A4 - POLYNOMIALS AND PARTIAL FRACTIONS

A4: POLYNOMIALS AND PARTIAL FRACTIONS ● Multiplication and division of polynomials ● Use of remainder and factor theorems, including factorising polynomials and solving cubic equations ● Use of: - 𝑎 3 + 𝑏 3 = ( 𝑎 + 𝑏 ) 𝑎 2 − 𝑎𝑏 + 𝑏 2 ( )- 𝑎 3 − 𝑏 3 = ( 𝑎 − 𝑏 ) 𝑎 2 + 𝑎𝑏 + 𝑏 2 ( )● Partial fractions with cases where the denominator is no more complicated than: - ( 𝑎𝑥 + 𝑏 ) ( 𝑐𝑥 + 𝑑 )- ( 𝑎𝑥 + 𝑏 ) ( 𝑐𝑥 + 𝑑 ) 2 - ( 𝑎𝑥 + 𝑏 ) 𝑥 2 + 𝑐 2 ( ) 1. Express in partial fractions. 2 + 𝑥 2 𝑥 + 2 ( ) 2 𝑥 + 4 ( ) [4] 2. The function is defined by for all 𝑓 ( 𝑥 ) 𝑓 ( 𝑥 ) = 3 𝑥 2 + ℎ 𝑥 2 + 𝑘𝑥 − 4 real values of . Given that is a factor of and that when 𝑥 3 𝑥 − 1 𝑓 ( 𝑥 ) is divided by , the remainder is , find the value of each 𝑓 ( 𝑥 ) 𝑥 + 1 − 4 of the constants and . ℎ 𝑘 [5] 3. (a) Factorise . 27 𝑥 3 + 125 [2] 3. (b) Explain why is the only real root of the operation . 𝑥 = − 5 3 27 𝑥 3 + 125 = 0 [2] 4. Express in partial fractions. 18 𝑥 + 7 𝑥 − 1 ( ) 4 𝑥 + 1 ( ) [4] 5. The polynomial is given by . 𝑓 ( 𝑥 ) 𝑓 ( 𝑥 ) = 2 𝑥 3 + 5 𝑥 2 − 4 𝑥 − 3 (a) Divide by . 𝑓 ( 𝑥 ) 𝑥 + 3 [2] (b) What can you deduce about ? 𝑥 + 3 [1] (c) Solve the equation . 𝑓 ( 𝑥 ) = 0 [2] 6. Factorise and explain why is the only real root of the equation 𝑥 3 − 8 𝑥 = 2 . 𝑥 3 − 8 = 0 [4]

7. By using long division, find the remainder when is 9 𝑥 4 − 13 𝑥 2 + 4 𝑥 − 2 divided by . 3 𝑥 2 + 2 𝑥 − 1 [3] 8. Given that for all real values of 20 𝑥 2 + 13 𝑥 + 5 = 𝐴𝑥 1 + 2 𝑥 ( ) + 𝐵𝑥 + 5 , find the value of and of . 𝑥 𝐴 𝐵 [3] 9. Express in partial fractions. 6 𝑥 𝑥 + 1 ( ) 𝑥 − 1 ( ) 2 [5] 10. Factorise . 125 𝑎 3 − 8 𝑏 3 [3] 11. The polynomial has a factor . 𝑓 ( 𝑥 ) = 6 𝑥 3 + 𝑎 𝑥 2 + 11 𝑥 − 6 𝑥 + 2( ) (a) Show that . 𝑎 = 19 [2] (b) Solve the equation Show your working clearly. 𝑓 ( 𝑥 ) = 0 . [4] (c) Hence solve the equation . 6 𝑦 3 + 19 𝑦 2 + 11 𝑦 − 6 = 0 [2] 12. Factorise . 8 𝑥 3 + 125 [3] 13. Express in partial fractions. 3 𝑥 2 − 2 𝑥 + 1 ( ) 2 2 𝑥 − 1 ( ) [5] 14. The polynomial is given by , where 𝑝 ( 𝑥 ) 𝑝 ( 𝑥 ) = 5 𝑥 3 + 𝑎 𝑥 2 + 𝑏𝑥 − 2 and are constants. It is given that is a factor of and when 𝑎 𝑏 𝑥 − 2( ) 𝑝 ( 𝑥 ) is divided by , the remainder is 5. 𝑝 ( 𝑥 ) 𝑥 − 1( ) (a) Show that and find the value of . 𝑎 = − 21 𝑏 [4] (b) Using the values from part (a), find the remainder when 𝑝 ( 𝑥 ) is divided by . 2 𝑥 + 1( ) [2] 15. (a) Divide by . 2 𝑥 3 − 𝑥 2 + 8 𝑥 − 4 2 𝑥 − 1 [1] 15. (b) Express in partial fractions. 2 + 5 𝑥 − 𝑥 2 2 𝑥 3 − 𝑥 2 + 8 𝑥 − 4 [5]

