Question: Use the long division method to find the result when $6x^3 - 7x^2 + 23x - 7$ is divided by $3x - 2$. If there is a remainder, express the result in the form $q(x) + \frac{r(x)}{b(x)}$.
1. **State the problem:** Divide the polynomial $$6x^3 - 7x^2 + 23x - 7$$ by the binomial $$3x - 2$$ using long division.
2. **Recall the formula:** When dividing polynomials, the result can be expressed as $$\text{Dividend} = (\text{Divisor}) \times (\text{Quotient}) + \text{Remainder}$$.
3. **Set up the division:**
Divide the leading term of the dividend $$6x^3$$ by the leading term of the divisor $$3x$$:
$$\frac{6x^3}{3x} = 2x^2$$
This is the first term of the quotient.
4. **Multiply and subtract:**
Multiply $$2x^2$$ by the divisor $$3x - 2$$:
$$2x^2 \times (3x - 2) = 6x^3 - 4x^2$$
Subtract this from the dividend:
$$\begin{aligned}
&(6x^3 - 7x^2 + 23x - 7) - (6x^3 - 4x^2) \\
&= 6x^3 - 7x^2 + 23x - 7 - 6x^3 + 4x^2 \\
&= \cancel{6x^3} - 7x^2 + 23x - 7 - \cancel{6x^3} + 4x^2 \\
&= (-7x^2 + 4x^2) + 23x - 7 = -3x^2 + 23x - 7
\end{aligned}$$
5. **Repeat the process:**
Divide the new leading term $$-3x^2$$ by $$3x$$:
$$\frac{-3x^2}{3x} = -x$$
Multiply $$-x$$ by the divisor:
$$-x \times (3x - 2) = -3x^2 + 2x$$
Subtract:
$$\begin{aligned}
&(-3x^2 + 23x - 7) - (-3x^2 + 2x) \\
&= -3x^2 + 23x - 7 + 3x^2 - 2x \\
&= \cancel{-3x^2} + 23x - 7 + \cancel{3x^2} - 2x \\
&= (23x - 2x) - 7 = 21x - 7
\end{aligned}$$
6. **Continue:**
Divide $$21x$$ by $$3x$$:
$$\frac{21x}{3x} = 7$$
Multiply $$7$$ by the divisor:
$$7 \times (3x - 2) = 21x - 14$$
Subtract:
$$\begin{aligned}
&(21x - 7) - (21x - 14) \\
&= 21x - 7 - 21x + 14 \\
&= \cancel{21x} - 7 - \cancel{21x} + 14 = 7
\end{aligned}$$
7. **Interpret the result:**
The quotient is $$2x^2 - x + 7$$ and the remainder is $$7$$.
Express the result as:
$$\frac{6x^3 - 7x^2 + 23x - 7}{3x - 2} = 2x^2 - x + 7 + \frac{7}{3x - 2}$$
**Final answer:** $$2x^2 - x + 7 + \frac{7}{3x - 2}$$