To find the quotient of \( (6n^2 - 11n + 3) \div (3n - 1) \), we can use polynomial long division.
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Divide the leading terms:
- The leading term of the dividend \( 6n^2 \) divided by the leading term of the divisor \( 3n \) gives: \[ \frac{6n^2}{3n} = 2n \]
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Multiply the entire divisor by this result: \[ 2n \cdot (3n - 1) = 6n^2 - 2n \]
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Subtract this from the original polynomial: \[ (6n^2 - 11n + 3) - (6n^2 - 2n) = -11n + 2n + 3 = -9n + 3 \]
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Repeat the process with the new polynomial \(-9n + 3\):
- Divide the leading term \(-9n\) by \(3n\): \[ \frac{-9n}{3n} = -3 \]
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Multiply the entire divisor by this result: \[ -3 \cdot (3n - 1) = -9n + 3 \]
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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., \( n \neq \frac{1}{3} \)).
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