Why Distinct Linear Factors Give Unique Decompositions

Why the Coefficients Are More Than a Calculation

A partial-fraction decomposition turns one rational function into a sum of simpler fractions. For a denominator made from distinct linear factors, coefficient methods can find the terms. But there is a deeper question: why is there a decomposition of the required form, and why can there be only one?

The key is to clear the denominators first. The resulting equation is a polynomial identity, so we can evaluate it at the roots of the original denominator. Each root isolates one coefficient. This gives a formula for the coefficients and, with a polynomial root argument, proves that the resulting decomposition is valid and unique.

Setting and Notation

Let \(r_1,\ldots,r_n\) be distinct real numbers, and let \(c\) be a nonzero real constant. Consider a proper rational function whose denominator has these distinct linear factors:

The condition \(\deg P<n\) is the properness condition here: the numerator has lower degree than the denominator. The distinctness condition means \(r_i\ne r_j\) whenever \(i\ne j\). In particular, for any fixed \(i\), every factor \(r_i-r_j\) with \(j\ne i\) is nonzero.

Multiplying both sides by \(c\prod_{i=1}^{n}(x-r_i)\) turns this into a polynomial identity. If we define \(B_i=cA_i\), the identity becomes

This cleared-denominator form is the central tool. At \(x=r_k\), every term on the right vanishes except the term with \(i=k\). That observation determines the only possible value of each \(B_k\), and therefore of each \(A_k\).

The Coefficient Formula

Set \(x=r_k\) in the cleared identity. In the term indexed by \(k\), the product contains every factor except \(x-r_k\). All other terms still contain \(x-r_k\), so they evaluate to zero. Thus

The factors in this product are nonzero because the roots are distinct. We can divide by them and by \(c\), obtaining the only possible coefficient:

This argument has so far shown that if a decomposition exists, its coefficients must have these values. To establish existence as well, we must show that choosing these values really produces an identity.

Why the Formula Gives Existence and Uniqueness

For existence, use the formula to define \(A_1,\ldots,A_n\), and put \(B_i=cA_i\). Consider the polynomial

Each product in this sum has degree \(n-1\), so \(\deg S\le n-1\). At any root \(r_k\), all terms except the one indexed by \(k\) vanish, giving

Therefore the polynomial \(P-S\), of degree at most \(n-1\), is zero at all \(n\) distinct numbers \(r_1,\ldots,r_n\). A nonzero polynomial of degree at most \(n-1\) cannot have \(n\) distinct roots: by the factor theorem, each distinct root contributes a distinct linear factor, so \(n\) such roots would require degree at least \(n\). Hence \(P-S\) is the zero polynomial. This proves \(P=S\), which is the cleared-denominator identity and therefore proves existence.

For uniqueness, suppose two sets of coefficients, \(A_i\) and \(C_i\), both give decompositions. Clear denominators in both identities and subtract them. Evaluating the resulting polynomial identity at \(x=r_k\) leaves only the term associated with \(r_k\):

The constant \(c\) and every factor in the product are nonzero. Consequently \(A_k-C_k=0\), or \(A_k=C_k\). This holds for each \(k\), so the two sets of coefficients are identical. The decomposition is unique.

Worked Examples

A Common Pitfall: Substituting Before Clearing Denominators

The original rational expression is undefined at each denominator root. It is therefore not valid to substitute \(x=r_i\) directly into an equality of rational functions and treat the fractions as ordinary finite values there. Instead, first multiply through by the full denominator. The result is a polynomial identity, and evaluating that identity at \(r_i\) is valid.

A related coefficient error is to omit the other denominator factor. In the first example, a mistaken calculation might say \(A_1=P(1)=5\), leading to the incorrect coefficient \(5\). But the cleared identity at \(x=1\) is \(P(1)=A_1(1-(-2))\), so

not \(5\). The other factor evaluated at the root is exactly what must divide the numerator value.

What the Conditions Tell Us

The theorem depends on both the degree condition and the distinctness of the roots. The degree condition ensures that the numerator can be represented using the degree-\((n-1)\) products in the cleared identity. If the numerator has degree at least \(n\), this proper-fraction form alone is not enough; a polynomial part must first be separated.

Distinctness is what makes the coefficient formula meaningful: none of the differences \(r_i-r_j\) in its denominator can be zero. It also ensures that evaluating at one root leaves a single term rather than several terms. When factors repeat, the appropriate decomposition form and its proof require different considerations; that case is not covered by this theorem.

The result matters because it explains why coefficient-finding methods are reliable, rather than merely successful in particular examples. For distinct linear factors, the root values force every coefficient, and the degree argument proves those forced values fit the entire numerator polynomial. There is no second valid set of coefficients to discover.