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

1. Set up polynomial long division: We want to divide the polynomial $3x^3 - 2x^2 + 4x - 3$ by $x^2 + 3x + 3$. We set this up like a standard long division problem.

$$ \begin{array}{c|cc cc} \multicolumn{2}{r}{3x} & -11 \\ x^2+3x+3 & 3x^3 & -2x^2 & +4x & -3 \\ \multicolumn{2}{r}{3x^3} & +9x^2 & +9x \\ \hline \multicolumn{2}{r}{0} & -11x^2 & -5x & -3 \\ \multicolumn{2}{r}{} & -11x^2 & -33x & -33 \\ \hline \multicolumn{2}{r}{} & 0 & 28x & +30 \end{array} $$

1. Divide the leading terms: Divide the leading term of the dividend ($3x^3$) by the leading term of the divisor ($x^2$). This gives $3x^3 / x^2 = 3x$. This is the first term of our quotient.

1. Multiply and subtract: Multiply the divisor ($x^2 + 3x + 3$) by the first term of the quotient ($3x$). This gives $3x(x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x$. Subtract this result from the dividend.

$(3x^3 - 2x^2 + 4x - 3) - (3x^3 + 9x^2 + 9x) = 3x^3 - 2x^2 + 4x - 3 - 3x^3 - 9x^2 - 9x = -11x^2 - 5x - 3$.

1. Bring down the next term: Bring down the next term of the dividend (-3) to form the new polynomial to work with: $-11x^2 - 5x - 3$.

1. Repeat the process: Divide the leading term of the new polynomial ($-11x^2$) by the leading term of the divisor ($x^2$). This gives $-11x^2 / x^2 = -11$. This is the next term of our quotient.

1. Multiply and subtract again: Multiply the divisor ($x^2 + 3x + 3$) by the new term of the quotient ($-11$). This gives $-11(x^2 + 3x + 3) = -11x^2 - 33x - 33$. Subtract this result from the current polynomial.

$(-11x^2 - 5x - 3) - (-11x^2 - 33x - 33) = -11x^2 - 5x - 3 + 11x^2 + 33x + 33 = 28x + 30$.

1. Determine the remainder: Since the degree of the resulting polynomial ($28x + 30$, degree 1) is less than the degree of the divisor ($x^2 + 3x + 3$, degree 2), this is our remainder.