Ex04 Polynomials Answer

Topic: Polynomials Question 1 [2008 EOY, Q11] marks (i) (a)  Show that  21x  is a factor of  3228 4 xx x .  Solution: Sub  1 2x :  11 1 1 128 44 4 084 2 4 4               Hence   is a factor of   [1] (i) (b)  Solve for x if  3228 4 0 xx x . Solution: By long division,  2(2 1)( 4) 0xx        Hence  2 1(2 1)( 4) 0 2xx x      or  2    [4] (ii) The expression  3242 0 1 2xx p x   leaves a remainder of  10x q   when divided by  223 1x x . Find the values of p and q.    Solution:                                 23242 0 1 2 ( 23 1 ) ( ) 1 0x xp x x xQ x x q          3242 0 1 2 ( 2 1 ) ( 1 ) ( ) 1 0x xp x x xQ x x q             Sub  1 2x ,  11 1 142 0 1 2 1 0 2 2 584 2 2 pq p q                  Sub  1x ,  42 0 1 2 1 0 6p qp q           Hence  6 2 25 19qq q     and  13p           [5] 21x 3228 4 xx x

Question 2 [2010 EOY, Q12] marks (a) Solve for x when  32671 4 8 0xx x  . Solution: By  trial  and  error,  when  2x ,  32671 4 8 0xx x    hence  (2 )x  is  a  factor.   By long division,  32 26 7 14 8 ( 2)(6 5 4)xx x x xx        Hence,  2x  or  5 25 4( 24) 12x    [M1]    14or 23    [3] (b) It is given that  32 212 15 (2 5 3) ( ) 158 83xa x x b x x Q x x       where  ()Qx  is  a polynomial.  (i) State the degree of  ()Qx . Solution: Degree of  ()Qx  is 1 as it is linear   [1] (ii) Find the values of a and b.    Solution: 32 212 15 (2 5 3) ( ) 158 83xa x x b x x Q x x          3212 15 (2 1)( 3) ( ) 158 83xa x x b x x Q x x        Sub  1 2x  to get  45 2ab      Sub  3x  to get  91 8 8ab      Solve simultaneously to get  20a  and  8b          [3]

Question 3 [2011 EOY, Q11] marks (i) Given that   32f( ) 2 3 2x xx A x    has a factor  2x and the other   two roots of the equation  f( ) 0x   are  2 and  k .  Find the value of  A  and   of  k .                                                                                                                Solution:      f2 0 3222 32 2 2 0 22 43 2 22 0 23 A A A           32 32 f( ) 2 3 2 2 3 23 2 43 x xx x xx x           f( ) 2 2x xxx k      24 3 2 3kk                       [4] (ii) Solve the equation  3232 7 2x xx  .                                                         Solution: 32327 2 0xxx     Found 1 linear factor                    Correct factorization into linear and quadratic                                      e.g.    223 4 1 0xx x      Correct factorization into 3 linear factors      23 1 1 0xx x       12, , 13x       [4]

Question 4 [2012 EOY, Q8] marks The cubic polynomial   f x  is such that the coefficient of  3x  is 2 and the roots of  f0 x   are −1, 2k  and  2k .  It is given that   f x  has a remainder of 168 when divided   by  2x .  (i) Show that  32 21 2 0kk k . Solution: 2f( ) 2 ( 1 ) ( 2 ) ( )x xx k x k     Ability to express f as a product of three linear factors and a constant 2.    2(2) 2(3)(2 2 )(2 ) 168fk k       23 32 42 4 2 2 8 21 2 0 kk k kk k    [3] (ii) Hence find a value for k and show that there are no other real values of k which satisfy this  equation. Solution: 32() 2 1 2fk k k k                 32(3) 3 3 6 12 0f     By trial and error, k = 3   2 2 () ( 3 ) ( ) 1 (3 ) ( 24 ) 1 f k k ak bk c M kk k M       22 24 (1 ) 3 0kk k            Hence, there is only one real value of k.          [4]

(iii) Use your answer to part (ii) to solve the equation  3212 2 1 0x xxee e    . Solution: 2312 1 2 0xx xee e                    32 21 2 0xx xee e            3l n 3xLet e x      [2] Question 5 [2013 EOY, Q5] marks Given that   2 6x x   is a factor of    43 22 1xxa xb x a b      , find the value   of  a  and of  b.    Solution: As 2 6( 2 ) ( 3 )xx x x   Let x = 2 and subs into the eqn. 13 (1)ab    Let 3x and subs into the eqn. 4 67 (2)ab    Solve the above two equations, a = 16 and b = 3 3x [4] Question 6 [2014 EOY, Q7] marks (i) Solve the equation 3225 2xx x  . Solution: 32 32 32 2 25 2 25 2 0 Let f( ) 2 5 2 f(1) 0, ( 1) is a factor of f( ). Using long division, (1 ) ( 2 32 ) 0 ( 1)(2 1)( 2) 0 11, or 2 2 xx x xx x xx x x x x xx x xx x x                      [4]

(ii) Find the remainder when 200 100(3 ) (2 )xx  is divided by 2 56xx  . Solution: 200 100 2 2 200 100 Let f( ) ( 3) ( 2) . When divided by 5 6, f( ) ( 5 6) Q( ) , where and are real constants. f( ) ( 2)( 3) Q( ) By Remainder Theorem, f(2) (2 3) (2 2) 1 2 1 f(3) (3 3 xx x xx x xx x a x b ab x xx x a x b ab                 200 100)( 3 2 ) 1 31 Solving the above two equations, 2, 5 The remainder is 2 5. ab ab x           [4] Question 7 [2015 EOY, Q9] marks (i) Solve for x when 32(e 2) e (7e 5)xx x  . Solution: 33 22(e 2) e (7e 5) 2 7 5 4 0xx x yyy      where  exy   1y  is a root so  1( 1)(2 1)( 4) 0 e 1 (rej.) or or 4 2 xyy y        Hence  ln 2 or ln 4x [5]

(ii) Given that 43 2f( ) 8 3 0x xp x xx q   , where p and q are unknown constants, leaves a remainder of 2 when divided by 2x and a remainder of 9 when divided by 23x , find the values of p and q and hence find the remainder when f( )x is divided by 226x x  . Solution: f( 2 ) 2 8 8 0 pq   and  3f 9 27 8 3002 pq       Solving,  4p  and  24q   Hence  43 2f( ) 8 4 3 0 2 4xxx x x     By long division, remainder is  62 x [6]