction form must account for that factor at every power. Writing only a term over the highest power generally leaves out part of the required structure.
The numerators also have a particular form. Because the factor is quadratic, the numerator above each power must have degree less than two: it can be linear, or it can be a constant as a special case. The resulting template is systematic, whether the quadratic occurs alone or alongside other factors.
Definition and Conditions
A real quadratic \(q(x)=ax^2+bx+c\), with \(a\ne 0\), is irreducible over the real numbers if it cannot be factored into real linear factors. Equivalently, its discriminant satisfies \(b^2-4ac<0\). For example, \(x^2+1\) is irreducible over the reals because its discriminant is \(-4\).
Suppose a rational function is proper and its denominator contains the factor \(q(x)^m\). “Proper” here means that the degree of its numerator is less than the degree of its denominator. The partial-fraction terms associated with this repeated quadratic have the form below.
Here \(m\) is a positive integer, and \(A_k\) and \(B_k\) are constants for each \(k=1,\ldots,m\). Some of these constants may be zero. A constant numerator is allowed: it is the case \(A_k=0\).
If the denominator has other factors as well, the displayed expression is the block for \(q(x)^m\); the other factors have their own partial-fraction terms. The template specifies the shape of the repeated-quadratic block. It does not determine the constants in that block.
Why the Template Has This Form
Two features need justification: why each numerator has degree less than two, and why the terms include every power from \(q(x)\) through \(q(x)^m\). The numerator-degree rule follows the standard partial-fraction rule for an irreducible quadratic: a numerator above a quadratic denominator factor must have degree less than the quadratic’s degree. The full sequence of powers can also be seen directly.
Polynomial division gives \(P(x)=q(x)S(x)+R(x)\), where \(\deg R<2\). Since \(\deg P<2m\) and \(\deg q=2\), the quotient satisfies \(\deg S<2m-2\), unless \(S\) is zero.
Divide the identity by \(q(x)^m\): \(\frac{P(x)}{q(x)^m}=\frac{R(x)}{q(x)^m}+\frac{S(x)}{q(x)^{m-1}}\). The remainder \(R(x)\) is linear or constant.
Because \(\deg S<2(m-1)\), the same division can be applied to \(S(x)/q(x)^{m-1}\). Repeating reduces the power by one each time and produces a numerator of degree less than two at each power, down to \(q(x)\).
This proves the stated form for a numerator of degree less than \(2m\). The order of the terms is sometimes written from the lowest power to the highest; either order describes the same template.
The coefficients in this form are also unique. To see why, suppose a sum of the stated terms is zero. Multiply the sum by \(q(x)^m\). The result is a polynomial identity whose term from the highest denominator power is \(A_mx+B_m\); every other term is divisible by \(q(x)\). Thus \(q(x)\) divides \(A_mx+B_m\). A nonzero polynomial of degree less than two cannot be divisible by a quadratic, so \(A_mx+B_m=0\). Removing that term and repeating the argument shows that every numerator is zero. This confirms that the form has no hidden freedom to replace the required terms with a different set of coefficients.
Worked Examples
Worked Example: A Quadratic Repeated Twice
Consider the rational function formed from the following two partial-fraction terms. Combine them and check the resulting numerator degree and template.
Set \(q(x)=x^2+1\). Its discriminant is \(0^2-4(1)(1)=-4<0\), so it is irreducible over the reals. The factor has multiplicity \(m=2\), and the full block must be
To check the given expression, put the two terms over the common denominator \(q(x)^2\):
Expand the numerator:
The combined fraction is therefore \(\frac{3x^3+4x^2+5x+3}{(x^2+1)^2}\). Its numerator has degree \(3<4\), as required for a proper fraction with a degree-four denominator. Checking against the template gives \(A_1=3\), \(B_1=4\), \(A_2=2\), and \(B_2=-1\); substituting these values reproduces the original two terms exactly.
Worked Example: A Shifted Irreducible Quadratic Repeated Three Times
Let \(q(x)=x^2-2x+5\), and combine the following terms:
The discriminant of \(q\) is \((-2)^2-4(1)(5)=-16<0\), so \(q\) is irreducible over the reals. Since its multiplicity is three, all three powers belong in the template.
First compute the square needed for a common denominator:
Putting the terms over \(q(x)^3\) gives the numerator \((x+2)q(x)^2+(2x-1)q(x)+(3-x)\). Expand the first product:
Expand the second product:
Now add both products and the remaining numerator \(3-x\):
Thus the combined fraction has denominator \(q(x)^3\) and numerator \(x^5-2x^4+8x^3+3x^2-4x+48\), whose degree is \(5<6\). Recombining that numerator as \((x+2)q(x)^2+(2x-1)q(x)+(3-x)\) verifies each of the three terms in the template.
Worked Example: A Different Irreducible Quadratic Repeated Three Times
Consider another block, this time using \(q(x)=x^2+9\):
The discriminant of \(x^2+9\) is \(0^2-4(1)(9)=-36<0\), so this is an irreducible quadratic. The multiplicity is three, so the template includes a linear numerator over each of \(q\), \(q^2\), and \(q^3\).
For a common denominator, compute \(q(x)^2=x^4+18x^2+81\). The numerator over \(q(x)^3\) is \((x+1)q(x)^2+(2x-3)q(x)+(1-x)\). The products are
Adding these expressions and \(1-x\) gives
The numerator degree is \(5<6\), and recombining it using the displayed products recovers the three original terms. This checks both the algebra and the required range of powers.
A Common Mistake: Leaving Out Lower Powers
A frequent error is to put a single linear numerator over the highest power and omit the lower powers. For example, the expression \(\frac{q(x)+1}{q(x)^2}\) cannot in general be represented by only \(\frac{Ax+B}{q(x)^2}\). If \(q(x)=x^2+1\), the numerator in the first expression is \(x^2+2\), which is quadratic, not linear.
The correction is to allow both powers. In this particular case, direct cancellation shows exactly what is missing:
With \(q(x)=x^2+1\), combining the right-hand side gives \(\frac{x^2+2}{(x^2+1)^2}\), which is the left-hand side. The lower-power term is essential to the full template, even though a particular rational function may have zero coefficients for some of its other terms.
What the Template Does—and Does Not—Tell You
The template determines which kinds of terms must be available: every successive power of the irreducible quadratic, with a numerator of degree at most one. It does not determine the values of \(A_k\) and \(B_k\). Those constants depend on the particular rational function and must be found from its numerator and denominator.
A useful check is to count the available constants. For multiplicity \(m\), there are two constants at each of \(m\) powers, for a total of \(2m\) constants. This agrees with the fact that a proper numerator over \(q(x)^m\) can have degree at most \(2m-1\), and therefore has \(2m\) polynomial coefficients available. This count is a structural check, not a way to determine the coefficients.
Writing the complete template matters because later calculations rely on having all the necessary terms available. A missing power or an overly restricted numerator can prevent a correct decomposition, even when the algebra that follows is performed accurately.
Check Your Understanding
Use the template and its conditions to answer the following questions.
- For an irreducible quadratic \(q(x)\) of multiplicity four, write the full repeated-quadratic block.
- Why must the numerator above each power of \(q(x)\) have degree less than two?
- What discriminant condition makes \(ax^2+bx+c\), with \(a\ne0\), irreducible over the reals?
- How many unknown constants appear in the repeated-quadratic block when the multiplicity is \(m\)?
- What structural information is missing if a proposed template includes only a term over \(q(x)^m\)?