16. The polynomial is given by , where 𝑝 ( 𝑥 ) 𝑝 ( 𝑥 ) = 𝑥 3 + 8 𝑥 2 + 21 𝑥 + 9 𝑎 is a constant. It is given that is a factor of . 𝑎 𝑥 + 2( ) 𝑝 ( 𝑥 ) (a) Find the value of the constant . 𝑎 [2] (b) Using the value from , find the remainder when is divided by ( 𝑎 ) 𝑝 ( 𝑥 ) . 𝑥 − 1( ) [1] 17. (a) Show that is a factor of . 2 𝑥 − 1 𝑓 ( 𝑥 ) = 2 𝑥 3 + 3 𝑥 2 − 8 𝑥 + 3 [1] 17. (b) Hence, factorise completely. 𝑓 ( 𝑥 ) [3] 18. (a) Factorise . 8 𝑥 3 − 1 [2] 18. (b) Explain why is the only real root of the equation . 𝑥 = 1 2 8 𝑥 3 − 1 = 0 [2] 19. The polynomial f is given by , ( 𝑥 ) 𝑓 ( 𝑥 ) = 𝑥 3 + 𝑎 𝑥 2 + 𝑏𝑥 + 24 where and are constants. is a factor of and when is 𝑎 𝑏 𝑥 + 4 𝑓 ( 𝑥 ) 𝑓 ( 𝑥 ) divided by , the remainder is . 𝑥 − 2 36 (a) Find the value of and . 𝑎 𝑏 [4] (b) Using the values from part , find the remainder when is divided by ( 𝑎 ) 𝑓 ( 𝑥 ) . 2 𝑥 + 1 [2] 20. Express in partial fractions. 3 𝑥 + 8 𝑥 − 1 ( ) 2 ( 𝑥 + 2 ) [4] 21. Express in partial fractions. 𝑥 2 + 18 𝑥 − 50 𝑥 𝑥 + 5 ( ) 2 [5]

