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 $(6n^2-11n+3)÷(3n-1)$, 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: $$\boxed{{\displaystyle 2n-3}}$$ for all values of $n$ where the expression is defined, except where the divisor $3n-1$ is zero (i.e., $n\ne \frac{1}{3}$).