Partial Fractions

Introduction to Rational Fractions

A Rational Fraction is the quotient of two polynomials, $\frac{P(x)}{Q(x)}$ where $Q(x) \neq 0$. Before Decomposition, you must classify it as proper or improper.

$$\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}$$

The Division Algorithm for polynomials converts improper fractions into a polynomial plus a proper remainder.

$P(x)$=Numerator polynomial(degree n)

$Q(x)$=Denominator polynomial(degree m)

$S(x)$=Quotient polynomial(degree n − m)

$R(x)$=Remainder polynomial(degree < m)

$\text{deg}(P) < \text{deg}(Q)$

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Proper fraction — ready for decomposition directly.

$\text{deg}(P) \ge \text{deg}(Q)$

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Improper fraction — requires Long Division first.

Degree Check: Always compare the highest power of $x$ in the numerator versus the denominator before starting.

Irreducible Factors: Ensure the denominator is fully factored into linear or irreducible quadratic terms before writing the decomposition form.

Long Division Example: $\frac{3x^2+1}{x-2} = 3x + 6 + \frac{13}{x-2}$ — the remainder $\frac{13}{x-2}$ is the proper fraction to decompose.

A partial fraction resolution produces an Identity — an equation that holds for all values of $x$, not just particular ones.

Identity vs Equation: $2x = 3$ is a conditional equation (true only for $x = \frac{3}{2}$), but $(x+3)(x+4) = x^2+7x+12$ is an identity (true for all $x$).

Key Theorem: If two polynomials are equal for all values of $x$, then they have the same degree and the coefficients of like powers must be equal.

Why It Matters: This theorem is the basis for the "equating coefficients" method — matching the $x^2$, $x$, and constant terms on both sides to find the unknown constants.

Case 1: Distinct Linear Factors

When the denominator factors into unique Linear Factors, each factor $(x-a)$ generates a single partial fraction with a constant numerator.

$$\frac{P(x)}{(x-a_1)(x-a_2)\cdots(x-a_n)} = \frac{A_1}{x-a_1} + \frac{A_2}{x-a_2} + \cdots + \frac{A_n}{x-a_n}$$

Each distinct linear factor in the denominator produces one partial fraction with an unknown constant numerator.

$A_i$=Unknown constant numerator for the $i$-th factor(Real number)

$x - a_i$=Linear factor with root at $x = a_i$

Substitution method

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Setting $x = a_i$ eliminates all other terms, giving $A_i$ directly.

Cover-up method

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Cover $(x-a_i)$ in the original fraction and evaluate the rest at $x = a_i$.

Uniqueness Required: Each factor must be distinct. If $(x-a)$ repeats, use Case 2 instead.

Cover-up Shortcut: To find $A$ for factor $(x-a)$, mentally 'cover' $(x-a)$ in the original denominator and substitute $x = a$ into what remains. This gives $A$ instantly.

Factoring First: Always factor the denominator completely — e.g. $x^2-5x+6 = (x-2)(x-3)$ before setting up the form.

Coefficient Factors: If the denominator has a factor like $(2x+3)$ instead of $(x-a)$, the partial fraction is still $\frac{B}{2x+3}$ with a constant numerator.

Case 2: Repeated Linear Factors

If a factor $(x-a)$ has Multiplicity $n$, it generates $n$ partial fractions with ascending powers in the denominator from 1 to $n$.

$$\frac{P(x)}{(x-a)^n} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n}$$

A repeated factor of multiplicity $n$ requires $n$ separate partial fractions with increasing powers.

$n$=Power (multiplicity) of the repeated factor(positive integer ≥ 2)

$A_1, A_2, \ldots, A_n$=Constants for each power level(Real numbers)

$n = 2$

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$\frac{A}{x-a} + \frac{B}{(x-a)^2}$ — two unknowns

$n = 3$

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$\frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3}$ — three unknowns

Staircase Rule: You cannot skip powers. For $(x-a)^3$, you must include terms for $(x-a)^1$, $(x-a)^2$, and $(x-a)^3$.

