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: D