Find the remainder when 3x3 - x2 + 7x + 5 is divided by 3x + 2.
 Given that 2x3 - x2 - 2x + 3 =   (Ax + B)(x - 1)(x + 2) + C(x - 1) + D, find the values of A, B, C and D.   Hence or otherwise, deduce the remainder when
2x3 - x2 - 2x + 3 is divided by x2 + x - 2.
 Solve the equation 2x3 + x2 - 19x = 6, giving your answers to two decimal places where necessary.    
 The expression x3 + px2 + qx + 6 has the same remainder when divided by
x + 1 and 2 - x. Given that the remainder when the expression is divided by x + 3 is -60, find the value of p and of q.
      remainder = f(-2/3)    
                        = 3(-2/3)3 - (-2/3)2 + 7(-2/3) + 5    
                        = -1
              2 - 1 - 2 + 3 = D  
       D = 2  
       Let x = -2  
       -16 - 4 + 4 + 3 = -3C + 2  
       3C = 15  
       C = 5  
       Let x = 0  
       3 = B (-1)(2) -5 + 2  
       -2B = 6  
       B = -3  
       Sub any value other than 1, 2 & 0 into x  
       A = 2  
    2x3 - x2 - 2x + 3 ≡ (Ax +   B)(x - 1)(x + 2) + C(x - 1) +   D  
   Since x2 + x - 2 =  (x - 1)(x + 2),        (note the similarity?)  
        thus, remainder = C(x - 1) + D  
                            = 5(x - 1) + 2  
                            = 5x - 3  
      
      let f(x) = 2x3 + x2 - 19x - 6  
       (x -   3)        f(3) = 0  
       let f(x) = (x - 3)(Ax2 + Bx + C)  
       A = 2  
       C = 2  
       let x = 1  
       2 + 1 - 19 - 6 = -2(2 + B + 2)  
       -22 = -8 - 2B  
       B = 7  
       2x2 + 7x + 2 = 0  
       x =   
          = -3.19 or -0.31 (3 s.f.)  
         by remainder theorem, f(-1) = f(2)  
       -1 + p - q + 6 = 8 + 4p + 2q + 6  
       p - q - 5 = 4p + 2q + 14  
       3p + 3q = 9 --------(1)  
       by remainder theorem, f(-3) = -60  
       -27 + 9p - 3q + 6 = -60  
       3q = 39 + 9p --------(2)  
       sub (2) into (1):  
       3p + 39 + 9p = -9  
       12p = -48  
       p = -4  
       q = 1  
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