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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)
$\text{deg}(P) < \text{deg}(Q)$
→Proper fraction — ready for decomposition directly.
$\text{deg}(P) \ge \text{deg}(Q)$
→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
→Setting $x = a_i$ eliminates all other terms, giving $A_i$ directly.
Cover-up method
→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)
$n = 2$
→$\frac{A}{x-a} + \frac{B}{(x-a)^2}$ — two unknowns
$n = 3$
→$\frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3}$ — 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 $k > 0$
→Simplest irreducible quadratic — e.g. $x^2+1$, $x^2+3$.
Two irreducible quadratics in denominator
→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
→$\frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$
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
1
Is the fraction improper? → Perform long division first.
2
Factor the denominator completely.
3
Identify factor types: distinct linear, repeated linear, irreducible quadratic, or repeated quadratic.
4
Write the decomposition form using the rules for each case.
5
Clear denominators and solve for constants using substitution and/or equating coefficients.
Application Strategy
Worked Examples
Problem Statement
Resolve $\frac{7x+25}{(x+3)(x+4)}$ into partial fractions.
Distinct Linear FactorsSubstitution Method
Strategic Toolkit
$$\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$
Step 1
Verify Proper Fraction
Degree of numerator (1) < degree of denominator (2). Proper — proceed directly.
$\text{deg}(7x+25) = 1 < \text{deg}((x+3)(x+4)) = 2$
Step 2
Set Up Form
Both factors are distinct linear, so each gets a constant numerator.
$$\frac{7x+25}{(x+3)(x+4)} = \frac{A}{x+3} + \frac{B}{x+4}$$
Step 3
Clear Denominators
Multiply both sides by $(x+3)(x+4)$.
$$7x + 25 = A(x+4) + B(x+3)$$
Step 4
Solve for A (substitute x = −3)
Setting $x = -3$ eliminates the $B$ term.
$$7(-3) + 25 = A(-3+4) \implies 4 = A$$
Step 5
Solve for B (substitute x = −4)
Setting $x = -4$ eliminates the $A$ term.
$$7(-4) + 25 = B(-4+3) \implies -3 = -B \implies B = 3$$
Step 6
Final Result
Substitute the constants back.
$$\frac{7x+25}{(x+3)(x+4)} = \frac{4}{x+3} + \frac{3}{x+4}$$
Problem Statement
Resolve $\frac{x^2+x-1}{(x+2)^3}$ into partial fractions.
Repeated Linear FactorsEquating Coefficients
Strategic Toolkit
$$\frac{P(x)}{(x-a)^3} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3}$$
Step 1
Set Up Form
Factor $(x+2)$ has multiplicity 3, so we need three terms with ascending powers.
$$\frac{x^2+x-1}{(x+2)^3} = \frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{(x+2)^3}$$
Step 2
Clear Denominators
Multiply both sides by $(x+2)^3$.
$$x^2+x-1 = A(x+2)^2 + B(x+2) + C$$
Step 3
Find C (substitute x = −2)
Setting $x=-2$ eliminates $A$ and $B$ terms.
$$(-2)^2+(-2)-1 = C \implies 4-2-1 = C \implies C = 1$$
Step 4
Find A (equate x² coefficients)
Expand: $A(x^2+4x+4) + B(x+2) + C$. The only $x^2$ term comes from $A$.
$$\text{Coefficient of } x^2: A = 1$$
Step 5
Find B (equate x coefficients)
Coefficient of $x$: $4A + B = 1$.
$$4(1) + B = 1 \implies B = -3$$
Step 6
Final Result
Substitute all constants.
$$\frac{1}{x+2} - \frac{3}{(x+2)^2} + \frac{1}{(x+2)^3}$$
Problem Statement
Resolve $\frac{3x-11}{(x^2+1)(x+3)}$ into partial fractions.
Irreducible Quadratic FactorsEquating CoefficientsSubstitution
Strategic Toolkit
$$\frac{P(x)}{(x^2+c)(x-a)} = \frac{Ax+B}{x^2+c} + \frac{C}{x-a}$$
Step 1
Set Up Form
The factor $x^2+1$ is irreducible (discriminant $0-4 = -4 < 0$), so it gets a linear numerator.
$$\frac{3x-11}{(x^2+1)(x+3)} = \frac{Ax+B}{x^2+1} + \frac{C}{x+3}$$
Step 2
Clear Denominators
Multiply both sides by $(x^2+1)(x+3)$.
$$3x-11 = (Ax+B)(x+3) + C(x^2+1)$$
Step 3
Find C (substitute x = −3)
Setting $x = -3$ eliminates the $(Ax+B)$ group.
$$3(-3)-11 = C((-3)^2+1) \implies -20 = 10C \implies C = -2$$
Step 4
Find A and B (equate coefficients)
Expand and compare: $(A+C)x^2 + (3A+B)x + (3B+C)$.
$$x^2: A+C = 0 \implies A = 2 \\ x: 3A+B = 3 \implies B = 3-6 = -3$$
Step 5
Final Result
Substitute all constants.
$$\frac{2x-3}{x^2+1} - \frac{2}{x+3}$$
Problem Statement
Resolve $\frac{2x^3+x^2-x-3}{x(2x+3)(x-1)}$ into partial fractions.
