1st Year Math Chapter 1 Exercise 1.3 Notes: Complex Polynomials as a Product of 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.
Exercise 1.3 connects complex numbers to polynomial equations, showing that every polynomial can be completely factored into linear factors once complex roots are allowed, and that complex roots of real-coefficient polynomials always come in conjugate pairs.
What Does This Exercise Cover?
This exercise uses the Fundamental Theorem of Algebra and the Factor Theorem together to break polynomials down completely, including cases where some or all of the roots are complex numbers.
Key Concepts
1. The Fundamental Theorem of Algebra
Every polynomial of degree n (n ≥ 1) with complex coefficients has exactly n roots, counting multiplicity, when complex numbers are allowed as solutions. This guarantees that any polynomial can be written as a product of exactly n linear factors.
2. Conjugate Root Pairs
If a polynomial has only real coefficients and a + bi (with b ≠ 0) is one of its roots, then a − bi must also be a root. This means non-real roots always appear in conjugate pairs for such polynomials — never alone.
3. The Factor Theorem
If x = r is a root of a polynomial p(x), then (x − r) is a factor of p(x). This applies equally whether r is real or complex, and is the main tool used to break a polynomial down once its roots are known.
4. Sum and Product of Roots
For a polynomial written as xⁿ + c₁xⁿ⁻¹ + … + cₙ, relationships between the coefficients and the sum/product of all roots can help find a missing root once the others are known, without fully dividing the polynomial.
Solved Examples
Example 1: Given that 1 + i is a root of x³ − 4x² + 6x − 4 = 0, find all the roots and write the polynomial as a product of linear factors.
Since the coefficients are real, the conjugate 1 − i must also be a root.
For x³ − 4x² + 6x − 4, the sum of all three roots equals 4 (from the coefficient of x²).
Sum of the two known roots: (1 + i) + (1 − i) = 2, so the third root = 4 − 2 = 2.
The three roots are 2, 1 + i, and 1 − i.
As linear factors: (x − 2)(x − (1+i))(x − (1−i)) = (x − 2)(x² − 2x + 2).
Answer: x³ − 4x² + 6x − 4 = (x − 2)(x² − 2x + 2), with roots 2, 1+i, 1−i
Example 2: Factor x² + 9 completely into linear factors over the complex numbers.
Set x² + 9 = 0, so x² = −9, giving x = ±3i.
The two roots, 3i and −3i, are a conjugate pair, as expected for a real-coefficient polynomial.
Answer: x² + 9 = (x − 3i)(x + 3i)
Sample MCQs
1. If a polynomial has real coefficients and a + bi is a root, another root must be:
a) a + bi b) a − bi c) −a + bi d) −a − bi
Answer: b) a − bi
2. According to the Fundamental Theorem of Algebra, a polynomial of degree n has exactly:
a) n − 1 roots b) n roots c) n + 1 roots d) only 1 root
Answer: b) n roots
3. The Factor Theorem states that if x = r is a root of p(x), then:
a) (x + r) is a factor b) (x − r) is a factor c) p(r) is undefined d) r must be real
Answer: b) (x − r) is a factor
4. Non-real (complex) roots of a real-coefficient polynomial always occur:
a) Singly b) In conjugate pairs c) As exactly three roots d) Never
Answer: b) In conjugate pairs
5. A polynomial of degree 4 always has, counting multiplicity:
a) 1 root b) 2 roots c) 3 roots d) 4 roots
Answer: d) 4 roots
Important Short Questions
- State the Fundamental Theorem of Algebra.
- Explain why complex roots of a real-coefficient polynomial occur in conjugate pairs.
- State the Factor Theorem in your own words.
- If 3 − 2i is a root of a real-coefficient polynomial, what other root must it have?
- How many linear factors does a degree-5 polynomial have over the complex numbers?
Important Long Questions
- Given that 1 + i is a root of x³ − 4x² + 6x − 4 = 0, find all roots and write the polynomial as a product of linear factors.
- Explain, with reasoning, why complex roots of a real-coefficient polynomial must occur in conjugate pairs.
- If 2 − 3i is one root of a real-coefficient cubic polynomial whose roots sum to 4, find the third (real) root.
- Factor x² + 9 completely into linear factors over the complex numbers.
How to Approach This Exercise Effectively
- Whenever a complex root is given for a real-coefficient polynomial, immediately write down its conjugate as a second root.
- Use sum-of-roots and product-of-roots relationships to find a missing real root quickly, instead of always dividing the full polynomial.
- Multiply out conjugate factor pairs like (x − (a+bi))(x − (a−bi)) into x² − 2ax + (a² + b²) — memorize this shortcut.
- Practice factoring simple expressions like x² + k (k > 0) into (x − i√k)(x + i√k) until it becomes automatic.
- Always double-check your final factorization by multiplying the factors back out.
FAQs
Q: What is the Fundamental Theorem of Algebra?
A: It states that every polynomial of degree n (n ≥ 1) has exactly n roots when complex numbers are allowed, guaranteeing it can be fully factored into n linear factors.
Q: Why must complex roots come in conjugate pairs?
A: Because when a real-coefficient polynomial is evaluated at a + bi and at its conjugate a − bi, the imaginary parts cancel in a mirrored way, so if one is a root the other automatically satisfies the equation too.
Q: Can a real-coefficient polynomial have an odd number of non-real roots?
A: No — since non-real roots always occur in conjugate pairs, the total count of non-real roots must always be even.
Q: How does the Factor Theorem help in this exercise?
A: Once a root (real or complex) is known, the Factor Theorem confirms that (x − root) is a factor, which can be divided out to reduce the polynomial’s degree and find the remaining roots.
Notes prepared for Punjab Board (PECTAA) 1st Year Mathematics, 2026-27 session, Chapter 1: Complex Numbers, Exercise 1.3.