Topical Revision Paper 4 Unit 4 Ploynomials

Topical Revision Paper 4 Term 2 Unit 4: Polynomials Question 1: It is given that f( ) ( 3)(2 1)g( )x x x x mx c     , where f( )x and g( )x are polynomials in terms of x, and m and c are constants. The polynomial f( )x has a remainder of 11 when divided by ( 3)x , and a remainder of 4 when divided by (2 1)x . (i) Find the value of m and of c . [4] (ii) Hence, find the remainder when f( )x is divided by 22 5 3xx . [1] Solution: (i) ( 3) 11 3 11( ) 422 7 3.5 2 5 f m c f m c m m c           (ii) Remainder is 2 5x Question 2: (i) Factorise 32 5 3 9x x x   completely. [4] (ii) Hence, solve the equation 6 4 28 20 6 9x x x   . [3] Solution: (i) 32 2 2 2 Let ( ) 5 3 9 ( 1) 1 5 3 9 0 ( ) ( 1)( 9) 51 6 f( ) ( 1)( 6 9) f( ) ( 1)( 3) f x x x x f f x x x bx b b x x x x x x x                        (ii) 2f( ) ( 1)( 3)x x x  

2( 1)( 3) 0 1 or 3 xx xx      6 4 28 20 6 9x x x   Let         322 2 2 2 32 22 2 5 2 3 2 9 2 5 3 9 0 1 or 3 2 1 n.a. or 2 3 3 or 2 6 or 1.222 x x x yx y y y yy xx x xx               Question 3: The polynomial P is given by   322 7 6 1P x x x x    . (i) Verify that 1 2x is a root of the equation   0Px  . [1] (ii) Express  Px in the form    221x Ax Bx C   where A , B and C are constants to be determined. [2] (iii) Find the remaining roots of   0Px  using completing square method, expressing your answer in exact form. [3] (iii) Hence solve the equation 232 7e 6e e 0x x x    . [3] Solution: (i) Substitute 1 2x :   32 1 1 12 7 6 12 2 2Px                    17 3 1 044     . (ii) Let    3 2 22 7 6 1 2 1x x x x Ax Bx C       By comparing coefficients of 2x and constant terms, A = 1, C = −1.

By comparing coefficients of x term, 7 2 4B A B    . (iii) The remaining roots come from solving 2 4 1 0xx     2 2 2 4 1 0 ( 2) 4 1 25 2 5 or 2 5 . xx x x xx                 (iv) Replace x by 1 e x : 32 1 1 12 7 6 1 0e e ex x x                    11 , 2 5, 2 5 (rej) e2 1e 2, 25 ln 2,1.44 or 0.693,1.44 x x xx         Question 4: Using long division, find the quotient when 3212 8 2x x x   is divided by 2(2 1) .x [3] Solution: Using long division,     2 3 2 32 2 2 31 4 4 1 12 8 2 12 12 3 4 2 2 4 4 1 21 x x x x x x x x x xx xx x                  Quotient = 31x Question 5: (i) Find the values of p and q for which 2 6xx is a factor of 322 17x px x q   . [4] (ii) Using the values of p and q found in part (i), solve the equation 322 17 0x px x q    . [3]

(iii) Hence, sketch the graph of 322 17y x px x q    . [3] Solution: (i) 2 6 ( 3)( 2) f (3) 54 9 51 0 9 3 f ( 2) 16 4 34 0 4 18 3 and 30 x x x x p q p q p q p q pq                         (ii)     322 3 17 30 0 3 2 2 5 0 3, 2 or 2.5 x x x x x x x            (iii)