1st Year Math Chapter 5 Exercise 5.1 Notes: Partial Fractions — Linear 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.
Chapter 5, Partial Fractions, follows Matrices and Determinants in the 14-unit Mathematics 11 (PECTAA) textbook. Exercise 5.1 introduces rational fractions and shows how to break a fraction apart into simpler pieces when its denominator has linear factors, distinct or repeated.
What Does This Exercise Cover?
This exercise teaches the reverse of combining fractions: starting from a single rational expression and splitting it back into a sum of simpler fractions, which is essential preparation for calculus topics later in the course.
Key Concepts
1. Rational Fractions
A rational fraction is an expression of the form P(x)/Q(x), where P(x) and Q(x) are polynomials and Q(x) ≠ 0.
2. Proper vs Improper Rational Fractions
A rational fraction is proper if the degree of P(x) is less than the degree of Q(x). It is improper if the degree of P(x) is greater than or equal to the degree of Q(x). An improper fraction must first be reduced by polynomial division into a polynomial plus a proper fraction before it can be resolved into partial fractions.
3. Distinct Linear Factors
When the denominator factors into distinct (non-repeating) linear factors, each factor gets its own single-term fraction with an unknown constant in the numerator.
4. Repeated Linear Factors
When a linear factor (x − a) is repeated n times in the denominator, the partial fraction expansion needs one term for every power of that factor from 1 up to n, each with its own unknown constant.
5. Finding the Unknown Constants
After clearing denominators, the unknown constants (A, B, C, …) can usually be found fastest by substituting the value of x that makes each linear factor zero.
Partial Fraction Forms for Linear Factors
| Denominator Type | Partial Fraction Form |
| Distinct linear factors (x − a)(x − b) | A/(x − a) + B/(x − b) |
| Repeated linear factor (x − a)² | A/(x − a) + B/(x − a)² |
| Repeated linear factor (x − a)³ | A/(x − a) + B/(x − a)² + C/(x − a)³ |
Solved Examples
Example 1: Resolve (x − 5) / [(x − 1)(x + 3)] into partial fractions.
Write (x − 5)/[(x − 1)(x + 3)] = A/(x − 1) + B/(x + 3).
Clear denominators: x − 5 = A(x + 3) + B(x − 1).
Let x = 1: 1 − 5 = A(4), so A = −1.
Let x = −3: −3 − 5 = B(−4), so B = 2.
Answer: −1/(x − 1) + 2/(x + 3)
Example 2: Resolve (3x + 1) / (x − 2)² into partial fractions.
Write (3x + 1)/(x − 2)² = A/(x − 2) + B/(x − 2)².
Clear denominators: 3x + 1 = A(x − 2) + B.
Let x = 2: 3(2) + 1 = B, so B = 7.
Comparing the coefficient of x: 3 = A.
Answer: 3/(x − 2) + 7/(x − 2)²
Example 3: Convert the improper fraction (x² + 1)/(x − 1) into a polynomial plus a proper fraction.
Divide x² + 1 by (x − 1): x² + 1 = (x − 1)(x + 1) + 2.
Dividing both sides by (x − 1) gives the polynomial part plus a remainder fraction.
Answer: (x² + 1)/(x − 1) = (x + 1) + 2/(x − 1)
Sample MCQs
1. A rational fraction P(x)/Q(x) is proper if:
a) degree(P) > degree(Q) b) degree(P) < degree(Q) c) degree(P) = degree(Q) d) Q(x) = 0
Answer: b) degree(P) < degree(Q)
2. Before resolving an improper fraction into partial fractions, you must first:
a) Factor Q(x) b) Perform polynomial division c) Find the roots of P(x) d) Multiply by Q(x)
Answer: b) Perform polynomial division
3. For a denominator with distinct linear factors (x − a)(x − b), the partial fraction form is:
a) A/(x − a)² + B/(x − b)² b) A/(x − a) + B/(x − b) c) (Ax + B)/[(x − a)(x − b)] d) A(x − a) + B(x − b)
Answer: b) A/(x − a) + B/(x − b)
4. For a repeated linear factor (x − a)², the partial fraction expansion includes:
a) A/(x − a) only b) A/(x − a) + B/(x − a)² c) A/(x − a)³ d) Ax/(x − a)
Answer: b) A/(x − a) + B/(x − a)²
5. The fastest way to find constants when the denominator has distinct linear factors is to:
a) Compare coefficients only b) Substitute the root of each factor c) Integrate both sides d) Square both sides
Answer: b) Substitute the root of each factor
Important Short Questions
- Define a proper rational fraction and an improper rational fraction.
- What must be done before resolving an improper fraction into partial fractions?
- Write the general partial fraction form for a denominator with two distinct linear factors.
- Write the general partial fraction form for a denominator with a repeated linear factor (x − a)².
- Resolve (x + 7)/[(x − 1)(x + 2)] into partial fractions.
Important Long Questions
- Resolve (x − 5)/[(x − 1)(x + 3)] into partial fractions, showing the substitution method in full.
- Resolve (3x + 1)/(x − 2)² into partial fractions.
- Convert (x² + 1)/(x − 1) into a polynomial plus a proper fraction using division, then resolve the remaining fraction into partial fractions.
- Resolve (2x − 1)/[(x + 1)(x − 2)²] into partial fractions, showing all steps for both the distinct and repeated linear factors.
How to Approach This Exercise Effectively
- Always check the degrees of the numerator and denominator first — this single check tells you whether division is needed before anything else.
- For distinct linear factors, substituting the root of each factor is much faster than comparing coefficients — use it whenever possible.
- For repeated factors, remember you need one fraction term for every power of the repeated factor, not just the highest one.
- After finding your constants, recombine the partial fractions over a common denominator as a check — it should simplify back to the original fraction.
- Practice a mix of distinct-factor and repeated-factor problems, since exam papers often combine both in the same question.
FAQs
Q: Why do improper fractions need division first?
A: Partial fraction decomposition only works cleanly on proper fractions; dividing first separates out the polynomial part, leaving a proper fraction to decompose.
Q: How do you know how many terms to include for a repeated factor?
A: You need one term for every power of the repeated factor, from the first power up to its full power in the denominator — for (x−a)³, that means three separate terms.
Q: What’s the fastest way to find the unknown constants?
A: Substituting the value of x that makes each linear factor zero eliminates all other unknowns at once, usually solving for one constant per substitution.
Q: Can this method be used for any denominator?
A: The linear-factor method covered here works whenever the denominator factors completely into real linear factors; denominators with irreducible quadratic factors need the approach covered in Exercise 5.2.
Notes prepared for Punjab Board (PECTAA) 1st Year Mathematics, 2026-27 session, Chapter 5: Partial Fractions, Exercise 5.1.