Lesson Notes By Weeks and Term v4 - SHS 2

Lesson Video

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Performance objectives

• Define and explain partial fractions.
• Resolve rational functions with linear factors.
• Resolve rational functions with irreducible quadratic factors.
• Apply these methods to simplify complex expressions.

Lesson summary

Welcome, students! Today, we are diving deeper into the world of algebraic fractions. We often learn how to combine several simple fractions into one complex fraction. But what if we need to do the reverse? What if we have a complex fraction and need to break it down into simpler, more manageable parts? This process is called decomposition into partial fractions. Why is this important? Imagine trying to understand a complex machine. It's easier to study each small component first before understanding the whole.

Lesson notes

Part 1: Recap of Basic Concepts

Before we tackle today's main topic, let's refresh our memory on what a rational function is and the simpler cases of partial fractions. Rational Function: A function that is a fraction where both the numerator and the denominator are polynomials. For example, `(3x + 1) / (x² - 4)`. Proper Rational Function: A rational function where the degree (highest power of x) of the numerator is *less than* the degree of the denominator. Example: `(2x + 5) / (x³ + x - 1)`. Degree of top is 1, degree of bottom is 3. Improper Rational Function: A rational function where the degree of the numerator is *greater than or equal to* the degree of the denominator. Example: `(x³ + 2x) / (x² - 1)`. Degree of top is 3, degree of bottom is 2.

Revision of Previous Cases: Distinct Linear Factors: If the denominator is `(ax + b)(cx + d)`, the form is: `A/(ax + b) + B/(cx + d)`. Repeated Linear Factors: If the denominator is `(ax + b)²`, the form is: `A/(ax + b) + B/(ax + b)²`. Part 2: New Concept - Denominators with Irreducible Quadratic Factors

An irreducible quadratic factor is a quadratic expression (like `ax² + bx + c`) that cannot be factorised into linear factors with real numbers. A quick way to check is to calculate its discriminant, `b² - 4ac`. If `b² - 4ac `A = 11/3` Let `x = -1`: `5(-1) + 1 = 0 + B(-1 - 2)` `-4 = -3B` => `B = 4/3` Combine Everything for the Final Answer: Remember the polynomial part from the long division! Final Answer: `(x + 3) + (11/3) / (x - 2) + (4/3) / (x + 1)` Or, written neatly: `x + 3 + 11 / (3(x - 2)) + 4 / (3(x + 1))`

Evaluation guide

• Resolve rational functions into partial fractions, specifically quadratic denominators and
• improper rational functions.
• Activity1: Working with Partial Fractions Collaborative Learning, Experiential Learning, Whole class discussion, Talk for Learning
• Approaches and Problem -based Learning