Why Coefficient Matching Works
When a quadratic factor is repeated in a denominator, its partial-fraction decomposition contains a term for each power of that factor. Finding the decomposition therefore means finding several unknown numerator coefficients at once. Clearing the denominator converts the rational equation into a polynomial equation, where the unknowns can be found by matching coefficients.
The key is to preserve every term while expanding. A missed cross term can produce plausible-looking but incorrect coefficients. In this tutorial, we will set up the equations, solve them in order, and verify each result by recombining the fractions.
For a proper rational function with denominator \(Q^m\), the decomposition form is
where “proper” means that the degree of \(P\) is less than the degree of the denominator. The coefficients \(A_i,B_i\) are the unknowns to determine. The method here is to multiply through by \(Q^m\), expand the resulting polynomial, and equate the coefficients of like powers of \(x\).
Clearing the Denominator for a Repeated Quadratic
For a squared quadratic, the form is
Multiplication by \(Q(x)^2\) gives a polynomial identity:
This identity is the working equation. Once both sides are written as sums of powers of \(x\), the coefficient of \(x^3\) on one side must equal the coefficient of \(x^3\) on the other, and likewise for every other power. This is valid because two polynomials are equal precisely when all their corresponding coefficients are equal.
For \(Q(x)=x^2+px+q\), expanding gives a useful general system:
Thus, if \(P(x)=u_3x^3+u_2x^2+u_1x+u_0\), matching coefficients produces
The equations are often easiest to solve from the highest power downward: first find \(A\), then \(B\), then \(C\), and finally \(D\). Each new equation uses coefficients already determined.
Worked Example: A Squared Quadratic with No Linear Term
Decompose
The quadratic \(x^2+4\) is irreducible over the reals, since its discriminant is \(0^2-4(1)(4)=-16<0\). Write the decomposition form:
Multiply by \((x^2+4)^2\):
Expand the product, keeping the contributions of \(Ax\) and \(B\) separate:
Therefore the right-hand side is
Match coefficients with \(x^3+2x^2+7x-3\):
Substitute the known values into the remaining equations:
The decomposition is
Check by recombining. The numerator on the right becomes
This is the original numerator, so the decomposition checks.
A Third-Power Quadratic Factor
With a cubic power of \(Q\), there are three numerator pairs. Clearing denominators gives
For \(Q(x)=x^2+1\), we have \(Q(x)^2=x^4+2x^2+1\). Expanding the right-hand side gives the coefficient equations
Here \([x^k]\) means “the coefficient of \(x^k\).” In practice, expand the products directly and read off these coefficients; the notation simply organizes the matching.
Worked Example: Solving Six Coefficients
Decompose
Use the three-term form and clear the denominator:
Expand each product:
After adding \(Ex+F\), the coefficients are \(A\) for \(x^5\), \(B\) for \(x^4\), \(2A+C\) for \(x^3\), \(2B+D\) for \(x^2\), \(A+C+E\) for \(x\), and \(B+D+F\) for the constant term. Matching them gives
Solve in order:
Therefore,
To check, recombine the numerators over the common denominator:
Adding these and \(2x+4\) gives
The recombined numerator matches the original exactly.
A Quadratic with a Linear Term
Coefficient matching works the same way when the irreducible quadratic is not centered at zero. The terms involving \(p\) in \(x^2+px+q\) must be included in the expansion; they affect more than one coefficient.
Worked Example: Matching a Non-Centered Quadratic
Decompose
The discriminant of \(x^2+2x+5\) is \(2^2-4(1)(5)=-16<0\), so it is irreducible over the reals. Set
Clearing denominators and expanding gives
Thus the coefficient equations are
Solve successively:
The result is
Check the numerator after recombination:
The numerator agrees with the one given, confirming the decomposition.
Common Pitfall: Losing a Cross Term
A frequent error is to expand a product such as \((Ax+B)(x^2+4)\) incompletely. For example, omitting the product \(Ax\cdot4\) leads to the incorrect expansion
This expression is missing \(4Ax\). In the first worked example, using the incorrect expansion would make the coefficient of \(x\) appear to be just \(C\), and would give \(C=7\). The correct expansion includes \(4Ax\), so the coefficient equation is \(4A+C=7\). Since \(A=1\), the correct value is \(C=3\).
The wrong value fails the recombination check: its numerator would have \(4x+7x=11x\), rather than the required \(7x\). Writing each product on its own line before collecting like powers helps prevent this mistake.
A Reliable Coefficient-Solving Procedure
For a factor \(Q^m\), include a linear numerator over each power \(Q,Q^2,\ldots,Q^m\).
Multiply both sides by \(Q^m\) and write the resulting polynomial identity.
Keep track of every contribution to each power of \(x\), including constant and cross terms.
Begin with the highest power, then substitute known coefficients into the remaining equations.
Multiply each numerator by the missing denominator factors and confirm that their sum is the original numerator.
This process matters because it turns a decomposition into a finite algebra problem. It also gives a direct way to detect errors: the proposed coefficients are correct only if recombination reproduces the original polynomial numerator exactly.
Check Your Understanding
Use coefficient matching to answer the following questions. Check each proposed decomposition by recombining its numerators.
- For a denominator \(Q(x)^2\), what is the degree of each numerator in the partial-fraction form, and why?
- After clearing denominators in \(\frac{P(x)}{Q(x)^2}=\frac{Ax+B}{Q(x)}+\frac{Cx+D}{Q(x)^2}\), what polynomial identity results?
- Set up the coefficient equations for \(\frac{3x^3+4x^2+9x+2}{(x^2+1)^2}\).
- For the quadratic \(x^2+2x+5\), what is its discriminant, and is it irreducible over the reals?
- Why is matching coefficients equivalent to checking that the cleared polynomial identity holds?
