Additional Mathematics Synthesis Guide 27: Partial Fractions — Repeated Linear Factors, Quadratic Factors and Reconstruction

Factor denominator → check properness → write complete template → solve constants → reconstruct → preserve domain.

1. Start with the denominator, not the numerator

For

(5x+1)/[(x−1)(x+2)],

the denominator has two distinct linear factors. The correct decomposition template is

A/(x−1)+B/(x+2).

The constants A and B are not guessed from the numerator. The factor pattern of the denominator determines the allowed pieces.

2. Properness must be checked first

A rational expression P(x)/Q(x) is proper when degree P<degree Q.

If it is improper, perform polynomial division first. For example,

(x²+3x+5)/(x+1)=x+2+3/(x+1).

The polynomial part is separated before any further partial-fraction decomposition.

Trying to decompose an improper expression directly into only proper simple fractions creates an incomplete representation.

3. Distinct linear factors

Decompose

(5x+1)/[(x−1)(x+2)].

Write

(5x+1)/[(x−1)(x+2)] = A/(x−1)+B/(x+2).

Multiply through:

5x+1=A(x+2)+B(x−1).

Set x=1: 6=3A, so A=2. Set x=−2: −9=−3B, so B=3.

Therefore

(5x+1)/[(x−1)(x+2)] = 2/(x−1)+3/(x+2).

4. Why strategic substitution works

After clearing denominators, choosing a root of one denominator factor makes one term survive while the others vanish. This is not a separate theorem; it is simply a convenient evaluation of the polynomial identity.

It works especially well for distinct linear factors. When repeated factors or irreducible quadratic factors appear, coefficient matching may still be needed.

5. Repeated linear factors need every power

If the denominator contains

(x−a)²,

the template must contain both

A/(x−a)+B/(x−a)².

For a power (x−a)³, include terms with denominators (x−a), (x−a)² and (x−a)³.

Skipping an intermediate power makes the decomposition space too small to represent the original numerator in general.

6. Worked repeated-factor example

Decompose

(3x+5)/[(x−1)²(x+2)].

Write

A/(x−1)+B/(x−1)²+C/(x+2).

Clearing denominators:

3x+5=A(x−1)(x+2)+B(x+2)+C(x−1)².

Set x=1: 8=3B, so B=8/3.

Set x=−2: −1=9C, so C=−1/9.

Use x=0:

5=−2A+2(8/3)−1/9.

This gives A=1/9.

Hence

(3x+5)/[(x−1)²(x+2)] = 1/[9(x−1)] + 8/[3(x−1)²] − 1/[9(x+2)].

7. A quadratic factor needs a linear numerator

If a quadratic factor such as x²+4 does not factor over the reals, its partial-fraction numerator must be one degree lower:

(Bx+C)/(x²+4).

Using only a constant numerator would generally be too restrictive because recombination must be able to produce both x and constant contributions from that quadratic block.

8. Worked linear-plus-quadratic example

Decompose

(2x²+3x+5)/[(x+1)(x²+4)].

Use

A/(x+1)+(Bx+C)/(x²+4).

Clearing denominators:

2x²+3x+5=A(x²+4)+(Bx+C)(x+1).

Set x=−1:

4=5A, so A=4/5.

Expand the remaining identity and compare coefficients. The result is B=6/5 and C=9/5.

Therefore

(2x²+3x+5)/[(x+1)(x²+4)] = 4/[5(x+1)] + (6x+9)/[5(x²+4)].

9. Coefficient matching is exact, not approximate

When two polynomials are equal for every x, their corresponding coefficients must match.

For an identity

2x²+3x+5=(A+B)x²+(B+C)x+(4A+C),

we obtain the exact system

• A+B=2
• B+C=3
• 4A+C=5

This is a polynomial identity, not a fit through a few selected data points.

10. Strategic substitution and coefficient matching can be combined

A good solution does not have to choose one technique exclusively. In the quadratic example, x=−1 immediately found A. Coefficient matching then found B and C efficiently.

Use substitutions that eliminate terms where possible, then compare coefficients for the remaining unknowns.

