Partial fractions Notes | Mathematics AA HL

Meaning and purpose of partial fractions

Partial-fraction decomposition reverses the process of adding rational expressions. It rewrites one complicated rational expression as a sum of simpler rational expressions.

For example, $\frac{5x+1}{(x-1)(x+2)}=\frac{2}{x-1}+\frac{3}{x+2}$ . The two forms are algebraically equivalent wherever the original denominator is non-zero.

The simpler pieces are easier to integrate, sum in a telescoping series, compare, simplify, or use inside a later algebraic argument.

A valid decomposition depends on the degree of the numerator, the complete factorization of the denominator, and the multiplicity and type of each factor.

Key terms

Rational expression: A quotient of two polynomials.
Partial-fraction decomposition: An equivalent sum of simpler rational expressions.
Factor: A polynomial expression that divides another polynomial exactly.
Identity: An equality true for every permitted value of the variable.
Domain restriction: A value excluded because it makes an original denominator zero.

Worked examples

Recognize a decomposition

Verify the structure $\frac{5x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}$ .

1The denominator has two distinct linear factors.
2Assign one constant numerator to each factor.
3After multiplying by $(x-1)(x+2)$ , obtain $5x+1=A(x+2)+B(x-1)$ .

Answer: Substitution gives $A=2$ and $B=3$ .

Check at one permitted value

Check the decomposition above at $x = 0$ .

1The original value is $\frac{1}{(-1)(2)}=-\frac12$ .
2The decomposed value is $-2+\frac32=-\frac12$ .

Answer: Both forms give the same value, which is a useful numerical check but not a complete proof.

Identify the benefit

Why is $\frac{2}{x-1}+\frac{3}{x+2}$ easier to integrate than the original fraction?

1Each term has a linear denominator whose derivative is constant.
2Each term matches a logarithmic integration pattern.

Answer: It integrates directly to $2\ln|x-1|+3\ln|x+2|+C$ .

Rules and formulas

Basic decomposition identity

\frac{P(x)}{Q(x)}=\sum\limits_{k=1}^{m}\frac{N_k(x)}{F_k(x)}

$P(x)$: numerator polynomial
$Q(x)$: factored denominator polynomial
$F_k(x)$: denominator factor or factor power
$N_k(x)$: numerator of degree lower than its associated irreducible factor

Proper and improper rational expressions

A rational expression is proper when the degree of its numerator is less than the degree of its denominator. It is improper when the numerator degree is greater than or equal to the denominator degree.

For an improper expression, polynomial division gives $\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)}$ , where the remainder satisfies $\operatorname{degree}R<\operatorname{degree}Q$ . Only the proper remainder is decomposed.

Use long division or synthetic division where suitable. Multiply the quotient by the divisor and add the remainder to verify the division.

Simplify common factors before decomposing, but record every restriction inherited from the original expression.

Key terms

Proper rational expression: A rational expression with numerator degree lower than denominator degree.
Improper rational expression: A rational expression with numerator degree at least the denominator degree.
Polynomial quotient: The polynomial part produced by division.
Remainder: The lower-degree polynomial left after division.

Worked examples

Classify two expressions

Classify $\frac{2x+1}{x^2-4}$ and $\frac{x^3+1}{x^2+1}$ .

1The first has degrees 1 and 2.
2The second has degrees 3 and 2.

Answer: The first is proper; the second is improper.

Divide before decomposing

Rewrite $\frac{x^2+3x+5}{x+1}$ .

1Divide $x^2+3x+5$ by $x+1$ .
2The quotient is $x+2$ , since $(x+1)(x+2)=x^2+3x+2$ .
3The remainder is 3.

Answer: $\frac{x^2+3x+5}{x+1}=x+2+\frac{3}{x+1}$ .

Long division with a quadratic divisor

Rewrite $\frac{2x^3+x^2-5}{x^2-1}$ .

1The first quotient term is $2x$ . Subtraction leaves $x^2+2x-5$ .
2The next quotient term is 1.
3The final remainder is $2x-4$ .

Answer: $\frac{2x^3+x^2-5}{x^2-1}=2x+1+\frac{2x-4}{x^2-1}$ .

Preserve a cancelled restriction

Simplify $\frac{x^2-1}{x^2-x}$ and state its domain restrictions.

1Factor to $\frac{(x-1)(x+1)}{x(x-1)}$ .
2Cancel $x-1$ to obtain $\frac{x+1}{x}$ .
3Keep the original excluded values.

Answer: The simplified form is $1+\frac1x$ , with $x\ne0$ and $x\ne1$ .

