6. Partial fractions and rational decomposition

A nonzero polynomial $p(x)=a_nx^n+\cdots+a_0$ has degree $n$ when $a_n\ne0$. The zero polynomial has no degree under our convention. For a nonzero divisor polynomial $d$, polynomial long division gives unique $q,r$ with $p=dq+r$ and either $r=0$ or $\deg r<\deg d$.

Proof of the division algorithm. If the degree of the current remainder is at least $\deg d$, subtract the multiple of $d$ that cancels its leading term. Each subtraction strictly reduces the degree, so the process terminates. If two representations exist, $d(q_1-q_2)=r_2-r_1$. A nonzero left side would have degree at least $\deg d$, whereas the right side has smaller degree. Therefore both sides are zero, proving uniqueness.

Learning goals

Perform long division first, choose a complete decomposition pattern, solve coefficients, and verify the identity on its original domain. The aim is structural algebra; integration is an important later application rather than part of the definition.

Patterns and justification

For a nonzero denominator polynomial $q$, the rational expression $p/q$ is proper if $p=0$ or $\deg p<\deg q$. Otherwise use $p/q=s+r/q$ with $r/q$ proper. Factor $q$ over the real numbers. A factor $(x-a)^m$ requires

$$\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_m}{(x-a)^m}.$$

An irreducible real quadratic $Q=x^2+bx+c$, with $b^2-4c<0$, repeated $m$ times requires $(B_jx+C_j)/Q^j$ for each $1\le j\le m$. A constant numerator alone over a quadratic is generally insufficient. All identities apply where the original denominator is nonzero.

Theorem: decomposition over distinct linear factors

If $q(x)=\prod_{j=1}^m(x-a_j)$ with distinct real $a_j$ and $\deg p<m$, then

$$\frac{p(x)}{q(x)}=\sum_{j=1}^m\frac{A_j}{x-a_j},\qquad A_j=\frac{p(a_j)}{\prod_{k\ne j}(a_j-a_k)}.$$

The coefficients are unique.

Proof. Multiply the proposed identity by $q$. Its right side is a polynomial of degree at most $m-1$. At $x=a_j$ only the $j$th term survives and equals $p(a_j)$ by the stated coefficient. The two polynomials agree at $m$ distinct points, so the root bound makes them identical. Any decomposition must have the same values of $A_j$ by the same substitutions, proving uniqueness. For repeated or quadratic factors, equating coefficients after clearing denominators is a reliable computational procedure; the calculation must still be checked.

Theorem: the complete repeated-factor pattern

Suppose a supplied real factorization is $q=c\prod_{i=1}^s F_i^{m_i}$, where $c\ne0$, the $F_i$ are distinct monic irreducible real polynomials of degree one or two, and $m_i\ge1$. If $p=0$ or $\deg p<\deg q$, there is a unique decomposition

$$\frac pq=\sum_{i=1}^s\sum_{j=1}^{m_i}\frac{R_{ij}}{F_i^j},\qquad \deg R_{ij}<\deg F_i,$$

where a zero numerator is allowed. Thus linear factors have constant numerators, while quadratic factors have numerators of degree at most one. Every power must be allowed, although its coefficient may turn out to be zero. This theorem assumes the displayed factorization; it does not prove that every real polynomial factors into linear and quadratic factors. All rational identities retain the original domain $q(x)\ne0$.

Proof. The polynomial division algorithm also gives the Euclidean algorithm: divide successive remainders until the last nonzero remainder, whose degree must strictly decrease at each step. Back-substitution writes that last remainder as a polynomial combination of the original pair. It divides both original polynomials, and every common divisor divides it. In particular, coprime polynomials $A,B$ admit $UA+VB=1$ after normalizing the last constant remainder. Consequently, if $A$ divides $BC$ and $A,B$ are coprime, multiplying this identity by $C$ shows that $A$ divides $C$.

Distinct monic irreducibles are coprime: a nonconstant divisor of either must be a constant multiple of that irreducible. An irreducible is prime by the preceding divisibility implication. Every nonconstant polynomial has an irreducible divisor: if it is reducible, repeatedly choose a nonconstant proper factor of smaller degree until this stops. Therefore the powers of distinct irreducibles are pairwise coprime. Put $D_i=F_i^{m_i}$, $D=\prod_iD_i=q/c$ and $E_i=D/D_i$. The same observation gives $\gcd(D_i,E_i)=1$, so choose $U_i,V_i$ with $U_iE_i+V_iD_i=1$. Let $P=p/c$, and let $H_i$ be the remainder of $PU_i$ on division by $D_i$. Then $\deg H_i<\deg D_i$ or $H_i=0$, and $D_i$ divides $P-H_iE_i$.

