College

What is the remainder when [tex]3x^3 - 2x^2 + 4x - 3[/tex] is divided by [tex]x^2 + 3x + 3[/tex]?

A. 30
B. [tex]3x - 11[/tex]
C. [tex]28x - 36[/tex]
D. [tex]28x + 30[/tex]

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]

Since the divisor is a quadratic polynomial, the remainder must be a linear polynomial of the form
[tex]$$
Ax + B.
$$[/tex]

We can use polynomial long division to find the quotient and the remainder. Here are the steps:

1. Divide the leading terms:

Divide the leading term of the dividend by the leading term of the divisor:

[tex]$$
\frac{3x^3}{x^2} = 3x.
$$[/tex]

This gives the first term of the quotient,
[tex]$$
Q(x) = 3x + \cdots.
$$[/tex]

2. Multiply and subtract:

Multiply the entire divisor by [tex]$3x$[/tex]:

[tex]$$
3x \cdot (x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.
$$[/tex]

Subtract this from the original dividend:

[tex]$$
\begin{aligned}
\left(3x^3 - 2x^2 + 4x - 3\right) - \left(3x^3 + 9x^2 + 9x\right)
&= (3x^3 - 3x^3) + (-2x^2 - 9x^2) + (4x - 9x) - 3 \\
&= -11x^2 - 5x - 3.
\end{aligned}
$$[/tex]

3. Divide the new leading term:

Divide the new leading term by the leading term of the divisor:

[tex]$$
\frac{-11x^2}{x^2} = -11.
$$[/tex]

Add this to the quotient:

[tex]$$
Q(x) = 3x - 11.
$$[/tex]

4. Multiply and subtract again:

Multiply the divisor by [tex]$-11$[/tex]:

[tex]$$
-11 \cdot (x^2 + 3x + 3) = -11x^2 - 33x - 33.
$$[/tex]

Subtract this product from the previous remainder:

[tex]$$
\begin{aligned}
\left(-11x^2 - 5x - 3\right) - \left(-11x^2 - 33x - 33\right)
&= (-11x^2 + 11x^2) + (-5x + 33x) + (-3 + 33) \\
&= 28x + 30.
\end{aligned}
$$[/tex]

This result, [tex]$28x + 30$[/tex], is the remainder.

5. Conclusion:

The polynomial division can be expressed as:

[tex]$$
3x^3 - 2x^2 + 4x - 3 = (x^2 + 3x + 3)(3x - 11) + (28x + 30).
$$[/tex]

Thus, the remainder when dividing the polynomial is

[tex]$$
28x + 30.
$$[/tex]