Rules and formulas

Polynomial division form

P(x)=Q(x)S(x)+R(x)

$S(x)$: quotient
$R(x)$: remainder

Improper rational expression

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

Remainder condition

\operatorname{degree}R<\operatorname{degree}Q

Factor structure and decomposition templates

Factor the denominator completely over the real numbers before writing the decomposition. The factor type and multiplicity determine every required term.

A distinct linear factor $x-a$ contributes $\frac{A}{x-a}$ . A repeated factor $(x-a)^m$ contributes a term for every power from 1 through $m$ .

An irreducible quadratic $q(x)$ contributes $\frac{Bx+C}{q(x)}$ , because its numerator must have degree less than 2. If $q(x)^m$ appears, include a separate linear numerator over every power $q(x)^j$ .

A real quadratic $ax^2+bx+c$ is irreducible over the reals when its discriminant satisfies $b^2-4ac<0$ .

Key terms

Distinct factor: A denominator factor appearing once.
Repeated factor: A denominator factor appearing with multiplicity greater than 1.
Irreducible quadratic: A real quadratic with no real linear factors.
Multiplicity: The number of times a factor occurs.
Template: The required partial-fraction form before the constants are found.

Worked examples

Distinct linear factors

Write the decomposition template for $\frac{P(x)}{(x-1)(x+4)}$ . Assume $P$ is a polynomial with $\deg P<2$ , so the fraction is proper.

1Both factors are linear and distinct.
2Assign one constant numerator to each.

Answer: $\frac{P(x)}{(x-1)(x+4)}=\frac{A}{x-1}+\frac{B}{x+4}$ . Retain the original denominator restrictions. For an improper fraction, divide first and add the polynomial quotient.

Repeated linear factor

Write the template for $\frac{P(x)}{x(x-2)^3}$ . Assume $P$ is a polynomial with $\deg P<4$ , so the fraction is proper.

1The factor $x$ contributes $\frac{A}{x}$ .
2The repeated factor requires denominators $x-2$ , $(x-2)^2$ , and $(x-2)^3$ .

Answer: $\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}+\frac{D}{(x-2)^3}$ . Retain the original denominator restrictions. For an improper fraction, divide first and add the polynomial quotient.

Linear and irreducible quadratic factors

Write the template for $\frac{P(x)}{(x+1)(x^2+4)}$ . Assume $P$ is a polynomial with $\deg P<3$ , so the fraction is proper.

1The linear factor contributes $\frac{A}{(x+1)}$ .
2Since $x^2+4$ is irreducible over the reals, use a linear numerator.

Answer: $\frac{A}{x+1}+\frac{Bx+C}{x^2+4}$ . Retain the original denominator restrictions. For an improper fraction, divide first and add the polynomial quotient.

Repeated irreducible quadratic

Write the template for $\frac{P(x)}{(x-3)^2(x^2+1)^2}$ . Assume $P$ is a polynomial with $\deg P<6$ , so the fraction is proper.

1Include both powers of each repeated factor.
2Use constants over linear factors and linear numerators over quadratic factors.

Answer: $\frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{Cx+D}{x^2+1}+\frac{Ex+F}{(x^2+1)^2}$ . Retain the original denominator restrictions. For an improper fraction, divide first and add the polynomial quotient.

Rules and formulas

Repeated linear factor template

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

$m$: multiplicity of the linear factor

Irreducible quadratic template

\frac{Bx+C}{ax^2+bx+c}

Quadratic irreducibility test

b^2-4ac<0

Distinct linear factors

For a proper expression with denominator $(x-a)(x-b)$ , write $\frac{A}{x-a}+\frac{B}{x-b}$ , clear the denominator, and solve $P(x)=A(x-b)+B(x-a)$ .

Substituting $x = a$ removes the term containing B, and substituting $x = b$ removes the term containing A. This is the fastest method when the roots are simple.

With three or more distinct linear factors, use one constant numerator per factor and substitute each root in turn.

After finding the constants, present the full decomposition and retain the original exclusions from the domain.

Key terms

Clearing denominators: Multiplying an identity by its common denominator.
Convenient substitution: Using a denominator root so all but one unknown term vanish.
Cover-up method: A shortcut for finding a coefficient associated with a distinct linear factor.

Worked examples

Two distinct factors

Decompose $\frac{7x-1}{(x-2)(x+3)}$ .

1Write $\frac{A}{x-2}+\frac{B}{x+3}$ .
2Clear denominators: $7x-1=A(x+3)+B(x-2)$ .
3Set $x=2$ : $13=5A$ , so $A=\frac{13}{5}$ .
4Set $x=-3$ : $-22=-5B$ , so $B=\frac{22}{5}$ .

