From a Partial Fraction Form to Its Coefficients
A partial fraction template specifies the shape of the terms in a decomposition, but it does not specify their coefficients. For an irreducible quadratic factor, the numerator must be linear: it has the form \(Ax+B\), with constants \(A\) and \(B\) to be determined.
The practical task is to turn the proposed decomposition into equations for those unknown constants. The reliable method is to multiply through by the common denominator. This produces a polynomial identity, which can be solved by matching coefficients or by evaluating it at judiciously selected values.
Because an irreducible quadratic has no real zeros, there is no real value of \(x\) to substitute that makes its factor vanish. This is why the cover-up method for distinct real linear factors does not directly give the coefficients here. Clearing denominators works without needing any real roots.
Why Clearing Denominators and Matching Coefficients Works
Suppose a rational function has already been written in the appropriate form, with distinct irreducible quadratic factors \(Q_1(x),\ldots,Q_m(x)\). Its unknown coefficients occur in numerators such as \(A_i x+B_i\). On any \(x\) where the denominators are nonzero, the proposed decomposition is an equality of rational expressions. Multiplying both sides by \(Q_1(x)\cdots Q_m(x)\) gives a polynomial identity.
After expanding, collect terms by powers of \(x\). Two polynomials are equal precisely when their corresponding coefficients are equal. In particular, if their difference is the zero polynomial, differentiating that difference \(j\) times and setting \(x=0\) gives \(j!\) times its coefficient of \(x^j\), which must be zero. Thus every coefficient in the difference is zero.
For two distinct quadratic factors, the standard form is \(\frac{Ax+B}{Q_1(x)}+\frac{Cx+D}{Q_2(x)}\). Multiplying through by \(Q_1(x)Q_2(x)\) gives \(P(x)=(Ax+B)Q_2(x)+(Cx+D)Q_1(x)\), where \(P(x)\) is the numerator of the original proper rational function. Expanding this equality creates a linear system in \(A,B,C,D\).
Worked Example: Matching Coefficients for Two Quadratics
Worked Example: Matching Coefficients for Two Quadratics
Find the constants in the proposed decomposition \(\frac{3x^3+x^2+6x+7}{(x^2+1)(x^2+4)} =\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4}\).
First multiply through by \((x^2+1)(x^2+4)\):
Expand each product separately:
Adding these expressions and matching the coefficients of \(x^3,x^2,x,\) and \(1\) gives
Subtract the first equation from the third: \(3A=3\), so \(A=1\). Then \(C=3-A=2\). Subtract the second equation from the fourth: \(3B=6\), so \(B=2\). Then \(D=1-B=-1\).
Therefore the decomposition is
Check by recombining the right-hand side's numerators over the common denominator:
The numerator matches the original, so the decomposition is verified.
Using Selected Substitutions
Coefficient matching is systematic, but expanding every coefficient equation is not always the shortest route. After clearing denominators, one may instead evaluate the resulting polynomial identity at selected values of \(x\). Each value gives an equation in the unknown coefficients. Values such as \(0,1,-1\) can be useful because signs and powers simplify.
These substitutions do not need to make a quadratic factor zero. In fact, an irreducible quadratic has no real zero to use. A set of evaluations is useful when it produces enough independent equations to determine the unknowns. If the cleared identity has degree at most \(n\), agreement at \(n+1\) distinct values is enough to establish equality of the two polynomials: their difference has degree at most \(n\), and a nonzero polynomial of degree at most \(n\) cannot have more than \(n\) distinct roots.
Worked Example: Solving by Selected Values
Worked Example: Solving by Selected Values
Determine \(A,B,C,D\) in \(\frac{x^3+2x^2+3x+1}{(x^2+1)(x^2+2)} =\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2}\).
Clearing denominators gives the polynomial identity
Evaluate at \(x=0\):
At \(x=1\), the left side is \(1+2+3+1=7\), so
At \(x=-1\), the left side is \(-1+2-3+1=-1\), so
Adding the equations from \(x=1\) and \(x=-1\) cancels the terms in \(A,C\): \(6=6B+4D\), or \(3B+2D=3\). Combine this with \(2B+D=1\). Doubling the latter gives \(4B+2D=2\); subtracting \(3B+2D=3\) gives \(B=-1\), and then \(D=1-2B=3\).
