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]
One method to solve this is through polynomial long division. The idea is to write
[tex]$$
3x^3 - 2x^2 + 4x - 3 = Q(x)\,(x^2 + 3x + 3) + R(x)
$$[/tex]
where [tex]$Q(x)$[/tex] is the quotient and [tex]$R(x)$[/tex] is the remainder. Since the divisor has degree [tex]$2$[/tex], the remainder must have degree strictly less than [tex]$2$[/tex], so we can write it in the form
[tex]$$
R(x) = Ax + B,
$$[/tex]
where [tex]$A$[/tex] and [tex]$B$[/tex] are constants to be determined.
Below are the steps of the division:
1. Divide the leading term of the dividend by the leading term of the divisor.
The leading term of the dividend is [tex]$3x^3$[/tex], and the leading term of the divisor is [tex]$x^2$[/tex]. Dividing, we get:
[tex]$$
\frac{3x^3}{x^2} = 3x.
$$[/tex]
Thus, the first term in the quotient is [tex]$3x$[/tex].
2. Multiply the divisor by [tex]$3x$[/tex].
Multiply each term in [tex]$x^2 + 3x + 3$[/tex] by [tex]$3x$[/tex]:
[tex]$$
3x\,(x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.
$$[/tex]
3. Subtract the result from the original dividend.
Subtract:
[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]
4. Divide the new leading term by the leading term of the divisor.
The new leading term is [tex]$-11x^2$[/tex], so we divide by [tex]$x^2$[/tex]:
[tex]$$
\frac{-11x^2}{x^2} = -11.
$$[/tex]
This gives the next (and last) term of the quotient: [tex]$-11$[/tex].
5. Multiply the divisor by [tex]$-11$[/tex].
Multiply:
[tex]$$
-11\,(x^2 + 3x + 3) = -11x^2 - 33x - 33.
$$[/tex]
6. Subtract this from the current polynomial.
Now subtract:
[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]
At this point, the degree of [tex]$28x+30$[/tex] (which is 1) is less than the degree of the divisor (which is 2), so we stop here. This [tex]$28x+30$[/tex] is the remainder.
Thus, the quotient is [tex]$3x - 11$[/tex] and the remainder is
[tex]$$
28x + 30.
$$[/tex]
So, the answer to the question is that the remainder is [tex]$$28x+30.$$[/tex]
[tex]$$
3x^3 - 2x^2 + 4x - 3
$$[/tex]
is divided by
[tex]$$
x^2 + 3x + 3.
$$[/tex]
One method to solve this is through polynomial long division. The idea is to write
[tex]$$
3x^3 - 2x^2 + 4x - 3 = Q(x)\,(x^2 + 3x + 3) + R(x)
$$[/tex]
where [tex]$Q(x)$[/tex] is the quotient and [tex]$R(x)$[/tex] is the remainder. Since the divisor has degree [tex]$2$[/tex], the remainder must have degree strictly less than [tex]$2$[/tex], so we can write it in the form
[tex]$$
R(x) = Ax + B,
$$[/tex]
where [tex]$A$[/tex] and [tex]$B$[/tex] are constants to be determined.
Below are the steps of the division:
1. Divide the leading term of the dividend by the leading term of the divisor.
The leading term of the dividend is [tex]$3x^3$[/tex], and the leading term of the divisor is [tex]$x^2$[/tex]. Dividing, we get:
[tex]$$
\frac{3x^3}{x^2} = 3x.
$$[/tex]
Thus, the first term in the quotient is [tex]$3x$[/tex].
2. Multiply the divisor by [tex]$3x$[/tex].
Multiply each term in [tex]$x^2 + 3x + 3$[/tex] by [tex]$3x$[/tex]:
[tex]$$
3x\,(x^2 + 3x + 3) = 3x^3 + 9x^2 + 9x.
$$[/tex]
3. Subtract the result from the original dividend.
Subtract:
[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]
4. Divide the new leading term by the leading term of the divisor.
The new leading term is [tex]$-11x^2$[/tex], so we divide by [tex]$x^2$[/tex]:
[tex]$$
\frac{-11x^2}{x^2} = -11.
$$[/tex]
This gives the next (and last) term of the quotient: [tex]$-11$[/tex].
5. Multiply the divisor by [tex]$-11$[/tex].
Multiply:
[tex]$$
-11\,(x^2 + 3x + 3) = -11x^2 - 33x - 33.
$$[/tex]
6. Subtract this from the current polynomial.
Now subtract:
[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]
At this point, the degree of [tex]$28x+30$[/tex] (which is 1) is less than the degree of the divisor (which is 2), so we stop here. This [tex]$28x+30$[/tex] is the remainder.
Thus, the quotient is [tex]$3x - 11$[/tex] and the remainder is
[tex]$$
28x + 30.
$$[/tex]
So, the answer to the question is that the remainder is [tex]$$28x+30.$$[/tex]