Answer: $\frac{7x-1}{(x-2)(x+3)}=\frac{13}{5(x-2)}+\frac{22}{5(x+3)}$ .

Three distinct factors

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

1Write $\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+2}$ .
2Set $x=0$ : $-1=-2A$ , so $A=\frac{1}{2}$ .
3Set $x=1$ : $4=3B$ , so $B=\frac{4}{3}$ .
4Set $x=-2$ : $1=6C$ , so $C=\frac{1}{6}$ .

Answer: $\frac{1}{2x}+\frac{4}{3(x-1)}+\frac{1}{6(x+2)}$ .

Use cover-up carefully

Find the coefficient over $x-4$ in $\frac{x+7}{(x+1)(x-4)}$ .

1Multiply by $x-4$ and substitute $x=4$ .
2Compute $B=\frac{4+7}{4+1}=\frac{11}{5}$ .

Answer: The term is $\frac{11}{5(x-4)}$ .

Rules and formulas

Two distinct linear factors

\frac{P(x)}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}

Cover-up coefficient

A=\left.\frac{P(x)}{\frac{Q(x)}{(x-a)}}\right|_{x=a}

$a$: simple root of the denominator

Repeated linear factors

If $(x-a)^m$ divides the denominator, include fractions over every power $(x-a),(x-a)^2,\ldots,(x-a)^m$ .

After clearing denominators, substituting $x=a$ directly finds the coefficient over the highest power $(x-a)^m$ . The lower-power coefficients can then be found by another substitution or by equating coefficients.

A repeated factor does not mean the same constant is used in every numerator. Each fraction has its own independent constant.

Check the completed result by recombining it over the least common denominator.

Key terms

Highest repeated-power term: The fraction whose denominator contains the full multiplicity of the repeated factor.
Independent constants: Separate unknowns assigned to the different denominator powers.
Multiplicity: The exponent on a repeated factor.

Worked examples

A repeated factor only

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

1Write $\frac{A}{x-3}+\frac{B}{(x-3)^2}$ .
2Clear denominators: $2x+1=A(x-3)+B$ .
3Set $x=3$ : $B=7$ .
4Compare x coefficients to obtain $A = 2$ .

Answer: $\frac{2x+1}{(x-3)^2}=\frac{2}{x-3}+\frac{7}{(x-3)^2}$ .

Linear and repeated linear factors

Decompose $\frac{3x+5}{x(x-1)^2}$ .

1Write $\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}$ .
2Clear denominators: $3x+5=A(x-1)^2+Bx(x-1)+Cx$ .
3Set $x=0$ : $A=5$ . Set $x=1$ : $C=8$ .
4Compare $x^2$ coefficients: $A+B=0$ , so $B=-5$ .

Answer: $\frac{3x+5}{x(x-1)^2}=\frac5x-\frac5{x-1}+\frac8{(x-1)^2}$ .

Verify the linear coefficient

Check the x coefficient in the cleared identity of the previous example.

1The x coefficient is $-2A-B+C$ .
2Substitute $A=5$ , $B=-5$ , and $C=8$ .
3Calculate $-10+5+8=3$ .

Answer: The coefficient matches the original numerator, providing an independent check.

Rules and formulas

Repeated linear decomposition

\frac{P(x)}{(x-a)^mR(x)}=\frac{A_1}{x-a}+\cdots+\frac{A_m}{(x-a)^m}+\text{terms from }R(x)

Highest-power coefficient

A_m=\left.\frac{P(x)}{R(x)}\right|_{x=a}

Irreducible quadratic factors

If a quadratic factor has no real roots, it cannot be split into real linear factors. A factor $q(x)=ax^2+bx+c$ therefore receives a linear numerator $Bx+C$ .

The linear numerator has the most general degree below 2. Both coefficients may be needed to reproduce the x squared, x, and constant terms of the original numerator.

For a repeated irreducible quadratic $q(x)^m$ , include $\frac{B_1x+C_1}{q(x)}+\frac{B_2x+C_2}{q(x)^2}+\cdots+\frac{B_mx+C_m}{q(x)^m}$ .

Substitution at real linear roots can still find some constants in a mixed denominator; equating coefficients usually completes the quadratic part.

Key terms

Linear numerator: A numerator of the form $Bx+C$ .
Real irreducibility: The absence of real roots or real linear factors.
Mixed denominator: A denominator containing more than one type of factor.

Worked examples

Linear plus irreducible quadratic

Decompose $\frac{2x^2+x+5}{(x-1)(x^2+4)}$ .