A4: POLYNOMIALS AND PARTIAL FRACTIONS (MARKING SCHEME) 1. Express in partial fractions. 2 + 𝑥 2 𝑥 − 2 ( ) 2 𝑥 + 4 ( ) 2 + 𝑥 2 𝑥 − 2 ( ) 2 𝑥 + 4 ( ) = 𝐴 ( 𝑥 − 2 ) + 𝐵 ( 𝑥 + 2 ) 2 + 𝐶 𝑥 + 4 2 + 𝑥 2 = 𝐴 ( 𝑥 − 2 ) ( 𝑥 + 4 ) + 𝐵 ( 𝑥 + 4 ) + 𝐶 ( 𝑥 − 2 ) 2 ★ Sub (A and C = 0, to find B) 𝑥 = 2 2 + 𝑥 2 = 𝐴 ( 2 − 2 ) ( 𝑥 + 4 ) + 𝐵 ( 𝑥 + 4 ) + 𝐶 ( 2 − 2 ) 2 2 + 4 = 0 + 6 𝐵 + 0 𝐵 = 1 ★ Sub 𝑥 = − 4 2 + 𝑥 2 = 𝐴 ( 𝑥 − 2 ) ( 𝑥 + 4 ) + 𝐵 ( 𝑥 + 4 ) + 𝐶 ( 𝑥 − 2 ) 2 2 + 16 = 0 + 0 + 36 𝐶 18 = 36 𝐶 𝐶 = 1 2 Sub ★ 𝑥 = 0 2 + 𝑥 2 = 𝐴 ( 𝑥 − 2 ) ( 𝑥 + 4 ) + 𝐵 ( 𝑥 + 4 ) + 𝐶 ( 𝑥 − 2 ) 2 2 + 0 = 𝐴 ( − 2 ) ( 4 ) + 4 𝐵 + 4 𝐶 2 = − 8 𝐴 + 4 𝐵 + 4 𝐶Sub 𝐵 = 1 , 𝐶 = 1 2 2 − 4 ( 1 ) − 4 1 2 ( )= − 8 𝐴 − 4 = − 8 𝐴 𝐴 = 1 2 2 + 𝑥 2 𝑥 − 2 ( ) 2 𝑥 + 4 ( ) = 1 2 ( 𝑥 − 2 ) + 1 ( 𝑥 + 2 ) 2 + 1 2 𝑥 + 4 ( ) [4]

2. The function is defined by for all 𝑓 ( 𝑥 ) 𝑓 ( 𝑥 ) = 3 𝑥 2 + ℎ 𝑥 2 + 𝑘𝑥 − 4 real values of . Given that is a factor of and that when 𝑥 3 𝑥 − 1 𝑓 ( 𝑥 ) is divided by , the remainder is , find the value of each 𝑓 ( 𝑥 ) 𝑥 + 1 − 4 of the constants and . ℎ 𝑘 Let 𝑓 ( 𝑥 ) = 3 𝑥 3 + ℎ 𝑥 2 + 𝑘𝑥 − 4 is a factor of No remainder since is a factor ( 3 𝑥 − 1 ) 𝑓 ( 𝑥 ) ★ ( 3 𝑥 − 1 ) 𝑓 1 3 ( )= 0 3 1 3 ( ) 3 + ℎ 1 3 ( ) 2 + 𝑘 1 3 ( )− 4 = 0 1 9 + 1 9 ℎ + 1 3 𝑘 − 4 = 0 1 + ℎ + 3 𝑘 − 36 = 0 ℎ + 3 𝑘 = 35 − − − ( 1 ) When is divided by 𝑓 ( 𝑥 ) 𝑥 + 1 , 𝑅 = − 4 𝑓 ( − 1 ) = − 4 3 ( − 1 ) 3 + ℎ ( − 1 ) 2 + 𝑘 ( − 1 ) − 4 = − 4 − 3 + ℎ − 𝑘 − 4 + 4 = 0 ℎ − 𝑘 = 3 − − − ( 2 ) ( 1 ) − ( 2 ) : ℎ + 3 𝑘 − ( ℎ − 𝑘 ) = 35 − 3 ℎ + 3 𝑘 − ℎ + 𝑘 = 32 4 𝑘 = 32 𝑘 = 8 ℎ = 11 [5]

