Maths Extension 1 Year 11 - Module 2 - Lesson 5
Finding Zeros of Polynomials
Name
Date
Class
1. Lesson focus
Rational root theorem and polynomial zeros. For an integer-coefficient polynomial, a rational root p/q in lowest terms has p dividing the constant term and q dividing the leading coefficient.
Success criteria
- Apply the lesson method accurately in a routine example.
- Show each algebraic or domain step and check the result.
- Explain why the method and restrictions fit the question.
2. Worked practice
2. Factor x³−2x²−5x+6 and state all zeros.
3. Use the rational root theorem for P(x)=2x³−5x+12 to list all distinct rational-root candidates.
3. Challenge
Use the rational root theorem to give the possible rational roots of 2x³−3x²−8x+12.
4. Multiple choice
For x³−2x²−5x+6, which is a zero?
A. 1
B. 4
C. −4
D. 6
Worked answer key
Compare each step with your working. Keep stated intervals, domains and restrictions in your final answer.
1. List the possible integer roots of x³−2x²−5x+6.
Integer candidates divide 6: ±1, ±2, ±3, ±6.
2. Factor x³−2x²−5x+6 and state all zeros.
It factors as (x−1)(x−3)(x+2); zeros are 1,3,−2.
3. Use the rational root theorem for P(x)=2x³−5x+12 to list all distinct rational-root candidates.
p divides 12 and q divides 2; after reducing duplicates the candidates are ±1/2, ±1, ±3/2, ±2, ±3, ±4, ±6 and ±12.
Challenge. Use the rational root theorem to give the possible rational roots of 2x³−3x²−8x+12.
p divides 12 and q divides 2, so candidates are ±1,±2,±3,±4,±6,±12 and their halves ±1/2,±3/2.
Multiple choice. For x³−2x²−5x+6, which is a zero?
Correct answer: A. Substitution gives 1−2−5+6=0; the factorisation is (x−1)(x−3)(x+2).
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