1Write $\frac{A}{x-1}+\frac{Bx+C}{x^2+4}$ .
2Clear denominators: $2x^2+x+5=A(x^2+4)+(Bx+C)(x-1)$ .
3Compare coefficients: $A+B=2$ , $C-B=1$ , and $4A-C=5$ .
4Solve to get $A=\frac{8}{5}$ , $B=\frac{2}{5}$ , and $C=\frac{7}{5}$ .

Answer: $\frac{8}{5(x-1)}+\frac{2x+7}{5(x^2+4)}$ .

Recognize an already simple quadratic term

Can $\frac{3x+2}{x^2+1}$ be split further over real linear factors?

1The discriminant of $x^2+1$ is $-4$ .
2The denominator has no real linear factors.
3The numerator is already linear.

Answer: No further real partial-fraction decomposition is required.

Repeated quadratic template

State the quadratic contribution for a denominator containing $(x^2+9)^2$ .

1Use a linear numerator over each power of the irreducible factor.

Answer: $\frac{Ax+B}{x^2+9}+\frac{Cx+D}{(x^2+9)^2}$ .

Rules and formulas

Irreducible quadratic contribution

\frac{Bx+C}{ax^2+bx+c}

Repeated quadratic contribution

\sum\limits_{j=1}^{m}\frac{B_jx+C_j}{q(x)^j}

$q(x)$: irreducible quadratic factor

Finding constants by substitution

After clearing denominators, choose values of x that make most terms vanish. Roots of distinct linear factors are the natural choices.

Substitution into the polynomial identity is valid even at values excluded from the original rational expression because clearing denominators produces an identity between polynomials.

For a repeated linear factor, substituting its root immediately gives the coefficient over the highest repeated power; use other values or coefficient comparison for the remaining constants.

A cover-up shortcut is efficient for simple linear factors, but write enough algebra to make the method traceable in an IB solution.

Key terms

Strategic value: A chosen x-value that removes one or more unknown terms.
Root substitution: Substituting a root of a denominator factor into the cleared identity.
Polynomial identity: An equality between polynomials that holds for all real x.

Worked examples

Find both constants by roots

Decompose $\frac{x+7}{(x+1)(x-4)}$ .

1Write $x+7=A(x-4)+B(x+1)$ .
2Set $x=-1$ : $6=-5A$ , so $A=\frac{-6}{5}$ .
3Set $x=4$ : $11=5B$ , so $B=\frac{11}{5}$ .

Answer: $\frac{x+7}{(x+1)(x-4)}=-\frac{6}{5(x+1)}+\frac{11}{5(x-4)}$ .

Find the highest repeated-power coefficient

In $\frac{3x+5}{x(x-1)^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}$ , find $C$ .

1Clear denominators to obtain $3x+5=A(x-1)^2+Bx(x-1)+Cx$ .
2Set $x=1$ .
3Then $8=C$ .

Answer: $C=8$ .

Use an ordinary value after root substitutions

Once two of three constants are known, how can a third be checked without fully expanding?

1Choose any convenient permitted value such as $x = 0$ or $x = 1$ that does not make every remaining coefficient vanish.
2Substitute known constants into the cleared identity and solve the resulting linear equation.

Answer: A non-root substitution can efficiently finish or verify the system.

Rules and formulas

Coefficient at a simple root

A=\lim_{x\to a}(x-a)\frac{P(x)}{Q(x)}

$a$: simple real root of Q

Highest repeated coefficient

A_m=\lim_{x\to a}(x-a)^m\frac{P(x)}{Q(x)}

$m$: multiplicity of the root

Finding constants by equating coefficients

Expand the cleared identity and collect like powers of x. Because two polynomials are identical, the coefficients of every matching power must be equal.

Coefficient comparison is especially useful for irreducible quadratic factors, repeated factors, or systems where root substitution does not determine every constant.

Include zero coefficients explicitly. If the original numerator has no x squared term, its x squared coefficient is 0 and provides an equation.

A hybrid method is usually fastest: use convenient roots first, then compare only the coefficients needed for the remaining unknowns.

Key terms

Like powers: Terms containing the same exponent of x.
Coefficient system: Simultaneous linear equations formed by matching polynomial coefficients.
Zero coefficient: A missing power interpreted as having coefficient 0.
Hybrid method: Combining strategic substitution with coefficient comparison.

Worked examples

Two constants from coefficients

Decompose $\frac{4x+7}{(x-1)(x+2)}$ by equating coefficients.

1Write $4x+7=A(x+2)+B(x-1)$ .
2Expand: $(A+B)x+(2A-B)$ .
3Equate coefficients: $A+B=4$ and $2A-B=7$ .
4Solve to obtain $A=\frac{11}{3}$ and $B=\frac{1}{3}$ .