11. Repeated quadratic factors follow the same completeness principle

If an enrichment problem contains an irreducible quadratic factor Q(x) repeated twice, a complete real decomposition has terms

(Ax+B)/Q(x) + (Cx+D)/[Q(x)]².

The principle is the same as repeated linear factors: include every power up to the multiplicity, with a numerator of degree lower than the factor.

If this form lies beyond the learner’s current assessed scope, treat it as a structural extension rather than an exam claim.

12. Distinct quadratic factors need separate linear numerators

For

P(x)/[(x²+1)(x²+4)],

a complete real template is

(Ax+B)/(x²+1)+(Cx+D)/(x²+4).

Each quadratic block receives its own general linear numerator.

13. Reconstruction is the strongest routine check

After finding a decomposition, recombine the fractions over the original common denominator.

If

2/(x−1)+3/(x+2)

is recombined, the numerator is

2(x+2)+3(x−1)=5x+1.

This restores the original expression exactly.

A decomposition is complete only when recombination returns the original rational function over its original domain.

14. Domain restrictions remain unchanged

If the original denominator is (x−1)(x+2), then x≠1 and x≠−2.

The decomposed form has the same restrictions. Partial fractions changes representation; it does not create values at excluded points.

When cancellation occurs before decomposition, retain any exclusions inherited from the original expression.

15. Cancellation should happen before decomposition

Suppose

(x²−1)/[(x−1)(x+2)].

Factor and cancel:

(x+1)/(x+2), with x≠1,−2.

Then divide:

1−1/(x+2).

Trying to decompose the uncancelled form hides the removable factor and creates unnecessary constants.

16. Improper expressions need polynomial division

Decompose

(x²+5x+8)/[(x+1)(x+2)].

The denominator expands to x²+3x+2, so degrees are equal. Divide:

(x²+5x+8)/(x²+3x+2)=1+(2x+6)/(x²+3x+2).

Then decompose the proper remainder:

(2x+6)/[(x+1)(x+2)] = 4/(x+1)−2/(x+2).

So the full expression is

1+4/(x+1)−2/(x+2).

17. Polynomial division can expose asymptotic structure

In the previous example, the rational expression is written as

1 + terms that tend to 0 as |x| becomes large.

This makes the large-|x| behaviour visible: the function approaches y=1. Partial fractions can therefore reveal graph structure as well as simplify algebra.

18. Repeated-factor coefficients have different roles

In

A/(x−a)+B/(x−a)²,

the B term dominates more strongly near x=a because 1/(x−a)² grows in magnitude faster than 1/(x−a).

The two terms are algebraically necessary and they also describe different local behaviours near the repeated denominator root.

19. A missing template term causes coefficient contradictions

If a learner tries to write

(3x+5)/[(x−1)²(x+2)] = A/(x−1)+C/(x+2)

without B/(x−1)², clearing denominators gives too few degrees of freedom to match the original numerator identity.

An inconsistent coefficient system is often evidence that the decomposition template was incomplete, not that arithmetic alone failed.

20. Partial fractions as representation choice

The original rational form is often best for identifying zeros, denominator restrictions or a compact model. The decomposed form is often best for local denominator behaviour, parameter comparison or later operations.

No representation is universally superior. The value of decomposition is that it exposes structure that the combined fraction hides.

21. Common failure patterns

• Decomposing before checking whether the rational expression is proper.
• Missing one of the powers for a repeated linear factor.
• Using a constant numerator over an irreducible quadratic factor.
• Forgetting to factor the denominator fully before writing the template.
• Assuming strategic substitution will find every unknown without coefficient matching.
• Losing original domain exclusions after cancellation.
• Accepting constants without recombining the decomposition.
• Expanding everything immediately and losing the denominator structure that tells you the correct form.

22. A reliable partial-fraction routine

1. Factor the denominator completely over the intended number system.
2. Record domain exclusions.
3. Cancel common factors if legal, while retaining original exclusions.
4. If the fraction is improper, divide first.
5. Write one term for each distinct linear factor and every power of repeated factors.
6. Use a general linear numerator over an irreducible quadratic factor.
7. Clear denominators and solve constants by substitutions and/or coefficient matching.
8. Recombine exactly to verify the result.