Solving Strategy: The highest-power constant (e.g. $A_n$) is found by substituting $x = a$. Lower-power constants require equating coefficients of like powers of $x$.

Mixed Denominators: When the denominator contains both repeated and non-repeated factors, e.g. $(x+1)^2(x-1)$, combine the rules: $\frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1}$.

Case 3: Irreducible Quadratic Factors

An Irreducible Quadratic factor $(ax^2+bx+c)$ with $b^2-4ac < 0$ requires a linear numerator $Ax+B$ in its partial fraction.

$$\frac{P(x)}{(x-r)(ax^2+bx+c)} = \frac{A}{x-r} + \frac{Bx+C}{ax^2+bx+c}$$

The numerator over an irreducible quadratic must be one degree less — a linear polynomial $Bx+C$.

$Bx+C$=Linear numerator for the quadratic part

$ax^2+bx+c$=Irreducible quadratic factor (no real roots)(Discriminant $b^2-4ac < 0$)

Factor is $x^2 + k$ where $k > 0$

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Simplest irreducible quadratic — e.g. $x^2+1$, $x^2+3$.

Two irreducible quadratics in denominator

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Each gets its own $\frac{Ax+B}{\text{quadratic}}$ term — e.g. $\frac{Ax+B}{x^2+2x+3} + \frac{Cx+D}{x^2-2x+3}$.

Why Linear Numerator?: The degree of the numerator must be exactly one less than the denominator factor. Degree 2 factor → degree 1 numerator.

Solving Method: Substitution eliminates the linear factor's constant first. Then equate coefficients of $x^2$, $x$, and the constant term to find $B$ and $C$.

Irreducibility Check: A quadratic is irreducible when its discriminant $b^2-4ac < 0$. For example, $x^2+x+1$ has $\Delta = 1-4 = -3 < 0$.

Case 4: Repeated Irreducible Quadratic Factors

When an irreducible quadratic factor is raised to power $n$, it generates $n$ partial fractions, each with a linear numerator and ascending powers of the quadratic.

$$\frac{P(x)}{(ax^2+bx+c)^n} = \frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + \cdots + \frac{A_nx+B_n}{(ax^2+bx+c)^n}$$

Analogous to repeated linear factors, but each term has a linear numerator instead of a constant.

$A_ix+B_i$=Linear numerator for the $i$-th power(Two unknowns per term)

$n$=Power of the repeated quadratic factor(positive integer ≥ 2)

$(x^2+1)^2$ in denominator

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$\frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$

Number of Unknowns: Each repeated quadratic power contributes 2 unknowns ($A_i$ and $B_i$), so $(ax^2+bx+c)^n$ produces $2n$ unknowns total.

Mixed Example: For $\frac{4x^2}{(x^2+1)^2(x-1)}$, the form is $\frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} + \frac{E}{x-1}$ — five unknowns total.

Solving Strategy: Substitute the linear root first (e.g. $x=1$) to find one constant, then equate coefficients of $x^4, x^3, x^2, x, x^0$ systematically.

Solving Methods Summary

Two primary methods exist for finding the unknown constants: the substitution method and the equating coefficients method.

Substitution Method: Plug in roots of the denominator factors to eliminate unknowns one at a time. Works best for distinct linear factors.

Equating Coefficients: Expand both sides, then match coefficients of $x^n, x^{n-1}, \ldots, x^0$. Essential when quadratic factors are present.

Hybrid Approach: Use substitution first to find as many constants as possible, then equate coefficients for the remaining unknowns.

Verification: Always check your result by recombining the partial fractions — the sum must equal the original expression.

Decision Flowchart

Is the fraction improper? → Perform long division first.

Factor the denominator completely.

Identify factor types: distinct linear, repeated linear, irreducible quadratic, or repeated quadratic.

Write the decomposition form using the rules for each case.

Clear denominators and solve for constants using substitution and/or equating coefficients.