Answer: $\frac{4x+7}{(x-1)(x+2)}=\frac{11}{3(x-1)}+\frac{1}{3(x+2)}$ .

Quadratic coefficient system

Form the coefficient equations for $\frac{2x^2+x+5}{(x-1)(x^2+4)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+4}$ .

1Expand the right side to $(A+B)x^2+(C-B)x+(4A-C)$ .
2Match the x squared, x, and constant coefficients.

Answer: $A+B=2$ , $C-B=1$ , and $4A-C=5$ .

Use a missing power

If a cleared right side is $(A+B)x^2+(2A-B+C)x+(A-C)$ and the left side is $5x+1$ , what equation comes from $x^2$ ?

1Write $5x+1=0x^2+5x+1$ .
2Equate the x squared coefficients.

Answer: $A+B=0$ .

Rules and formulas

Polynomial coefficient identity

a_nx^n+\cdots+a_0=b_nx^n+\cdots+b_0\Rightarrow a_k=b_k

$k$: each matching power index

Reconstructing and verifying the original fraction

Verification reverses the decomposition: place all partial fractions over the least common denominator, expand the new numerator, and check that it equals the original numerator.

Also compare domain restrictions. Recombining algebraically does not restore any value excluded by the original denominator.

A numerical substitution at one or two permitted x-values is a useful arithmetic check, but it cannot prove a polynomial identity by itself.

For an improper expression, verify both the polynomial quotient and the reconstructed proper remainder.

Key terms

Least common denominator: The smallest denominator containing every required factor at its highest multiplicity.
Reconstruction: Combining the simpler fractions back into one rational expression.
Structural check: Checking degree, factor multiplicity, and numerator form in addition to arithmetic.

Worked examples

Reconstruct two fractions

Combine $\frac{2}{x-1}+\frac{3}{x+2}$ .

1Use common denominator $(x-1)(x+2)$ .
2The numerator is $2(x+2)+3(x-1)$ .
3Simplify to $5x+1$ .

Answer: $\frac{2}{x-1}+\frac{3}{x+2}=\frac{5x+1}{(x-1)(x+2)}$ .

Reconstruct a repeated-factor result

Verify $\frac5x-\frac5{x-1}+\frac8{(x-1)^2}$ .

1Use common denominator $x(x-1)^2$ .
2The numerator is $5(x-1)^2-5x(x-1)+8x$ .
3Expand to obtain $3x+5$ .

Answer: The expression reconstructs $\frac{3x+5}{x(x-1)^2}$ .

Numerical confidence check

Check $\frac{4x+7}{(x-1)(x+2)}=\frac{11}{3(x-1)}+\frac1{3(x+2)}$ at $x=0$ .

1The original value is $\frac{7}{(-2)}=\frac{-7}{2}$ .
2The decomposed value is $\frac{-11}{3}+\frac{1}{6}=\frac{-7}{2}$ .

Answer: The values agree at $x = 0$ .

Rules and formulas

Reconstruction check

\sum\limits_{k=1}^{m}\frac{N_k(x)}{F_k(x)}=\frac{\frac{\sum\limits_{k=1}^{m}N_k(x)\,Q(x)}{F_k(x)}}{Q(x)}

Applications of partial fractions

Partial fractions prepare rational functions for integration. Fractions over linear factors integrate to logarithms, while irreducible quadratic parts may lead to logarithmic or inverse-tangent forms in later calculus.

They also create telescoping sums. Consecutive terms cancel after a rational term in n is decomposed into shifted simpler fractions.

In algebra, a decomposition can expose poles, signs, asymptotic contributions, or recurrence patterns that are hidden in the original combined fraction.

The application does not replace the decomposition method: first check properness, factor, choose the complete template, solve the constants, and only then integrate or sum.

Key terms

Telescoping sum: A series whose intermediate terms cancel after decomposition.
Pole: A denominator zero associated with an unbounded rational function value.
Logarithmic integral: An integral matching $\frac{\int f'(x)}{f(x)\,dx}=\ln|f(x)|+C$ .

Worked examples

Integrate using a decomposition

Evaluate $\int\frac{5x+1}{(x-1)(x+2)}\,dx$ .

1Use $\frac{5x+1}{(x-1)(x+2)}=\frac2{x-1}+\frac3{x+2}$ .
2Integrate term by term.

Answer: $2\ln|x-1|+3\ln|x+2|+C$ .

Create a telescoping sum

Find $\sum\limits_{n=1}^{N}\frac1{n(n+1)}$ .