23. Practice set

1. Decompose (3x+4)/[(x−1)(x+2)].
2. Decompose (7x−1)/[(x+1)(x−3)].
3. State the correct template for P(x)/[(x−2)²(x+1)].
4. State the correct template for P(x)/[(x−2)³(x+1)].
5. Decompose (3x+5)/[(x−1)²(x+2)].
6. State the correct template for P(x)/[(x+1)(x²+9)].
7. Why is the numerator over x²+9 linear?
8. Decompose (2x²+3x+5)/[(x+1)(x²+4)].
9. What must be done first to (x²+5x+8)/[(x+1)(x+2)]?
10. Decompose the expression in Question 9 fully.
11. Simplify (x²−1)/[(x−1)(x+2)] before decomposition and state original exclusions.
12. Why must B/(x−a)² appear when (x−a)² is a denominator factor?
13. Give one reason strategic substitution is efficient for distinct linear factors.
14. Give one reason coefficient matching is still needed in some quadratic-factor problems.
15. What does reconstruction mean in partial fractions?
16. If a decomposition recombines to the wrong x² coefficient, what should you inspect?
17. State the correct enrichment template for P(x)/[(x²+1)²(x−2)].
18. Why should improper fractions be divided before decomposition?
19. What happens to the domain after partial-fraction decomposition?
20. Give one graph-related insight polynomial division can reveal.

Answers

1. Let A/(x−1)+B/(x+2). Then x=1 gives A=7/3 and x=−2 gives B=−2/3.
2. A/(x+1)+B/(x−3): x=−1 gives A=2; x=3 gives B=5.
3. A/(x−2)+B/(x−2)²+C/(x+1).
4. A/(x−2)+B/(x−2)²+C/(x−2)³+D/(x+1).
5. 1/[9(x−1)] + 8/[3(x−1)²] − 1/[9(x+2)].
6. A/(x+1)+(Bx+C)/(x²+9).
7. Because the numerator degree must be lower than the quadratic denominator degree while remaining general enough to match x and constant coefficients.
8. 4/[5(x+1)] + (6x+9)/[5(x²+4)].
9. Polynomial division, because numerator and denominator have equal degree.
10. 1+4/(x+1)−2/(x+2).
11. (x+1)/(x+2)=1−1/(x+2), with original x≠1,−2.
12. A repeated factor requires a separate term for every denominator power up to its multiplicity.
13. Substituting a factor root eliminates all terms containing that factor, isolating one coefficient.
14. The quadratic factor has no real root to isolate both numerator coefficients, so comparing polynomial coefficients supplies the remaining equations.
15. Combining the partial fractions over the original common denominator and recovering the original numerator exactly.
16. The template, signs, arithmetic and coefficient equations.
17. (Ax+B)/(x²+1)+(Cx+D)/(x²+1)²+E/(x−2).
18. A pure sum of proper simple fractions cannot represent the polynomial part of an improper rational expression.
19. It remains the same as the original function’s legal domain, including any inherited exclusions.
20. It can expose the polynomial or constant asymptote approached as the proper remainder tends to zero.

24. What mastery looks like

Mastery means the learner can read a factored denominator and immediately write a complete decomposition template, distinguish proper from improper rational expressions, use repeated-factor and quadratic-factor numerator rules correctly, combine strategic substitution with coefficient matching, and verify the final representation by exact reconstruction.

The transfer test is to change only the denominator architecture. If the learner can adapt the template correctly when a factor repeats or a quadratic block appears, rather than recycling the distinct-linear formula, partial fractions has become structural reasoning.

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Continue through Batch 07

• Guide 25: Calculus Rule Selection — Rational Powers, Product, Quotient and Chain Rules
• Guide 26: Connected Rates of Change — Geometry, Variables, Units and Constraints
• Guide 28: Principal Values, Inverse Trigonometric Functions, Graphs, Symmetry and Interval Solutions
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