Raffles Institution Y4 WS4 Partial Fractions

Page 1 of 26 RAFFLES INSTITUTION RAFFLES PROGRAMME 2023 YEAR 4 MATHEMATICS TOPIC 2: REMAINDER & FACTOR THEOREMS AND PARTIAL FRACTIONS (MATH 1) WORKSHEET 4 Name: ( ) Class: 4 ( ) Date: WORKSHEET 4: PARTIAL FRACTIONS think! Add Math Textbook A Chapter 4 p.70 (1) INTRODUCTION (1.1) Proper and Improper Algebraic Fractions An algebraic fraction P( ) Q( ) x x , where P( )x and Q( )x are polynomials such that Q( ) 0x ≠ , is called a rational function. The table shows some examples of proper and improper algebraic fractions. KEY UNDERSTANDING(S) Students will understand that  Rational functions (algebraic fractions) can be expressed as partial fractions.  Partial fraction decomposition , which breaks down an algebraic fraction into simpler partial fractions, is useful in the integration process. LEARNER OUTCOMES At the end of this worksheet, students will be able to  Express proper algebraic fractions in partial fractions using an appropriate form based on what the denominator contains. (i) Distinct linear factors (ii) Repeated linear factors (iii) Quadratic factor (cannot be factorized)  Identify improper fractions and express it as a sum of a polynomial and a proper algebraic fraction first.

Page 2 of 26 Proper algebraic fractions 1 5 x− , 2 8 23 x xx − −+ , 4 5 42 3 27 xx x −+− + Improper algebraic fractions 3 2 8 46 x xx +− −+ , 43 2 95 xx x x −+ − , 2 x x+ , ( )( ) 269 31 2 xx xx ++ −+ Is P( ) Q( ) x x proper or improper if: (a) degree of P( ) degree of Q( )xx< ? (b) degree of P( ) degree of Q( )xx> ? (c) degree of P( ) degree of Q( )xx= ? (1.2) What are Partial Fractions? Early in Algebra, you learn how to combine “simple” algebraic fractions into single algebraic fraction. For example, ( ) ( ) ( )( ) ( )( ) 2 13 223 21 21 54 21 xx x x xx x xx ++ −+=−+ −+ −= −+ The Method of Partial Fractions does the opposite. It dissects a single algebraic fraction into a sum of single proper fractions. While this is a little more complicated than going the other direction, it is also more useful. Applications of the method of partial fractions include: • Integration of rational functions in Calculus • Integration of the secant function in Mercator map projection (widely used in navigation) • Fractional radioactive decay law and Bateman equations in nuclear engineering • Minimum payments due on credit card bills

Page 3 of 26 (2) PROPER FRACTION WITH DISTINCT LINEAR FACTORS IN DENOMINATOR We first consider proper algebraic fraction whose denominator can be factorised completely into n distinct linear factors. EG 1 Express ( )( ) 95 32 5 x xx − −+ in partial fractions. Steps: 1. Express the fraction as a sum of its components in the form A ax b+ for each distinct linear factor in the denominator. 2. Find the values of unknown constants by “removing” the denominators on both sides and solve the identity using the methods we learnt in Worksheet 1. 3. Express the original algebraic fraction in partial fractions. Computational Thinking - Opportunities for Algorithmic Thinking - Write out the general approach as a sequence of steps - Work out the rules for partial fractions Rule 1: For every linear factor of the form ( )ax b+ in the denominator, there will be a component of the form A ax b+ , where A is a constant.

Page 4 of 26 EG 2 Express each of the following in partial fractions. (a) ( )( )( ) 21 123 x xx x + −++ (b) 2 1 12 xx−−

Page 5 of 26 Note: 1. Remember to factorise the denominator completely. 2. Do not expand the denominator in the answer. EG 3 think! Add Math Textbook A p.71 Practice Now 13 Q2 (i) Factorise completely the cubic polynomial 322 3 17 12xx x+−+ . (ii) Express 2 32 7 25 8 2 3 17 12 xx xx x −+ +−+ as a sum of 3 partial fractions.

Page 6 of 26 HOMEWORK 1 LEVEL 2 1. Express each of the following in partial fractions. (a) 2 8 51 9 64 x x + − (b) 2 7 23xx+ [Ans: (a) 89 38 38xx++− (b) ( ) 7 14 3 32 3xx− + ]

Page 7 of 26 (3) COVER-UP METHOD The cover-up method is a faster technique in finding unknown constants in partial fraction s. We can only apply this method when the denominator is a product of linear factors. For example, if the denominator has three distinct linear factors, we have ( )( )( ) f( )x ABC xaxbxc xa xb xc ≡+−−−− −−− Then by cover -up method, A can be computed by covering up the term ( )xa− in the denominator on the LHS and substituting xa= in the remaining expression: ( )( ) f( )aA abac= −− This works because the computation is equivalent to multiplying the expression throughout by the term ( )xa− and then making the substitution xa= . Similarly, by substituting xb= and xc= , we can compute B and C : ( )( ) f( )bB babc= −− , ( )( ) f( )cC cacb= −− Using EG 3: ( )( )( ) ( )( ) ( ) ( )( ) ( ) 2 2 2 2 7 25 8 1 4 23 1 4 23 By cover-up method: 7(1) 25(1) 8 10Sub. 1: 21423 51 7( 4) 25( 4) 8 220Sub. 4: 44 1 8 3 5 11 337 25 83 22Sub. : 332 122 x x AB C xx x x x x xA xB xC −+ ≡+−−+ − −+ − −+−= = = =+− − − − −+= −= = =−− −− − −  −+  = =  − 2 32 55 4 51 114 22 7 25 8 2 4 5Hence 2 3 17 12 1 4 2 3 xx x x xx xx − = =−  +     −+ ≡+−+ − +− +−

Page 8 of 26 EG 4 Express each of the following in partial fractions using the cover-up method. (a) ( )( ) 1 5 14 1xx−+ (b) 2 10 11 2 35 x xx − −−

Page 9 of 26 EG 5 (i) Express ( ) 1 2xx + in partial fractions. (ii) Hence, find the exact value of 111 1 ... 1 3 2 4 3 5 18 20+ + ++××× × .

Page 10 of 26 HOMEWORK 2 LEVEL 1 1. Express each of the following in partial fractions. (a) ( )( ) 34 22 1 x xx − +− (b) ( )( ) 3 21 2xx−+ (c) 2 41 34 x xx + +− [Ans: (a) 21 22 1xx −+− (b) ( ) ( ) 63 52 1 5 2xx −−+ (c) 31 41xx ++− ] LEVEL 2 1. Express each of the following in partial fractions. (a) 2 43 2 53 x xx − −− (b) 32 2 43x xx−− [Ans: (a) ( ) ( ) 10 9 72 1 7 3xx ++− (b) ( ) ( ) 2 32 2 54 1 5 1xx x−+ + +− ]