Answer :
We want to find the remainder when
$$
3x^3 - 2x^2 + 4x - 3
$$
is divided by
$$
x^2 + 3x + 3.
$$
Since the divisor is a quadratic polynomial (degree 2), the remainder must be of degree less than 2; that is, it will be a linear polynomial of the form
$$
ax + b.
$$
We can determine the quotient and remainder using polynomial long division. Here are the detailed steps:
1. **Divide the Leading Terms:**
Divide the leading term of the dividend, $3x^3$, by the leading term of the divisor, $x^2$:
$$
\frac{3x^3}{x^2} = 3x.
$$
This $3x$ is the first term of the quotient.
2. **Multiply and Subtract:**
Multiply the entire divisor by $3x$:
$$
3x \cdot (x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.
$$
Subtract this product from the original dividend:
$$
\begin{aligned}
\bigl(3x^3 - 2x^2 + 4x - 3\bigr) - \bigl(3x^3 + 9x^2 + 9x\bigr)
&= (3x^3 - 3x^3) + (-2x^2 - 9x^2) + (4x - 9x) - 3 \\
&= -11x^2 - 5x - 3.
\end{aligned}
$$
This new polynomial, $-11x^2 - 5x - 3$, is the intermediate remainder.
3. **Second Division Step:**
Now, divide the leading term of the intermediate remainder, $-11x^2$, by the leading term of the divisor, $x^2$:
$$
\frac{-11x^2}{x^2} = -11.
$$
This $-11$ is the next term of the quotient.
4. **Multiply and Subtract Again:**
Multiply the divisor by $-11$:
$$
-11 \cdot (x^2 + 3x + 3) = -11x^2 - 33x - 33.
$$
Subtract this from the intermediate remainder:
$$
\begin{aligned}
\bigl(-11x^2 - 5x - 3\bigr) - \bigl(-11x^2 - 33x - 33\bigr)
&= (-11x^2 + 11x^2) + (-5x + 33x) + (-3 + 33) \\
&= 28x + 30.
\end{aligned}
$$
Since the degree of $28x + 30$ (which is 1) is less than the degree of the divisor (which is 2), this is the final remainder.
5. **Conclusion:**
The quotient obtained is
$$
3x - 11,
$$
and the remainder is
$$
28x + 30.
$$
Thus, the remainder when $\left(3x^3 - 2x^2 + 4x - 3\right)$ is divided by $\left(x^2 + 3x + 3\right)$ is
$$
28x + 30.
$$
$$
3x^3 - 2x^2 + 4x - 3
$$
is divided by
$$
x^2 + 3x + 3.
$$
Since the divisor is a quadratic polynomial (degree 2), the remainder must be of degree less than 2; that is, it will be a linear polynomial of the form
$$
ax + b.
$$
We can determine the quotient and remainder using polynomial long division. Here are the detailed steps:
1. **Divide the Leading Terms:**
Divide the leading term of the dividend, $3x^3$, by the leading term of the divisor, $x^2$:
$$
\frac{3x^3}{x^2} = 3x.
$$
This $3x$ is the first term of the quotient.
2. **Multiply and Subtract:**
Multiply the entire divisor by $3x$:
$$
3x \cdot (x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.
$$
Subtract this product from the original dividend:
$$
\begin{aligned}
\bigl(3x^3 - 2x^2 + 4x - 3\bigr) - \bigl(3x^3 + 9x^2 + 9x\bigr)
&= (3x^3 - 3x^3) + (-2x^2 - 9x^2) + (4x - 9x) - 3 \\
&= -11x^2 - 5x - 3.
\end{aligned}
$$
This new polynomial, $-11x^2 - 5x - 3$, is the intermediate remainder.
3. **Second Division Step:**
Now, divide the leading term of the intermediate remainder, $-11x^2$, by the leading term of the divisor, $x^2$:
$$
\frac{-11x^2}{x^2} = -11.
$$
This $-11$ is the next term of the quotient.
4. **Multiply and Subtract Again:**
Multiply the divisor by $-11$:
$$
-11 \cdot (x^2 + 3x + 3) = -11x^2 - 33x - 33.
$$
Subtract this from the intermediate remainder:
$$
\begin{aligned}
\bigl(-11x^2 - 5x - 3\bigr) - \bigl(-11x^2 - 33x - 33\bigr)
&= (-11x^2 + 11x^2) + (-5x + 33x) + (-3 + 33) \\
&= 28x + 30.
\end{aligned}
$$
Since the degree of $28x + 30$ (which is 1) is less than the degree of the divisor (which is 2), this is the final remainder.
5. **Conclusion:**
The quotient obtained is
$$
3x - 11,
$$
and the remainder is
$$
28x + 30.
$$
Thus, the remainder when $\left(3x^3 - 2x^2 + 4x - 3\right)$ is divided by $\left(x^2 + 3x + 3\right)$ is
$$
28x + 30.
$$
