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

The remainder is 28x+30

Step-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+30

We can not solve it further because remainder has two terms and divisor has 3 terms. Therefore the remainder is 28x+30

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