1Decompose $\frac1{n(n+1)}=\frac1n-\frac1{n+1}$ .
2Write several terms and cancel adjacent opposites.
3Only $1-\frac{1}{(N+1)}$ remains.

Answer: $\sum\limits_{n=1}^{N}\frac1{n(n+1)}=1-\frac1{N+1}$ .

Integrate across two linear factors

Decompose and integrate $\frac1{1-x^2}$ .

1Factor $1-x^2=(1-x)(1+x)$ .
2Decompose as $\frac1{2(1-x)}+\frac1{2(1+x)}$ .
3Account for the derivative of 1 − x when integrating.

Answer: $\int\frac{dx}{1-x^2}=\frac12\ln\left|\frac{1+x}{1-x}\right|+C$ .

Rules and formulas

Linear-factor integral

\int\frac{A}{x-a}\,dx=A\ln|x-a|+C

Basic telescoping decomposition

\frac1{n(n+1)}=\frac1n-\frac1{n+1}

Mixed and advanced decompositions

A complete HL problem may combine polynomial division, distinct or repeated linear factors, and irreducible quadratics. Handle these stages in a fixed order rather than trying to solve everything at once.

First divide if improper. Then factor the denominator of the proper remainder. Next write every required term, clear denominators, use convenient roots, and finish with coefficients.

Large coefficient systems are still linear. Organize them by descending powers of x and use exact simultaneous-equation methods.

The decomposition is unique once the denominator factorization and correct numerator degrees are fixed, so two different verified answers must simplify to the same constants.

Key terms

Mixed decomposition: A decomposition involving more than one factor type or an additional polynomial quotient.
Uniqueness: The property that a complete partial-fraction template has one set of constants for the given rational expression.
Workflow: Divide, factor, template, clear, solve, reconstruct.

Worked examples

Improper expression with two linear factors

Decompose $\frac{x^3+2x^2+3}{x^2-1}$ .

1Divide to obtain $x+2+\frac{x+5}{x^2-1}$ .
2Factor $x^2-1=(x-1)(x+1)$ .
3Write $\frac{x+5}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$ .
4Substitution gives $A=3$ and $B=-2$ .

Answer: $x+2+\frac3{x-1}-\frac2{x+1}$ .

Full mixed template

State the template for $\frac{5x^3-5x^2-3x+1}{(x-1)^2(x^2+1)}$ .

1The expression is proper.
2Include both powers of x − 1.
3Use a linear numerator over x squared plus 1.

Answer: $\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}$ .

Solve the mixed example

Find the constants in the previous template.

1Clear denominators and substitute $x = 1$ to obtain $B = -1$ .
2Expand the remaining identity and compare coefficients.
3Solve the linear system to obtain $A = 2$ , $C = 3$ , and $D = 4$ .

Answer: $\frac2{x-1}-\frac1{(x-1)^2}+\frac{3x+4}{x^2+1}$ .

Rules and formulas

Complete workflow

\text{divide}\to\text{factor}\to\text{template}\to\text{clear}\to\text{solve}\to\text{verify}

Calculator methods and IB HL exam strategy

A GDC can perform polynomial division, factor some polynomials, solve the resulting simultaneous linear equations, and numerically check a decomposition. Availability depends on the calculator model and exam mode.

A CAS partial-fraction command is useful for verification, but an IB solution should still show the correct template and enough working to establish the constants.

Enter exact rational coefficients where possible. If the GDC displays decimals, convert simple results back to fractions and verify them in the identity.

For a numerical check, choose permitted x-values away from denominator zeros and compare the original and decomposed values to full precision.

A reliable exam workflow is: degree check, division, factorization, complete template, cleared identity, strategic substitutions, coefficient comparison, reconstruction, application.

Key terms

CAS: Computer algebra system capable of symbolic manipulation.
Exact mode: Calculator output retained as fractions or radicals rather than rounded decimals.
Numerical verification: Comparing both forms at safe input values.
Method marks: Credit for the correct mathematical setup and process.

Worked examples

Use a GDC to solve constants

How can a GDC solve $A+B=2$ , $C-B=1$ , and $4A-C=5$ ?

1Open the simultaneous-equation solver for three unknowns.
2Enter coefficients in a consistent A, B, C order.
3Read exact fractions if supported, then substitute them back.

Answer: $A=\frac{8}{5}$ , $B=\frac{2}{5}$ , and $C=\frac{7}{5}$ .

Check without forbidden values

Choose a safe numerical test value for a denominator $(x-1)^2(x^2+1)$ .

1Avoid $x = 1$ .
2The quadratic factor is never zero over the reals.
3Choose a simple value such as $x = 0$ .

Answer: $x = 0$ is a convenient permitted test value.

