What is the remainder when (3x^3-2x^2+4x-3) is divided by (x^2+3x+3)?

Accepted Solution

Answer: The remainder is 28x+30Step-by-step explanation:We will divide  3x^3-2x^2+4x-3 by x^2+3x+3.To get the first term of the dividend we will multiply x^2+3x+3 by 3x  3x(x^2+3x+3)=3x^3+9x^2+9x It means          x^2+3x+3) 3x^3-2x^2+4x-3(  3x -11                          3x^3+9x^2+9x                    -           -         -                   _________________                                    11x^2-5x-3       (Now multiply x^2+3x+3 by -11 to get the                                    11x^2-33x-33     first term)                               -            +      +                              _______________                                           28x+30We can not solve it further because remainder has two terms and divisor has 3 terms. Therefore the remainder is 28x+30....