For each $i$, all the other $H_kE_k$ are divisible by $D_i$. Hence every $D_i$ divides $P-\sum_iH_iE_i$. Pairwise coprimeness and the divisibility implication show, inductively, that their product $D$ divides this polynomial. If it were nonzero, its degree would be smaller than $\deg D$, contradicting divisibility by $D$. Hence it is zero. Thus $P/D=\sum_iH_i/D_i$.

Repeated division of $H_i$ by $F_i$ gives a unique expansion $H_i=T_{i0}+T_{i1}F_i+\cdots+T_{i,m_i-1}F_i^{m_i-1}$ with each remainder $T_{ik}$ zero or of degree smaller than $\deg F_i$. Dividing by $F_i^{m_i}$ gives the displayed pattern with $R_{ij}=T_{i,m_i-j}$.

For uniqueness, subtract two decompositions and combine the terms over each $D_i$ into $H_i/D_i$, with $\deg H_i<\deg D_i$. After multiplication by $D$, reduction modulo $D_i$ gives $H_iE_i\equiv0\pmod{D_i}$. Since $E_i$ has the inverse $U_i$ modulo $D_i$, $D_i$ divides $H_i$, forcing $H_i=0$. Uniqueness in repeated polynomial division then forces every $R_{ij}=0$. This proves both existence and uniqueness.

Worked examples

Repeated factors. Decompose $(3x+1)/(x(x+1)^2)$. Set

$$\frac{3x+1}{x(x+1)^2}=\frac A x+\frac B{x+1}+\frac C{(x+1)^2}.$$

Clearing denominators gives $3x+1=A(x+1)^2+Bx(x+1)+Cx$. Substitution of $x=0$ gives $A=1$; $x=-1$ gives $C=2$; the $x^2$ coefficient gives $B=-1$. Substitution back yields the original numerator. Omitting the $1/(x+1)$ term would make the decomposition impossible.

A quadratic and an improper expression. Long division gives

$$\frac{x^3+2x+1}{x(x^2+1)}=1+\frac{x+1}{x(x^2+1)}.$$

For the proper part write $A/x+(Bx+C)/(x^2+1)$. The numerator identity is $x+1=A(x^2+1)+(Bx+C)x$, so $A=1$, $B=-1$, $C=1$. Hence the answer is $1+1/x+(-x+1)/(x^2+1)$ for $x\ne0$. The quadratic has no real roots but still contributes two numerator coefficients.

Graded practice

1. Foundation. Decompose $1/((x-1)(x+2))$.
2. Core. Decompose $x/(x-1)^2$ and state its domain.
3. Core. Decompose $1/(x(x^2+4))$ over the reals.
4. Proof and diagnosis. A student proposes $1/(x^2+1)=A/(x-i)+B/(x+i)$ as a real-factor decomposition. Explain the issue, then prove that a real expression $(Bx+C)/(x^2+1)$ is identically zero only if $B=C=0$.

Full solutions

1. Write $1=A(x+2)+B(x-1)$. At $x=1$, $A=1/3$; at $x=-2$, $B=-1/3$. Thus the expression is $\frac13(1/(x-1)-1/(x+2))$ on $\mathbb R\setminus\{-2,1\}$.
2. Since $x=(x-1)+1$, the result is $1/(x-1)+1/(x-1)^2$, for $x\ne1$. This is the full repeated-factor pattern.
3. Clear denominators in $A/x+(Bx+C)/(x^2+4)$ to obtain $1=A(x^2+4)+(Bx+C)x$. Therefore $A=1/4$, $B=-1/4$, $C=0$. The answer is $1/(4x)-x/(4(x^2+4))$, for $x\ne0$.
4. The proposed factors use nonreal numbers and describe a complex, not real, decomposition; they are not wrong over $\mathbb C$. Over $\mathbb R$, $x^2+1$ is irreducible. Since it is never zero on the real line, the identity $(Bx+C)/(x^2+1)=0$ implies $Bx+C=0$ for every real $x$. Evaluating at $0$ gives $C=0$ and at $1$ gives $B=0$.

Summary and term definitions

A proper rational expression has numerator degree smaller than denominator degree. An irreducible quadratic has no real linear factor. A partial-fraction decomposition is an identity of rational expressions on their common domain. Include every lower power of a repeated factor, allow a linear numerator over each quadratic power, and verify the cleared polynomial identity rather than trusting a coefficient guess.