Answer :
We want to find the remainder when
[tex]$$
3x^3 - 2x^2 + 4x - 3
$$[/tex]
is divided by
[tex]$$
x^2 + 3x + 3.
$$[/tex]
We will use polynomial long division.
1. First Division Step:
The leading term of the dividend is [tex]$3x^3$[/tex] and of the divisor is [tex]$x^2$[/tex]. Dividing these gives
[tex]$$
\frac{3x^3}{x^2} = 3x.
$$[/tex]
Multiply the divisor by [tex]$3x$[/tex]:
[tex]$$
3x \cdot (x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.
$$[/tex]
Subtract this result from the dividend:
[tex]$$
\begin{aligned}
(3x^3 - 2x^2 + 4x - 3) &- (3x^3 + 9x^2 + 9x) \\
&= 3x^3 - 2x^2 + 4x - 3 - 3x^3 - 9x^2 - 9x \\
&= (-2x^2 - 9x^2) + (4x - 9x) - 3 \\
&= -11x^2 - 5x - 3.
\end{aligned}
$$[/tex]
2. Second Division Step:
Now, the new dividend is [tex]$-11x^2 - 5x - 3$[/tex]. Divide the leading term [tex]$-11x^2$[/tex] by [tex]$x^2$[/tex]:
[tex]$$
\frac{-11x^2}{x^2} = -11.
$$[/tex]
Multiply the divisor by [tex]$-11$[/tex]:
[tex]$$
-11 \cdot (x^2 + 3x + 3) = -11x^2 - 33x - 33.
$$[/tex]
Subtract this from the current dividend:
[tex]$$
\begin{aligned}
(-11x^2 - 5x - 3) &- (-11x^2 - 33x - 33) \\
&= -11x^2 - 5x - 3 + 11x^2 + 33x + 33 \\
&= ( -11x^2 + 11x^2) + (-5x + 33x) + (-3 + 33) \\
&= 28x + 30.
\end{aligned}
$$[/tex]
Since the degree of the remainder [tex]$28x + 30$[/tex] is less than the degree of the divisor (which is 2), the division is complete.
3. Conclusion:
The remainder when dividing
[tex]$$
3x^3 - 2x^2 + 4x - 3
$$[/tex]
by
[tex]$$
x^2 + 3x + 3
$$[/tex]
is
[tex]$$
28x + 30.
$$[/tex]
Thus, the correct answer is [tex]$\boxed{28x+30}$[/tex].
[tex]$$
3x^3 - 2x^2 + 4x - 3
$$[/tex]
is divided by
[tex]$$
x^2 + 3x + 3.
$$[/tex]
We will use polynomial long division.
1. First Division Step:
The leading term of the dividend is [tex]$3x^3$[/tex] and of the divisor is [tex]$x^2$[/tex]. Dividing these gives
[tex]$$
\frac{3x^3}{x^2} = 3x.
$$[/tex]
Multiply the divisor by [tex]$3x$[/tex]:
[tex]$$
3x \cdot (x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.
$$[/tex]
Subtract this result from the dividend:
[tex]$$
\begin{aligned}
(3x^3 - 2x^2 + 4x - 3) &- (3x^3 + 9x^2 + 9x) \\
&= 3x^3 - 2x^2 + 4x - 3 - 3x^3 - 9x^2 - 9x \\
&= (-2x^2 - 9x^2) + (4x - 9x) - 3 \\
&= -11x^2 - 5x - 3.
\end{aligned}
$$[/tex]
2. Second Division Step:
Now, the new dividend is [tex]$-11x^2 - 5x - 3$[/tex]. Divide the leading term [tex]$-11x^2$[/tex] by [tex]$x^2$[/tex]:
[tex]$$
\frac{-11x^2}{x^2} = -11.
$$[/tex]
Multiply the divisor by [tex]$-11$[/tex]:
[tex]$$
-11 \cdot (x^2 + 3x + 3) = -11x^2 - 33x - 33.
$$[/tex]
Subtract this from the current dividend:
[tex]$$
\begin{aligned}
(-11x^2 - 5x - 3) &- (-11x^2 - 33x - 33) \\
&= -11x^2 - 5x - 3 + 11x^2 + 33x + 33 \\
&= ( -11x^2 + 11x^2) + (-5x + 33x) + (-3 + 33) \\
&= 28x + 30.
\end{aligned}
$$[/tex]
Since the degree of the remainder [tex]$28x + 30$[/tex] is less than the degree of the divisor (which is 2), the division is complete.
3. Conclusion:
The remainder when dividing
[tex]$$
3x^3 - 2x^2 + 4x - 3
$$[/tex]
by
[tex]$$
x^2 + 3x + 3
$$[/tex]
is
[tex]$$
28x + 30.
$$[/tex]
Thus, the correct answer is [tex]$\boxed{28x+30}$[/tex].