Subtracting the \(x=-1\) equation from the \(x=1\) equation gives \(8=6A+4C\), or \(3A+2C=4\). Now evaluate at \(x=2\). The left side is \(8+8+6+1=23\), giving
Substitute \(B=-1\) and \(D=3\): \(23=6(2A-1)+5(2C+3)\), so \(12A+10C=14\), or \(6A+5C=7\). Subtract \(3A+2C=4\), doubled, namely \(6A+4C=8\). This gives \(C=-1\), and then \(3A-2=4\), so \(A=2\).
The resulting decomposition is
Check by recombining: \((2x-1)(x^2+2)=2x^3-x^2+4x-2\), and \((-x+3)(x^2+1)=-x^3+3x^2-x+3\). Their sum is \(x^3+2x^2+3x+1\), as required.
Separating Even and Odd Parts
When the quadratic factors involve \(x^2\) alone, it can be efficient to write \(y=x^2\) after clearing denominators. Terms involving \(x\) form the odd part of the identity, while terms without a factor of \(x\) form the even part. Each part can then be solved separately. This is still coefficient solving: the substitution reorganizes the equations rather than changing the required partial fraction form.
Worked Example: Three Quadratics and a Reduced System
Worked Example: Three Quadratics and a Reduced System
Find the coefficients in \(\frac{2x^4+28x^2+50}{(x^2+1)(x^2+4)(x^2+9)} =\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4}+\frac{Ex+F}{x^2+9}\).
Set \(y=x^2\) and clear the denominators:
The terms containing \(x\) must sum to zero, and the terms without \(x\) must equal \(2y^2+28y+50\). The odd part, after removing its common factor \(x\), is
This is a polynomial identity in \(y\): it holds for all \(y=x^2\geq 0\), so the polynomial must be identically zero. We can therefore evaluate this polynomial identity at convenient values, including negative \(y\). At \(y=-1\), only the term in \(A\) remains, and \(24A=0\), so \(A=0\). At \(y=-4\), only the term in \(C\) remains, and \(-15C=0\), so \(C=0\). At \(y=-9\), only the term in \(E\) remains, and \(40E=0\), so \(E=0\).
The even part is
At \(y=-1\), \(24B=2-28+50=24\), so \(B=1\). At \(y=-4\), \(-15D=32-112+50=-30\), so \(D=2\). At \(y=-9\), \(40F=162-252+50=-40\), so \(F=-1\).
Thus the decomposition is
Check by recombining the numerators with \(y=x^2\):
Replacing \(y\) by \(x^2\) recovers the original numerator, verifying the result.
A Common Expansion Error
A frequent mistake is to multiply a coefficient by the wrong power of \(x\). In the first example, an incorrect expansion might claim
The term \(4Ax\) comes from multiplying \(Ax\) by \(4\), not \(4Ax^2\). The correct expansion is
Using the incorrect version would assign the wrong powers to terms and therefore produce false coefficient equations. A dependable check is to distribute each product in two separate lines before collecting like powers. After solving, recombine the numerators over the common denominator; this catches errors in both expansion and algebra.
Choosing a Method and Checking the Result
Coefficient matching is dependable for any of these systems and is often clearest when there are only a few unknowns. Selected substitutions can reduce arithmetic when the cleared identity has a manageable degree and chosen values simplify the factors. A hybrid approach is also useful: expand enough to understand the structure, then use evaluations to obtain simpler equations.
In every case, substitutions are applied to the cleared polynomial identity, not used as a substitute for clearing denominators. Once the coefficients are found, recombine the proposed partial fractions. The resulting numerator must equal the original numerator, while the denominator remains the common product of the quadratic factors.
Check Your Understanding
Use the cleared polynomial identity as the starting point for each question.
- Why does an irreducible real quadratic have no real value of \(x\) that can be used to make its factor zero?
- After clearing denominators in \(\frac{P(x)}{(x^2+1)(x^2+4)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4}\), write the polynomial identity.
- What equations result from matching the coefficients of \(x^3\) and \(x^2\) in that identity?
- When can evaluating a cleared identity at selected values determine the polynomial equality?
- Why is recombining the numerators a useful final check?