Interpret a CAS result

A CAS gives a decomposition immediately. What working should still be shown?

1State whether division is required.
2Show the denominator factorization and complete template.
3Show the cleared identity or equations used for the constants.
4State original domain restrictions.

Answer: Technology may verify arithmetic, but it should not replace the modelling and algebra required for method marks.

Rules and formulas

Reasonableness identity

P(x)\stackrel{?}{=}\sum\limits_{k=1}^{m}N_k(x)\frac{Q(x)}{F_k(x)}

IB HL-style practice questions with worked solutions

Attempt each question from the factor structure onward before reading the solution. The questions progress from classification and division to repeated factors, irreducible quadratics, applications, and an HL parameter problem.

Show the decomposition template and cleared identity before calculator work. Preserve exact fractions and verify difficult answers by reconstruction.

Key terms

Setup mark: Credit for choosing the correct decomposition before finding constants.
Exact coefficient: A fraction or integer rather than a rounded decimal.
Verification: Recombining or substituting to check the decomposition.

Worked examples

1. Proper or improper

Classify $\frac{2x+1}{x^2-4}$ and $\frac{x^3+1}{x^2+1}$ .

1Compare numerator and denominator degrees.

Answer: The first is proper and the second is improper.

2. Polynomial division

Rewrite $\frac{2x^3-3x^2+x+5}{x^2-1}$ as a polynomial plus a proper fraction.

1Divide the leading terms to obtain $2x$ .
2After subtraction, the next quotient term is $-3$ .
3The remainder is $3x+2$ .

Answer: $2x-3+\frac{3x+2}{x^2-1}$ .

3. Two distinct linear factors

Decompose $\frac{5x+7}{(x-1)(x+2)}$ .

1Write $\frac{A}{(x-1)}+\frac{B}{(x+2)}$ .
2Set $x=1$ to get $A=4$ .
3Set $x=-2$ to get $B=1$ .

Answer: $\frac4{x-1}+\frac1{x+2}$ .

4. Three distinct linear factors

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

1Use $\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x+2)}$ .
2Substitute $x=0,1,-2$ to obtain $A=\frac{1}{2}$ , $B=\frac{4}{3}$ , and $C=\frac{1}{6}$ .

Answer: $\frac1{2x}+\frac4{3(x-1)}+\frac1{6(x+2)}$ .

5. Repeated linear factor

Decompose $\frac{3x+5}{x(x-1)^2}$ .

1Use $\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x-1)^2}$ .
2Substitution gives $A=5$ and $C=8$ .
3The $x^2$ coefficient gives $B=-5$ .

Answer: $\frac5x-\frac5{x-1}+\frac8{(x-1)^2}$ .

6. Irreducible quadratic factor

Decompose $\frac{2x^2+x+5}{(x-1)(x^2+4)}$ .

1Use $\frac{A}{(x-1)}+\frac{(Bx+C)}{(x^2+4)}$ .
2Equate coefficients to obtain $A=\frac{8}{5}$ , $B=\frac{2}{5}$ , and $C=\frac{7}{5}$ .

Answer: $\frac8{5(x-1)}+\frac{2x+7}{5(x^2+4)}$ .

7. Mixed improper expression

Decompose $\frac{x^3+2x^2+3}{x^2-1}$ .

1Divide to get $x+2+\frac{(x+5)}{(x^2-1)}$ .
2Factor $x^2-1$ and decompose the remainder.
3Substitution gives constants 3 and $-2$ .

Answer: $x+2+\frac3{x-1}-\frac2{x+1}$ .

8. Write a complete repeated template

Write the decomposition form for $\frac{2x+1}{(x-3)^2(x^2+1)^2}$ .

1Include both powers of the repeated linear factor.
2Include both powers of the irreducible quadratic with linear numerators.

Answer: $\frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{Cx+D}{x^2+1}+\frac{Ex+F}{(x^2+1)^2}$ .

9. Reconstruct a fraction

Combine $\frac2{x-1}-\frac1{x+2}$ .

1Use denominator $(x-1)(x+2)$ .
2Simplify $2(x+2)-(x-1)=x+5$ .

Answer: $\frac{x+5}{(x-1)(x+2)}$ .

10. Integration application

Evaluate $\int\frac{5x+1}{(x-1)(x+2)}\,dx$ .

1Decompose as $\frac{2}{(x-1)}+\frac{3}{(x+2)}$ .
2Integrate each logarithmic term.

Answer: $2\ln|x-1|+3\ln|x+2|+C$ .

11. Telescoping application

Find $\sum\limits_{n=1}^{N}\frac1{n(n+2)}$ .

