1st Year Math Chapter 5 Exercise 5.2 Notes: Partial Fractions — Quadratic Factors (Punjab Board 2026-27)
Complete concept notes, formulas, and solved examples for ICS/FSc Part 1 Mathematics, Punjab Board (PECTAA), Single National Curriculum 2026-27 session.Exercise 5.2 extends the partial fraction techniques from Exercise 5.1 to denominators containing irreducible quadratic factors — quadratics that cannot be broken down into real linear factors, distinct or repeated.
What Does This Exercise Cover?
This exercise teaches the correct partial fraction form to use when a quadratic factor can’t be factored further, and how to find the resulting constants by comparing coefficients.
Key Concepts
1. Irreducible Quadratic Factors
A quadratic factor is irreducible if it cannot be factored into real linear factors — equivalently, its discriminant is negative. For example, x² + 1 is irreducible, since x² + 1 = 0 has no real solutions.
2. Distinct Irreducible Quadratic Factor
When the denominator contains an irreducible quadratic factor, its numerator must be a linear expression (Ax + B), not just a constant, since a constant alone cannot match all the necessary terms when denominators are cleared.
3. Repeated Irreducible Quadratic Factor
When an irreducible quadratic factor is repeated, one (Ax + B)-style term is needed for each power of that factor, similar to how repeated linear factors are handled.
4. Finding the Unknown Constants
Since an irreducible quadratic factor has no real roots, the substitution shortcut from Exercise 5.1 cannot fully solve for its constants. Instead, constants are found by expanding both sides and comparing the coefficients of matching powers of x.
Partial Fraction Forms for Quadratic Factors
| Denominator Type | Partial Fraction Form |
| Distinct irreducible quadratic (x² + 1) | (Ax + B)/(x² + 1) |
| Repeated irreducible quadratic (x² + 1)² | (Ax + B)/(x² + 1) + (Cx + D)/(x² + 1)² |
Solved Examples
Example 1: Resolve (x + 3) / [(x − 1)(x² + 1)] into partial fractions.
Write (x + 3)/[(x − 1)(x² + 1)] = A/(x − 1) + (Bx + C)/(x² + 1).
Clear denominators: x + 3 = A(x² + 1) + (Bx + C)(x − 1).
Let x = 1: 1 + 3 = A(2), so A = 2.
Expand the right side: Ax² + A + Bx² − Bx + Cx − C = (A+B)x² + (C−B)x + (A−C).
Comparing x² coefficients: A + B = 0, so B = −2.
Comparing constants: A − C = 3, so C = A − 3 = −1.
Answer: 2/(x − 1) + (−2x − 1)/(x² + 1)
Example 2: Resolve (2x² + 3) / (x² + 1)² into partial fractions.
Write (2x² + 3)/(x² + 1)² = (Ax + B)/(x² + 1) + (Cx + D)/(x² + 1)².
Clear denominators: 2x² + 3 = (Ax + B)(x² + 1) + (Cx + D).
Expand: Ax³ + Ax + Bx² + B + Cx + D = Ax³ + Bx² + (A + C)x + (B + D).
Comparing x³: A = 0. Comparing x²: B = 2.
Comparing x: A + C = 0, so C = 0. Comparing constants: B + D = 3, so D = 1.
Answer: 2/(x² + 1) + 1/(x² + 1)²
Sample MCQs
1. A quadratic factor is called irreducible if:
a) It has two real roots b) It cannot be factored into real linear factors c) It equals zero d) It is always positive
Answer: b) It cannot be factored into real linear factors
2. For a distinct irreducible quadratic factor, the numerator in its partial fraction is of the form:
a) A b) Ax + B c) Ax² + B d) A/x
Answer: b) Ax + B
3. The most reliable way to find constants for a quadratic factor is to:
a) Substitute real roots only b) Compare coefficients c) Square both sides d) Take the derivative
Answer: b) Compare coefficients
4. A repeated irreducible quadratic factor (x² + 1)² requires how many partial fraction terms?
a) 1 b) 2 c) 3 d) 4
Answer: b) 2
5. Which of these is an irreducible quadratic?
a) x² − 4 b) x² − 1 c) x² + 1 d) x² − 9
Answer: c) x² + 1
Important Short Questions
- Define an irreducible quadratic factor, and give one example.
- Write the general partial fraction form for a denominator with a distinct irreducible quadratic factor.
- Write the general partial fraction form for a denominator with a repeated irreducible quadratic factor.
- Why can’t the substitution method alone be used to find constants for a quadratic factor?
- Explain how comparing coefficients helps find unknown constants in partial fractions.
Important Long Questions
- Resolve (x + 3)/[(x − 1)(x² + 1)] into partial fractions, showing the coefficient comparison in full.
- Resolve (2x² + 3)/(x² + 1)² into partial fractions.
- Resolve (5x² − 3x + 2)/[(x + 2)(x² + 4)] into partial fractions.
- Explain the difference between resolving a fraction with a repeated linear factor versus a repeated irreducible quadratic factor.
How to Approach This Exercise Effectively
- Check the discriminant of any quadratic factor first — a negative value confirms it’s irreducible and needs the (Ax+B) form.
- Combine substitution and coefficient comparison when a denominator has both a linear and a quadratic factor: substitute for the linear factor’s root, then compare coefficients for the rest.
- Write out the full expansion carefully before comparing coefficients — a lost term here is the most common source of errors in this exercise.
- Double-check by matching the highest power of x first (this often gives one constant immediately, like A = 0 in some cases).
- Practice recombining your partial fractions to confirm they simplify back to the original expression.
FAQs
Q: What makes a quadratic factor ‘irreducible’?
A: A quadratic factor is irreducible when its discriminant (b² − 4ac) is negative, meaning it cannot be broken down into two real linear factors.
Q: Why do quadratic factors need Ax + B instead of just A in the numerator?
A: A single constant can’t provide enough flexibility to match all the terms produced when denominators are cleared; the linear term Ax is necessary for the equation to balance.
Q: Can substitution be combined with comparing coefficients in the same problem?
A: Yes — if a denominator has both linear and quadratic factors, substitution can quickly solve for the linear factor’s constant, and comparing coefficients handles the rest.
Q: How can I tell a quadratic can’t be factored further without a discriminant check?
A: If no combination of factors of the constant term adds up to the middle coefficient, the quadratic likely doesn’t factor over the reals — though computing the discriminant directly is the reliable way to confirm this.
Notes prepared for Punjab Board (PECTAA) 1st Year Mathematics, 2026-27 session, Chapter 5: Partial Fractions, Exercise 5.2.