SHS1 Additional Mathematics · Semester 1, Week 17
Applications of Algebra
Lesson notes
Learning Objectives
Indicator: 1.1.2.LI.18 - Apply partial fraction decomposition up to factors with exponents and irreducible quadratic factors.
By the end of the lesson, learners can:
- Resolve a proper rational function with distinct linear factors in the denominator into partial fractions and determine the constants A, B and C.
- Decompose rational functions whose denominators contain repeated linear factors of the form (x - p)ⁿ into partial fractions with the correct form for each power.
- Express rational functions with irreducible quadratic factors in the denominator as partial fractions of the form (Ax + B)/(ax² + bx + c).
- Convert an improper rational function into a polynomial plus a proper rational function before decomposing into partial fractions.
- Apply partial fraction decomposition to solve problems involving equivalent expressions and verify results by comparing numerators.
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Sign in with phone numberCurriculum details
- Strand
- Modelling with Algebra (Strand 1)
- Sub-strand
- Applications of Algebra (1.2)
- Content standard
- 1.1.2.CS.1 - Demonstrate knowledge and understanding of applying algebraic processes and reasoning involving sequence, functions, and linear programming. 1.1.2.LO.1 Examine, analyse, determine and predict other terms in a pattern/sequence. 1.1.2.LO.2 Distinguish among various types of relations, find the domain and range of, and evaluate functions. 1.1.2.LO.3 Show that a function is injective (into) and/or surjective (onto). Find the inverse, describe the relationship between two variables and establish composite functions. 1.1.2.LO.4 Graph linear and quadratic functions and determine the intercepts. 1.1.2.LO.5 Find graphical and algebraic solutions to a system of three linear equations in three variables and apply them to solve real life problems. 1.1.2.LO.6 Perform algebraic manipulations on polynomial functions and graph polynomial functions. 1.1.2.LO.7 Find the domain, range, and zero of a rational function and state when it is undefined. 1.1.2.LO.8 Identify and describe the order of a matrix, the identity matrix and the zero matrix; find the determinant and perform basic arithmetic operations on 2 by 2 matrices (addition and subtraction).
- Indicator
- 1.1.2.LI.18 - Apply partial fraction decomposition up to factors with exponents and irreducible quadratic factors.
- Suggested placement
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Semester 1, Week 17
(Week 17 of the year)
Our suggestion, laid out in curriculum order across three terms of twelve weeks. NaCCA does not fix the week, so follow your school's scheme of learning.
- Source document note
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The official curriculum document has a fault in this entry, so the curriculum text above is transcribed exactly as printed:
- exemplars - p.120: adjacent duplicated CambriaMath characters were collapsed, but this PDF also maps some equation glyphs to the wrong letter or operator; verify every mathematical expression in this row against the rendered page
- Curriculum reference
- NaCCA curriculum document, p. 120
Exemplars (from the NaCCA curriculum)
Activity 1: Partial Fractions. Collaborative Learning, Experiential Learning, Whole class discussion, Talk for Learning Approaches and Problem-based Learning: - Learners work in groups/pairs to split rational functions into partial fractions. - Learners apply knowledge of equivalence to determine values of constants of functions. Example: Find the values of the constants A, B and C such that 𝑥 ) − 4𝑥 + 5 = 𝐴(𝑥 − 1)(𝑥 + 2) + 𝐵(𝑥 + 2)(𝑥 − 4) + 𝐶(𝑥 − 4)(𝑥 − 1) Learners resolve rational functions into partial fractions in which: - The denominator is made up of linear factors: 𝑎 − 𝑏 𝐴 𝐵 𝐶 = + + ... (𝑥 − 𝑝)(𝑥 + 𝑞)(𝑥 − 𝑟 ) ... 𝑥 − 𝑝 𝑥 + 𝑞 𝑥 − 𝑟 - The denominator is made up of repeated factors: 𝑎 − 𝑏 𝐴 𝐵 𝐶 = + + (𝑥 − 𝑝) " (𝑥 − 𝑝) (𝑥 − 𝑝) ) (𝑥 − 𝑝) " - The denominator contains irreducible quadratic factors: 𝑎 − 𝑏 𝐴 + 𝐵 = ) ) 𝑎 + 𝑏 + 𝑐 𝑎 + 𝑏 + 𝑐 - The function is an improper rational function. )0 ! $ ! Example: resolve (0*!)(0$)) in partial fractions. )0 ! $ ! ! " Solution: (0*!)(0$)) = 2 + (0*!) = (0$)) Teaching and Learning Resources: - Cut out shapes of different colours and orientation - Maths Technology learning apps, tools and devices - Chalkboard illustrations - Worksheets - Cut-out geometrical shapes of different colours and orientation Assessment (1.1.2.AS.18). The document marks these depth-of-knowledge levels for this indicator: Level 1 Recall; Level 3 Strategic reasoning.