1Use $\frac1{n(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right)$ .
2Write the first and last terms and cancel the middle terms.

Answer: $\frac34-\frac1{2(N+1)}-\frac1{2(N+2)}$ .

12. Parameter from a partial-fraction coefficient

In the decomposition of $\frac{x^2+kx+1}{(x-1)(x+1)(x-2)}$ , the coefficient of $\frac{1}{(x-2)}$ is 5. Find $k$ .

1The coefficient is $C=\left.\frac{x^2+kx+1}{(x-1)(x+1)}\right|_{x=2}$ .
2Thus $C=\frac{(5+2k)}{3}$ .
3Set $\frac{(5+2k)}{3}=5$ and solve.

Answer: $k=5$ .

Rules and formulas

Distinct linear template

\frac{A}{x-a}+\frac{B}{x-b}

Repeated linear template

\sum\limits_{j=1}^{m}\frac{A_j}{(x-a)^j}

Quadratic template

\frac{Bx+C}{ax^2+bx+c}

Topic summary and cheat sheet

Partial fractions reverse rational-expression addition. The denominator factorization determines the decomposition, while the numerator degree determines the form above each factor.

Always divide an improper expression first. For the proper remainder, factor fully, include every repeated power, clear denominators, and solve constants by strategic substitution and coefficient comparison.

Use a constant numerator over a linear factor and a linear numerator over an irreducible quadratic factor. Repeated factors require a term for every power up to their multiplicity.

Finish by reconstructing the original fraction, retaining domain restrictions, and then applying the simpler fractions to integration, telescoping sums, or later analysis.

Key terms

Six-step workflow: Divide, factor, template, clear, solve, verify.
Linear-factor rule: Use a constant numerator.
Quadratic-factor rule: Use a linear numerator.
Repeated-factor rule: Include every denominator power from 1 through the multiplicity.

Worked examples

Rapid template check

What terms are required for denominator $x(x-2)^2(x^2+1)$ ?

1One constant over x.
2Two constants over the first and second powers of x − 2.
3One linear numerator over x squared plus 1.

Answer: $\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}+\frac{Dx+E}{x^2+1}$ .

Rapid verification check

What should be checked after all constants are found?

1Recombine over the least common denominator.
2Match the original numerator.
3Confirm excluded values and, if useful, compare at a safe numerical input.

Answer: The reconstructed identity and domain should agree with the original rational expression.

Rules and formulas

Improper-expression step

\frac{P}{Q}=S+\frac{R}{Q},\qquad \operatorname{degree}R<\operatorname{degree}Q

Distinct linear factors

\frac{P(x)}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}

Repeated linear factor

\sum\limits_{j=1}^{m}\frac{A_j}{(x-a)^j}

Repeated irreducible quadratic

\sum\limits_{j=1}^{m}\frac{B_jx+C_j}{q(x)^j}

Topic summary

Partial fractions rewrite one rational expression as a sum of simpler rational expressions.
The decomposition is used in integration, telescoping sums, simplification, and rational-function analysis.
A rational expression is proper only when the numerator degree is lower than the denominator degree.
Use polynomial division before decomposing an improper rational expression.
Simplify common factors where appropriate but retain restrictions from the original denominator.
Factor the denominator completely over the real numbers before choosing a template.
A distinct linear factor $x-a$ contributes $\frac{A}{(x-a)}$ .
A repeated factor $(x-a)^m$ requires terms over every power from 1 to $m$ .
An irreducible quadratic contributes $\frac{(Bx+C)}{(ax^2+bx+c)}$ .
A repeated irreducible quadratic requires a separate linear numerator over every factor power.
Clear denominators to create a polynomial identity before solving constants.
Substitute roots of distinct linear factors into the cleared identity, not the original fraction.
For a repeated root, direct substitution finds the coefficient over the highest repeated power.
Equate coefficients after fully expanding and collecting like powers.
Treat a missing polynomial power as having coefficient zero.
A hybrid of root substitution and coefficient comparison is often the fastest method.
Use exact fractions for constants and delay decimal approximation.
Reconstruct the original numerator over the least common denominator to verify the decomposition.
A numerical input is a useful check but does not prove the identity.
Retain all original domain restrictions in the final result.
Linear partial fractions integrate using $\frac{\int A}{(x-a)\,dx}=A\ln|x-a|+C$ .
The identity $\frac{1}{[n(n+1)]}=\frac{1}{n}-\frac{1}{(n+1)}$ creates a telescoping sum.
A GDC or CAS may verify division, constants, and reconstruction but should not replace the setup.
The reliable workflow is: divide, factor, template, clear, solve, reconstruct, then apply.