Question

Use the long division method to find the result when 3, x, cubed, plus, 2, x, squared, minus, 18, x, plus, 93x
3
+2x
2
−18x+9 is divided by x, plus, 3x+3.

Answer 1

To perform long division for the polynomial \(3x^3 + 2x^2 - 18x + 9\) divided by \(x + 3\), we follow these steps:

1. **Set up the long division**:
We write \(3x^3 + 2x^2 - 18x + 9\) under the long division symbol and \(x + 3\) outside.

2. **Divide the leading terms**:
Divide the leading term of the dividend (\(3x^3\)) by the leading term of the divisor (\(x\)):
\[
\frac{3x^3}{x} = 3x^2
\]
This is the first term of the quotient.

3. **Multiply and subtract**:
Multiply the entire divisor \(x + 3\) by \(3x^2\):
\[
3x^2(x + 3) = 3x^3 + 9x^2
\]
Now subtract this from the original polynomial:
\[
(3x^3 + 2x^2 - 18x + 9) - (3x^3 + 9x^2) = 2x^2 - 9x^2 - 18x + 9 = -7x^2 - 18x + 9
\]

4. **Repeat the process**:
Divide the new leading term (\(-7x^2\)) by the leading term of the divisor (\(x\)):
\[
\frac{-7x^2}{x} = -7x
\]
Multiply the divisor by \(-7x\):
\[
-7x(x + 3) = -7x^2 - 21x
\]
Subtract again:
\[
(-7x^2 - 18x + 9) - (-7x^2 - 21x) = -18x + 21x + 9 = 3x + 9
\]

5. **Repeat again**:
Divide \(3x\) by \(x\):
\[
\frac{3x}{x} = 3
\]
Multiply the divisor by \(3\):
\[
3(x + 3) = 3x + 9
\]
Subtract:
\[
(3x + 9) - (3x + 9) = 0
\]

At this point, there are no more terms to bring down, and the remainder is zero.

6. **Final quotient**:
Thus, the result of the long division is:
\[
\boxed{3x^2 - 7x + 3}
\]
The polynomial \(3x^3 + 2x^2 - 18x + 9\) divided by \(x + 3\) gives a quotient of \(3x^2 - 7x + 3\) with a remainder of 0.