3. (a) Factorise . 27 𝑥 3 + 125 Factorise cubic equation: 27 𝑥 3 + 125★ 𝑎 3 + 𝑏 3 = ( 𝑎 + 𝑏 ) 𝑎 2 − 𝑎𝑏 + 𝑏 2 ( ) = 3 𝑥 ( ) 3 + 5 ( ) 3 = ( 3 𝑥 + 5 ) 9 𝑥 2 − 15 𝑥 + 25 ( ) [2] 3. (b) Explain why is the only real root of the operation . 𝑥 = − 5 3 27 𝑥 3 + 125 = 0 Consider 9 𝑥 2 − 15 𝑥 + 25 ( ) 𝑏 2 − 4 𝑎𝑐 = ( − 15 ) 2 − 4 ( 9 ) ( 25 ) = − 675Since , therefore cannot be factorised. 𝑏 2 − 4 𝑎𝑐 < 0 9 𝑥 2 − 15 𝑥 + 25Thus, 𝑥 = − 5 3 [2] 4. Express in partial fractions. 18 𝑥 + 7 𝑥 − 1 ( ) 4 𝑥 + 1 ( ) 18 𝑥 + 7 𝑥 − 1 ( ) 4 𝑥 + 1 ( ) = 𝐴 𝑥 − 1 + 𝐵 4 𝑥 + 1 18 𝑥 + 7 = 𝐴 ( 4 𝑥 + 1 ) + 𝐵 ( 𝑥 − 1 )Let 𝑥 = 1 18 + 7 = 5 𝐴 𝐴 = 5Let 𝑥 = − 1 4 18 − 1 4 ( ) + 7 = 𝐵 − 1 4 − 1 ( ) 5 2 = − 5 4 𝐵 𝐵 = − 2 18 𝑥 + 7 𝑥 − 1 ( ) 4 𝑥 + 1 ( ) = 5 𝑥 − 1 − 2 4 𝑥 + 1 [4]

5. The polynomial is given by . 𝑓 ( 𝑥 ) 𝑓 ( 𝑥 ) = 2 𝑥 3 + 5 𝑥 2 − 4 𝑥 − 3 (a) Divide by . 𝑓 ( 𝑥 ) 𝑥 + 3 Do long division ★ [2] (b) What can you deduce about ? 𝑥 + 3 is a factor of . 𝑥 + 3 𝑓 ( 𝑥 ) [1] (c) Solve the equation . 𝑓 ( 𝑥 ) = 0 𝑥 + 3 ( ) 2 𝑥 2 − 𝑥 − 1 ( )= 0 ( 𝑥 + 3 ) ( 2 𝑥 + 1 ) ( 𝑥 − 1 ) = 0 or or 𝑥 = − 3 − 1 2 1 [2] 6. Factorise and explain why is the only real root of the equation 𝑥 3 − 8 𝑥 = 2 . 𝑥 3 − 8 = 0 Factorise cubic equation: ★ 𝑎 3 − 𝑏 3 = ( 𝑎 − 𝑏 ) 𝑎 2 + 𝑎𝑏 + 𝑏 2 ( ) 𝑥 3 − 8 = 𝑥 − 2 ( ) ( 𝑥 2 + 2 𝑥 + 4 )Consider ( 𝑥 2 + 2 𝑥 + 4 ) 𝑏 2 − 4 𝑎𝑐 = 2 ( ) 2 − 4 ( 1 ) ( 4 ) =− 12 therefore cannot be factorised. 𝑏 2 − 4 𝑎𝑐 < 0 , 𝑥 2 + 2 𝑥 + 4Thus, 𝑥 = 2 [4]

7. By using long division, find the remainder when is 9 𝑥 4 − 13 𝑥 2 + 4 𝑥 − 2 divided by . 3 𝑥 2 + 2 𝑥 − 1 Remainder is 6 𝑥 − 4 [3] 8. Given that for all real values of 20 𝑥 2 + 13 𝑥 + 5 = 𝐴𝑥 1 + 2 𝑥 ( ) + 𝐵𝑥 + 5 , find the value of and of . 𝑥 𝐴 𝐵 Sub 𝑥 = − 1 2 20 − 1 2 ( ) 2 + 13 − 1 2 ( ) + 5 = 0 + 𝐵 − 1 2 ( )+ 5 7 2 − 5 = − 1 2 𝐵 − 3 2 = − 1 2 𝐵 𝐵 = 3 Sub 𝑥 = 1 2