Answer :
To find the remainder when
[tex]$$3x^3 - 2x^2 + 4x - 3$$[/tex]
is divided by
[tex]$$x^2 + 3x + 3,$$[/tex]
we can perform polynomial long division.
Step 1. Divide the leading term of the dividend by the leading term of the divisor:
[tex]$$\frac{3x^3}{x^2} = 3x.$$[/tex]
Multiply the entire divisor by [tex]$3x$[/tex]:
[tex]$$3x(x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.$$[/tex]
Subtract this product from the original dividend:
[tex]\[
\begin{aligned}
& (3x^3 - 2x^2 + 4x - 3) - (3x^3 + 9x^2 + 9x) \\
=\, & 3x^3 - 3x^3 - 2x^2 - 9x^2 + 4x - 9x - 3 \\
=\, & -11x^2 - 5x - 3.
\end{aligned}
\][/tex]
Step 2. Now, divide the leading term of the new polynomial by the leading term of the divisor:
[tex]$$\frac{-11x^2}{x^2} = -11.$$[/tex]
Multiply the divisor by [tex]$-11$[/tex]:
[tex]$$-11(x^2 + 3x + 3) = -11x^2 - 33x - 33.$$[/tex]
Subtract this from the previous result:
[tex]\[
\begin{aligned}
& (-11x^2 - 5x - 3) - (-11x^2 - 33x - 33) \\
=\, & -11x^2 + 11x^2 - 5x + 33x - 3 + 33 \\
=\, & 28x + 30.
\end{aligned}
\][/tex]
Since the degree of the remainder ([tex]$28x + 30$[/tex], which is linear) is less than the degree of the divisor (which is quadratic), the division process stops here.
Thus, the remainder when
[tex]$$3x^3 - 2x^2 + 4x - 3$$[/tex]
is divided by
[tex]$$x^2 + 3x + 3$$[/tex]
is
[tex]$$\boxed{28x+30}.$$[/tex]
[tex]$$3x^3 - 2x^2 + 4x - 3$$[/tex]
is divided by
[tex]$$x^2 + 3x + 3,$$[/tex]
we can perform polynomial long division.
Step 1. Divide the leading term of the dividend by the leading term of the divisor:
[tex]$$\frac{3x^3}{x^2} = 3x.$$[/tex]
Multiply the entire divisor by [tex]$3x$[/tex]:
[tex]$$3x(x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.$$[/tex]
Subtract this product from the original dividend:
[tex]\[
\begin{aligned}
& (3x^3 - 2x^2 + 4x - 3) - (3x^3 + 9x^2 + 9x) \\
=\, & 3x^3 - 3x^3 - 2x^2 - 9x^2 + 4x - 9x - 3 \\
=\, & -11x^2 - 5x - 3.
\end{aligned}
\][/tex]
Step 2. Now, divide the leading term of the new polynomial by the leading term of the divisor:
[tex]$$\frac{-11x^2}{x^2} = -11.$$[/tex]
Multiply the divisor by [tex]$-11$[/tex]:
[tex]$$-11(x^2 + 3x + 3) = -11x^2 - 33x - 33.$$[/tex]
Subtract this from the previous result:
[tex]\[
\begin{aligned}
& (-11x^2 - 5x - 3) - (-11x^2 - 33x - 33) \\
=\, & -11x^2 + 11x^2 - 5x + 33x - 3 + 33 \\
=\, & 28x + 30.
\end{aligned}
\][/tex]
Since the degree of the remainder ([tex]$28x + 30$[/tex], which is linear) is less than the degree of the divisor (which is quadratic), the division process stops here.
Thus, the remainder when
[tex]$$3x^3 - 2x^2 + 4x - 3$$[/tex]
is divided by
[tex]$$x^2 + 3x + 3$$[/tex]
is
[tex]$$\boxed{28x+30}.$$[/tex]
