Question

Use long division to divide 3x3−9x2−72/3x2−2 .(1 point) Responses x−3 x minus 3 x−3+2x−78 x minus 3 plus 2 x minus 78 x−3+2x−723x2−2 x minus 3 plus Start Fraction 2 x minus 72 over 3 x squared minus 2 End Fraction x−3+2x−783x2−2

Answer 1

To perform long division with the polynomial \(3x^3 - 9x^2 - 72\) divided by \(3x^2 - 2\), follow these steps:

1. **Set up the division**:
Write \(3x^3 - 9x^2 - 72\) under the long division symbol and \(3x^2 - 2\) outside.

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

3. **Multiply and subtract**:
Multiply the entire divisor \(3x^2 - 2\) by \(x\) and subtract from the original polynomial:
\[
x(3x^2 - 2) = 3x^3 - 2x
\]
Now, subtract this from the original polynomial:
\[
(3x^3 - 9x^2 - 72) - (3x^3 - 2x) = -9x^2 + 2x - 72
\]

4. **Bring down the next term (if needed)**:
There are no additional terms to bring down because we are still working with the remainder \(-9x^2 + 2x - 72\).

5. **Repeat the process**:
Divide the leading term of the new polynomial by the leading term of the divisor:
\[
\frac{-9x^2}{3x^2} = -3
\]
Add this to the quotient: \(x - 3\).

6. **Multiply and subtract again**:
Multiply the entire divisor \(3x^2 - 2\) by \(-3\):
\[
-3(3x^2 - 2) = -9x^2 + 6
\]
Subtract this from the previous result:
\[
(-9x^2 + 2x - 72) - (-9x^2 + 6) = 2x - 66
\]

7. **Final remainder**:
Now we continue with \(2x - 66\). The division is complete, but you can still divide:
\[
\frac{2x}{3x^2 - 2}
\]
This gives you the remainder.

So, the result of the division is:
\[
\text{Quotient: } x - 3 \quad \text{Remainder: } \frac{2x - 66}{3x^2 - 2}
\]

Alternatively, you can write the full expression as:
\[
x - 3 + \frac{2x - 66}{3x^2 - 2}
\]

Putting it all together, the final answer is:
\[
\boxed{x - 3 + \frac{2x - 66}{3x^2 - 2}}
\]