Question

what is the quotient of (6n^2 - 11n +3) / (3n - 1) for all values of n where the expression is defined

Answer 1

To find the quotient of \( (6n^2 - 11n + 3) \div (3n - 1) \), we can use polynomial long division.

1. **Divide the leading terms**:
- The leading term of the dividend \( 6n^2 \) divided by the leading term of the divisor \( 3n \) gives:
\[
\frac{6n^2}{3n} = 2n
\]

2. **Multiply the entire divisor by this result**:
\[
2n \cdot (3n - 1) = 6n^2 - 2n
\]

3. **Subtract this from the original polynomial**:
\[
(6n^2 - 11n + 3) - (6n^2 - 2n) = -11n + 2n + 3 = -9n + 3
\]

4. **Repeat the process with the new polynomial \(-9n + 3\)**:
- Divide the leading term \(-9n\) by \(3n\):
\[
\frac{-9n}{3n} = -3
\]

5. **Multiply the entire divisor by this result**:
\[
-3 \cdot (3n - 1) = -9n + 3
\]

6. **Subtract this from the previous remainder**:
\[
(-9n + 3) - (-9n + 3) = 0
\]

At this point, our remainder is \(0\). Thus, we can conclude that:
\[
\frac{6n^2 - 11n + 3}{3n - 1} = 2n - 3
\]

The quotient is:
\[
\boxed{2n - 3}
\]
for all values of \( n \) where the expression is defined, except where the divisor \( 3n - 1 \) is zero (i.e., \( n \neq \frac{1}{3} \)).