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Answer :
To find the remainder when
$$
3x^3 - 2x^2 + 4x - 3
$$
is divided by
$$
x^2 + 3x + 3,
$$
we perform polynomial long division.
**Step 1. Divide the leading term of the dividend by the leading term of the divisor.**
The leading term of the dividend is $3x^3$ and that of the divisor is $x^2$. Dividing these gives:
$$
\frac{3x^3}{x^2} = 3x.
$$
So, the first term of the quotient is $3x$.
**Step 2. Multiply the divisor by $3x$ and subtract the result from the dividend.**
Multiply:
$$
3x \cdot (x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.
$$
Subtract this from the original dividend:
\[
\begin{aligned}
(3x^3 - 2x^2 + 4x - 3) &- (3x^3 + 9x^2 + 9x) \\
&= (3x^3 - 3x^3) + (-2x^2 - 9x^2) + (4x - 9x) - 3 \\
&= -11x^2 - 5x - 3.
\end{aligned}
\]
**Step 3. Divide the new leading term by the leading term of the divisor.**
Now, divide $-11x^2$ by $x^2$:
$$
\frac{-11x^2}{x^2} = -11.
$$
This is the next term of the quotient.
**Step 4. Multiply the divisor by $-11$ and subtract.**
Multiply:
$$
-11 \cdot (x^2 + 3x + 3) = -11x^2 - 33x - 33.
$$
Subtract this product from the current polynomial:
\[
\begin{aligned}
(-11x^2 - 5x - 3) &- (-11x^2 - 33x - 33) \\
&= (-11x^2 + 11x^2) + (-5x + 33x) + (-3 + 33) \\
&= 28x + 30.
\end{aligned}
\]
Since the degree of the remainder $28x + 30$ is less than the degree of the divisor $x^2+3x+3$, we have completed the division.
**Final Answer:**
The quotient is
$$
3x - 11,
$$
and the remainder is
$$
28x + 30.
$$
Thus, the remainder when $$3x^3-2x^2+4x-3$$ is divided by $$x^2+3x+3$$ is $\boxed{28x+30}$.
$$
3x^3 - 2x^2 + 4x - 3
$$
is divided by
$$
x^2 + 3x + 3,
$$
we perform polynomial long division.
**Step 1. Divide the leading term of the dividend by the leading term of the divisor.**
The leading term of the dividend is $3x^3$ and that of the divisor is $x^2$. Dividing these gives:
$$
\frac{3x^3}{x^2} = 3x.
$$
So, the first term of the quotient is $3x$.
**Step 2. Multiply the divisor by $3x$ and subtract the result from the dividend.**
Multiply:
$$
3x \cdot (x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.
$$
Subtract this from the original dividend:
\[
\begin{aligned}
(3x^3 - 2x^2 + 4x - 3) &- (3x^3 + 9x^2 + 9x) \\
&= (3x^3 - 3x^3) + (-2x^2 - 9x^2) + (4x - 9x) - 3 \\
&= -11x^2 - 5x - 3.
\end{aligned}
\]
**Step 3. Divide the new leading term by the leading term of the divisor.**
Now, divide $-11x^2$ by $x^2$:
$$
\frac{-11x^2}{x^2} = -11.
$$
This is the next term of the quotient.
**Step 4. Multiply the divisor by $-11$ and subtract.**
Multiply:
$$
-11 \cdot (x^2 + 3x + 3) = -11x^2 - 33x - 33.
$$
Subtract this product from the current polynomial:
\[
\begin{aligned}
(-11x^2 - 5x - 3) &- (-11x^2 - 33x - 33) \\
&= (-11x^2 + 11x^2) + (-5x + 33x) + (-3 + 33) \\
&= 28x + 30.
\end{aligned}
\]
Since the degree of the remainder $28x + 30$ is less than the degree of the divisor $x^2+3x+3$, we have completed the division.
**Final Answer:**
The quotient is
$$
3x - 11,
$$
and the remainder is
$$
28x + 30.
$$
Thus, the remainder when $$3x^3-2x^2+4x-3$$ is divided by $$x^2+3x+3$$ is $\boxed{28x+30}$.
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