Question

The expression 4x^3 - bx^2 + x - c leaves remainders 0 and 30 when divided by ( x + 1 ) and ( 2x - 3 ) respectively. Calculate the values of b and c and hence factorise the expression completely.

Answer 1

4x^3-bx^2 +x -c leaves remainder 0 when divided by ( x + 1)
it means ( x + 1) is factor of 4x^3 -bx^2+x-c
so , x = -1 is zero of given expression .
4(-1)^3-b(-1)^2+(-1)-c = 0
-4 -b-1 -c =0
b+ c = -5 --------------------(1)

now ,
when divided by (2x -3) leaves remainder is 30
hence,
4x^3-bx^2+x -c -30 completely divisible by (2x -3 )

=> 4(3/2)^3-b(3/2)^2 +(3/2) -c-30=0
=> 27/2 -9/4 b +3/2 -c-30 =0
=> -9/4 b -c -15 =0
=> 9b + 4 c +60 =0 ----------------(2)
solve equations (1) and (2)
b = -8 and c = 3

hence,
4x^3 +8x^2 +x -3

Answer 2

P(x) = 4 x³ - b x² + x - c 

Given (x+1) divides P(x).   So  P(-1) = 0
     So   -4 - b - 1 - c = 0    =>  b + c = -5       ---- (1)

Given P(x) gives a remainder of 30 when divided by (2x - 3).
     P(3/2) = 30     =>  4 (3/2)³  - b (3/2)² + 3/2 - c = 30
            Simplify to get    9 b + 4 c = - 60       --- (2)

Solving (1) and (2) , we get:   c = 3,  b = - 8

P(x) = 4 x³ + 8 x² + x - 3 
        = (x+1) (4 x² + m x - 3)      LET
        = 4 x³ + (m +4) x² + x (m -3) - 3 
  Comparing coefficients, we get:  m = 4

Hence:  P(x) = (x+1) (4x² + 4 x - 3) = (x+1) (2x + 3) (2x - 1)
                 follow usual method for factorization of a quadratic.