Improper FractionsLong DivisionDistinct Linear Factors
Strategic Toolkit
$$\text{Improper} \to \text{Polynomial} + \text{Proper Fraction}$$
Step 1
Check Degrees
Numerator degree = 3. Denominator: $x(2x+3)(x-1) = 2x^3+x^2-3x$, also degree 3. Since deg(P) = deg(Q), this is improper.
$\text{deg}(2x^3+x^2-x-3) = 3 = \text{deg}(2x^3+x^2-3x)$
Step 2
Perform Long Division
Divide numerator by denominator. Quotient = 1, Remainder = $2x-3$.
$$\frac{2x^3+x^2-x-3}{2x^3+x^2-3x} = 1 + \frac{2x-3}{x(2x+3)(x-1)}$$
Step 3
Decompose the Proper Part
Set up the form for distinct linear factors.
$$\frac{2x-3}{x(2x+3)(x-1)} = \frac{A}{x} + \frac{B}{2x+3} + \frac{C}{x-1}$$
Step 4
Solve Constants by Substitution
Use $x=0$, $x=-\frac{3}{2}$, and $x=1$.
$$x=0: A=1 \quad x=-\tfrac{3}{2}: B=-\tfrac{8}{5} \quad x=1: C=-\tfrac{1}{5}$$
Step 5
Final Result
Combine the polynomial quotient with the decomposed remainder.
$$1 + \frac{1}{x} - \frac{8}{5(2x+3)} - \frac{1}{5(x-1)}$$
Memorize formulas
Distinct Linear Factors
Each unique linear factor gets its own constant numerator.
Formula
$$\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$
Repeated Linear Factors
Repeated factor of multiplicity $n$ produces $n$ terms with ascending powers.
Formula
$$\frac{P(x)}{(x-a)^n} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n}$$
Irreducible Quadratic Factor
Quadratic denominator requires a linear numerator $Ax+B$.
Formula
$$\frac{P(x)}{(x-r)(ax^2+bx+c)} = \frac{A}{x-r} + \frac{Bx+C}{ax^2+bx+c}$$
Repeated Quadratic Factor
Repeated irreducible quadratic produces terms with linear numerators and ascending powers.
Formula
$$\frac{P(x)}{(x^2+c)^2} = \frac{Ax+B}{x^2+c} + \frac{Cx+D}{(x^2+c)^2}$$
Review
1
The partial fraction decomposition of $\frac{x^2-5x+2}{(x-1)(x-2)(x-3)}$ is:
Ans
$\frac{-1}{x-1} + \frac{4}{x-2} - \frac{2}{x-3}$
Review
Tap to reveal
Set up the cover-up form
Because the denominator is a product of three distinct linear factors, write the fraction as $\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$ and solve for each constant by substituting the corresponding root.
$$x^2-5x+2=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$$
Evaluate at the roots
Substitute $x=1,2,3$ one at a time so all but one term vanish, then divide by the remaining factor product.
$$A=-1,\quad B=4,\quad C=-2$$
2
The decomposition of $\frac{x^2+x-1}{(x+2)^3}$ gives the coefficient of $\frac{1}{(x+2)^2}$ as:
Ans
$-3$
Review
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Write the repeated-factor form
A repeated linear factor of multiplicity 3 needs three terms, one for each power of $(x+2)$. The coefficient we want is the constant attached to $\frac{1}{(x+2)^2}$.
$$\frac{x^2+x-1}{(x+2)^3}=\frac{A}{x+2}+\frac{B}{(x+2)^2}+\frac{C}{(x+2)^3}$$
Match coefficients
After clearing denominators, substitute $x=-2$ to get $C=1$, then compare coefficients of $x^2$ and $x$ to isolate $A$ and $B$.
$$A=1,\quad B=-3,\quad C=1$$
3
To decompose $\frac{3x^2+1}{x-2}$, the first step is to perform long division. The result is:
Ans
$3x+6+\frac{13}{x-2}$
Review
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Divide the polynomials first
The numerator has higher degree than the linear denominator, so rewrite the rational expression as a polynomial plus a proper remainder term.
$$\frac{3x^2+1}{x-2}=3x+6+\frac{13}{x-2}$$
Check the remainder
Multiplying back gives $(x-2)(3x+6)=3x^2-12$, and adding the remainder 13 restores $3x^2+1$, so the quotient and remainder are consistent.
