Raffles Institution Y4 WS3 Cubic Equations

Page 1 of 18 RAFFLES INSTITUTION RAFFLES PROGRAMME 2023 YEAR 4 MATHEMATICS TOPIC 2: REMAINDER & FACTOR THEOREMS, PARTIAL FRACTIONS (MATH 1) WORKSHEET 3 Name: ( ) Class: 4 ( ) Date: WORKSHEET 3: CUBIC EQUATIONS think! Add Math Textbook A Chapter 4 p.65 (1) INTRODUCTION In general, a cubic equation 32 0 ( 0)ax bx cx d a+ ++= ≠ has at least one real root and at most 3 real roots as shown in the diagram below. No of real roots = 1 No of real roots = 2 No of real roots = 3 Eg: 3 20x −= has 1 real root as the graph of 3 2yx= − intersects the x-axis at only 1 point Eg: 322 8 80xxx− += has 1 repeated real root and 1 real root as the graph of 32288yx x x=−+ touches the x-axis at one point and intersects at another point Eg: 32 2 5 60xxx− − += has 3 real roots as the graph of 32 2 56yx x x=− −+ intersects the x-axis at 3 points Question: Is it possible for a cubic equation to have no real roots? KEY UNDERSTANDING(S) Students will understand that • Remainder and Factor Theorems can be used to factorise cubic expressions and hence solve cubic equations. LEARNER OUTCOMES At the end of the worksheet, students will be able to • Factorize cubic expressions or solve cubic equations using the Remainder and Factor Theorems

Page 2 of 18 (2) CUBIC EQUATIONS Factor Theorem can be used to factorise quadratic expressions with integer coefficients so as to solve the corresponding quadratic equations as follow: Eg To solve quadratic equation 2 2 30xx− −= , we factorise the quadratic expression and obtain two linear factors which has integer coefficients: 2 2 3 = ( 3)( 1)xx x x−− − + Hence, ( )( ) 2 2 30 3 10 3 or 1 xx xx x − −= + −= = − Similarly, we can apply Factor Theorem to solve cubic equations. Eg To solve cubic equation 32 2 5 60xxx− − += , we first find the 1st factor: 1st factor is ( 1)x− We factorise the cubic expression into a linear and a quadratic factors, 32 22 5 6 ( 1)( 6)x x x x xx− − += − −− We then further factorise the quadratic expression and obtain two linear factors which has integer coefficients if it is factorisable, i.e., ( )( ) 2( 6) 2 3xx x x−− = + − Hence, 32 2 5 60 ( 1)( 2)( 3) 0 1, 2 or 3 xxx xx x x − − += − + −= = − Question: How to find the first factor? EG 1 Solve the cubic equation 32 4 60x xx− ++= Step 1: Use trial and error and the factor theorem to determine a linear factor 32() 4 6fx x x x= − ++ .

Page 3 of 18 Interesting point to note: For the trial factor xk− where k is integer, k is always a factor of the constant term of the cubic expression. For example, if ()xk− is a factor of 32() 4 6fx x x x= − ++ , then 2( ) ( )( )f x xkx b xc=− ++ and 6kc−= (constant term). Thus, the possible values of k are 1, 2, 3 and 6±± ± ± i.e. k is a factor of 6. (Integral Root Theorem) Step 2: Determine the other factor(s) of ()fx by division. Note: A cubic polynomial can be expressed as a product of 1 linear and 1 quadratic factor or as a product of 3 linear factors. In other words, all cubic polynomials have at least one linear factor. Use Desmos or other graph plotting apps to visualise this in a graphical form and convince yourself of its validity. Step 3: Factorise ()fx completely and solve () 0fx = . Note: The above method works only when you are given the first root / factor or you can use trial and error to find the first integral root.

Page 4 of 18 EG 2 Factorise the expression 322 5 4 12xxx+ −− completely. EG 3 The expression 32 5 10px x qx− ++ has factor 2x – 1 but leaves a remainder of –20 when divided by x + 2. Find the values of p and q and factorise the expression completely.

Page 5 of 18 EG 4 Solve the equation 322 5 7 12 0xxx+ −−= . Hence sketch the graph of 322 5 7 12yx x x= + −− , showing the intercepts with the axes and solve the inequality 322 5 7 12 0xxx+ −−> . Blended Learning Online Activity (OPTIONAL): Students to access HeyMath! online lesson: “Year 4 Maths – Remainder and Factor Theorem: “Solving Cubic Equations – Summary and EG 1” [6:46] to consolidate learning before doing Homework 1. HOMEWORK 1 LEVEL 1 1 Solve the equation (a) 322 7 7 30 0xxx− −+= (b) 322 9 20 12xx x+=+ (c) 322 5 40xx− += 2 The expression 3224x ax bx+ ++ is divisible by 1x− but leaves a remainder of 3 when divided by 2x+ . Find the values of a and b and factorise the expression. 3 Find the x-coordinate of each of the three points of intersection of the curves 265yx= − and 617yx x= − . [Ans: (1a) 12, 2 or 32− (b) 16, or 22−− (c) 112 or 1744± (2) 11, 622ab= =− (3) 21, or 332− ]

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Page 7 of 18 LEVEL 2 1 Given that 4 3 224 12 5 1x x a x ax− − −− is exactly divisible by 2xa+ . (a) show that 323 5 20aa+ −= , (b) find the possible values of a.

Page 8 of 18 2 The functions ( )fx and ( )gx are defined as 43 2 4 32 ( ) 3 12 2 4, ( ) 2 8 2. fx x x x x gx x x x x =+ − ++ = + − +− (i) Solve completely the equation ( ) ( ) 0f x gx−= . (ii) ( )fx and ( )gx have a common factor xa− . Find the value of a. [Ans: (1b) 171 or 3a −±=− (2) 1, 2 or 3x=− , 2a= ]

Page 9 of 18 (3) SUM AND DIFFERENCE OF CUBES Recall: Difference of two Squares 22 ( )( )a b a ba b−=+ − How about difference of two cubes and sum of two cubes? Are you able to visualise how a3 - b3 and a3+b3 look like geometrically? Blended Learning Online Activity (SDL): Students to access HeyMath! online lesson: “Year 4 Maths – Remainder and Factor Theorem: Sum and Difference of Cubes – Geometric Method” [5:24] and then complete EG 5. EG 5 Use Remainder and Factor Theorem to prove (a) ( )( ) 33 2 2a b a b a ab b+=+ −+ , Hence, prove (b) ( )( ) 33 2 2a b aba a bb−=− ++ .

Page 10 of 18 (4) FURTHER EQUATIONS Blended Learning Online Activity (SDL): Students to access HeyMath! online lesson: “Year 4 Maths – Remainder and Factor Theorem: “Solving Cubic Equations – EG 2” [3:18] and then complete EG 6. EG 6 Given that 2 32xx−+ is a factor of 4 322 12x px x qx+ ++− , find the values of p and q. Hence solve the equation 4 322 12 0x px x qx+ ++−= .