Solving Coefficients by Polynomial Identity

Why Polynomial Identities Determine the Coefficients

Once the form of a partial-fraction decomposition is known, its unknown constants can be found by algebra. The essential move is to multiply both sides by the common denominator. The resulting equality is an equality between polynomials, so the coefficients of matching powers of \(x\) must agree.

This method applies to proper rational functions whose denominators have been factored into linear factors and irreducible quadratic factors. For repeated factors, the complete form includes a term for each power of the factor. The numerator over a linear factor is a constant; the numerator over an irreducible quadratic is linear. These bounded numerator forms are the starting point, not something to guess during the coefficient calculation.

For example, if the denominator contains \((x-a)^2(x+b)\), the two linear factors are distinct only when \(a+b\ne 0\). Under that condition, a complete form for a proper remainder is

If \(a+b=0\), the factors coincide instead: the denominator is \((x-a)^3\), so the corresponding form has terms for all three powers of that single factor. The distinction matters because the denominators in the proposed decomposition must match the actual factorization.

From Rational Equality to Coefficient Equations

Suppose a decomposition has been proposed and its denominator factors are all included. Multiply both sides by the common denominator. Each fraction becomes a polynomial term: its numerator is multiplied by the parts of the common denominator that were not under that fraction’s original denominator. Expand those terms, collect like powers of \(x\), and equate coefficients.

This is valid on every value of \(x\) where the original rational expressions are defined. After denominators have been cleared, the two sides are polynomials. If they agree for all such \(x\), they agree at infinitely many values. Two polynomials that agree at infinitely many values are identical, so their coefficients match. Conversely, if the coefficients match, the cleared polynomial identity holds, and dividing by the common denominator recovers the rational equality wherever that denominator is nonzero.

Worked Example: A Repeated Linear Factor

Why Matching Coefficients Works

The coefficient comparison is not an approximation or a convenient pattern. It follows from the meaning of polynomial equality. Here is the key fact and its justification.

To justify the direction used in coefficient solving, subtract one polynomial from the other. If the polynomials agree at infinitely many values, their difference has infinitely many roots. A nonzero polynomial of degree \(n\) has at most \(n\) roots, so the difference must be the zero polynomial. A polynomial is the zero polynomial exactly when all of its coefficients are zero. Therefore the coefficients in the original polynomials agree power by power. In the other direction, equal coefficients give equal sums of powers, so the polynomials are identical.

In a partial-fraction calculation, the cleared equality holds for every \(x\) in the domain of the original rational function. Only finitely many values are excluded by its denominator, leaving infinitely many values. The fact therefore applies. This explains why a coefficient system encodes the entire rational identity rather than merely checking it at a few chosen inputs.

Worked Example: A Repeated Irreducible Quadratic

Worked Example: Mixed Repeated Factors

A Common Error: Clearing the Denominator Incompletely

Each fraction must be multiplied by every factor of the common denominator that remains after canceling its own denominator. Omitting one of those factors changes the polynomial equation and can produce a false system.

For the first worked example, the term \(B/(x-1)^2\) must be multiplied by \(x+2\) when the common denominator is cleared. A mistaken calculation might write

The \(B\) term is wrong: its denominator is \((x-1)^2\), so multiplying it by the common denominator leaves the factor \(x+2\). The corrected identity is

With the incorrect expression, the constant \(B\) contributes no \(x\)-term; in the correct expression, \(B(x+2)=Bx+2B\) contributes to both the \(x\)-coefficient and the constant coefficient. Keeping the uncanceled factors visible until after clearing denominators is a reliable way to avoid this error.

Using the Method Reliably

Coefficient matching is especially useful when repeated factors create several unknowns and when the denominator contains both linear and quadratic factors. A sound calculation keeps the algebra organized: write the full cleared identity first, expand each product separately, then collect powers. Do not skip directly from the rational expression to a coefficient system, because that makes omitted factors and sign errors harder to spot.

After solving, check the result at the polynomial level by substituting the constants into the cleared identity and expanding. This confirms that the proposed numerators reproduce the original numerator. Matching coefficients establishes the identity; the explicit reconstruction is a practical way to catch arithmetic mistakes in the expansion or system-solving steps.