$$(x-2)(3x+6)+13=3x^2+1$$
4
The denominator $(x-2)^2(x^2-4)$ simplifies to:
Ans
$(x+2)(x-2)^3$
Review
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Factor the quadratic
Use the difference of squares to split $x^2-4$ into linear factors, then combine the repeated $(x-2)$ factors.
$$(x-2)^2(x^2-4)=(x-2)^2(x-2)(x+2)=(x+2)(x-2)^3$$
5
In the decomposition $\frac{5x-5}{(x^2+1)(x-2)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-2}$, the value of $C$ is:
Ans
$1$
Review
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Use the root that kills the other term
Substitute $x=2$ after clearing denominators so the $(Ax+B)(x-2)$ term vanishes and only the constant $C$ remains.
$$5(2)-5=C(2^2+1)\Rightarrow 5=5C$$
Solve for the coefficient
Divide both sides by 5 to isolate $C$.
$$C=1$$
6
The correct partial fraction form for $\frac{P(x)}{(x-1)(x^2+x+1)^2}$ is:
Ans
$\frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} + \frac{Dx+E}{(x^2+x+1)^2}$
Review
Tap to reveal
Assign one term per factor power
A distinct linear factor gets one constant numerator, while each power of a repeated irreducible quadratic gets its own linear numerator.
$$\frac{P(x)}{(x-1)(x^2+x+1)^2}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}+\frac{Dx+E}{(x^2+x+1)^2}$$
Count the unknowns
There is 1 constant from the linear factor and 2 linear numerators, each contributing 2 constants, so the total is 5.
$$1+2+2=5$$
7
Given $\frac{1}{(x-1)(2x-1)(3x-1)}$, the constant $A$ in $\frac{A}{x-1}$ equals:
Ans
$\frac{1}{2}$
Review
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Apply the cover-up rule
For the coefficient of $\frac{1}{x-1}$, set $x=1$ after removing the $(x-1)$ factor from the denominator.
$$A=\frac{1}{(2\cdot 1-1)(3\cdot 1-1)}=\frac{1}{2}$$
8
The partial fraction decomposition of $\frac{x^2+3x+4}{(x+1)(x+2)}$ is:
Ans
$1 + \frac{2}{x+1} - \frac{2}{x+2}$
Review
Tap to reveal
Remove the improper part first
Because the numerator and denominator have the same degree, divide to separate the whole-number term from the proper fraction.
$$\frac{x^2+3x+4}{(x+1)(x+2)}=1+\frac{2}{(x+1)(x+2)}$$
Split the proper fraction
Decompose the remainder into simple linear factors and use the roots to determine the constants quickly.
$$\frac{2}{(x+1)(x+2)}=\frac{2}{x+1}-\frac{2}{x+2}$$
9
In the decomposition of $\frac{4x^2}{(x^2+1)^2(x-1)}$, the constant $E$ in $\frac{E}{x-1}$ equals:
Ans
$1$
Review
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Isolate the linear factor coefficient
After clearing denominators, substitute the root $x=1$ so every term except the one with $E$ disappears.
$$4(1)^2=E(1^2+1)^2$$
Solve the resulting equation
Simplify the remaining arithmetic to get the coefficient on $\frac{1}{x-1}$.
$$4=4E\Rightarrow E=1$$
10
Which fraction is a proper rational fraction?
Ans
$\frac{9x^2}{x^3-1}$
Review
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Compare degrees, not factors
A rational fraction is proper only when the numerator’s degree is strictly less than the denominator’s degree.
$$\deg(9x^2)=2<\deg(x^3-1)=3$$
Select the only proper option
Only the first fraction satisfies the degree condition; the others are equal-degree or higher-degree cases.
$$\frac{9x^2}{x^3-1}$$
11
For the identity $7x+25 = A(x+4)+B(x+3)$, the value of $A+B$ is:
Ans
$7$
Review
Tap to reveal
Expand and match coefficients
Since the identity holds for all $x$, the coefficients of like powers of $x$ must match on both sides.
$$A(x+4)+B(x+3)=(A+B)x+(4A+3B)$$
Read off the x-coefficient
The coefficient of $x$ on the left is 7, so the sum of the constants must be 7.
$$A+B=7$$
12
The number of unknown constants needed to decompose $\frac{P(x)}{(x-1)^2(x+3)(x^2+4)}$ is:
Ans
$5$
Review
Tap to reveal
Count each factor contribution
A repeated linear factor of power 2 contributes two constants, a distinct linear factor contributes one, and an irreducible quadratic contributes two through a linear numerator.
$$\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+3}+\frac{Dx+E}{x^2+4}$$
Add the constants
Combine the counts from each factor type to get the total number of unknowns.
$$2+1+2=5$$
Question 1 of 12
0:00
1
The partial fraction decomposition of $\frac{x^2-5x+2}{(x-1)(x-2)(x-3)}$ is: