Coefficient Matching For Distinct Linear Factors

Why Match Coefficients?

Partial fractions rewrite a rational function as a sum of simpler rational functions. When the denominator has distinct linear factors, each fraction in the standard partial-fraction form has a constant numerator. The constants can be found by clearing denominators and comparing coefficients of powers of \(x\).

For a proper rational function whose denominator is a product of \(n\) distinct linear factors, root substitution can determine the \(n\) fraction constants directly. Coefficient matching is still useful: it gives a systematic way to solve for them and check the result. It is especially important when the full decomposition also has a polynomial part, because evaluating at the denominator's roots does not determine that part.

From a Rational Equation to a Polynomial Identity

Suppose the denominator factors into distinct linear factors, and consider a proper rational function with three such factors. Its proposed form is

where \(P(x)\) is a polynomial of degree less than \(3\), and \(A,B,C\) are constants to be found. For \(x\ne a,b,c\), multiply both sides by \((x-a)(x-b)(x-c)\). The result is

Both sides are now polynomials. Expand the products, collect like powers, and equate the coefficients. For instance, if each side has degree at most \(2\), compare the coefficients of \(x^2\), \(x\), and \(1\). This gives three equations for the three unknowns.

Why is this comparison valid? The cleared equation holds for every \(x\) except possibly the three roots of the original denominator, so the difference between its two sides is a polynomial that vanishes at infinitely many real numbers. A nonzero polynomial of degree \(d\) has at most \(d\) roots: if \(r\) is a root, the factor theorem writes the polynomial as \((x-r)\) times a polynomial of degree one less, and this argument can be repeated. Therefore the difference polynomial must be the zero polynomial. Its coefficients are all zero, which is exactly why like-power coefficients must agree.

A Reliable Coefficient-Matching Procedure

Worked Examples

When Root Substitution Is Not Enough

For a proper fraction with only distinct linear factors, setting \(x\) equal to each denominator root isolates the corresponding fraction constant. That is the basis of the cover-up method. But if the function is improper, its decomposition also contains a polynomial part. At the roots, that part is multiplied by the denominator and disappears, so root substitution cannot determine its coefficients. Matching all powers of \(x\) supplies the additional equations.

A Common Coefficient-Matching Mistake

When expanding the cleared identity, it is easy to lose a minus sign in a constant term. For example, in the second worked example, the product \(B(x-2)\) is \(Bx-2B\), not \(Bx+2B\). The incorrect expansion would give

and would lead to the wrong constant equation \(A+2B=-5\). The correction is to distribute \(B\) to both terms, including the negative term: \(B(x-2)=Bx-2B\). Thus the correct expansion is \((A+B)x+(A-2B)\), and the correct equations give \(A=3\), \(B=4\). Recombining the fractions is a useful check because it exposes sign errors in the numerator.

How Coefficient Matching and Root Substitution Work Together

These methods are complementary, but they do not provide exactly the same information in every setup. In the proper three-factor example, matching coefficients produced a system for the three fraction constants. Root substitution could also find those constants directly. In the improper example, root substitution could find the two fraction constants, but only coefficient matching determined the polynomial part.

Whenever the constants have been found, check the decomposition by combining its terms over the original denominator and simplifying the numerator. This verifies that the proposed expression reproduces the rational function wherever its denominator is nonzero.