Question: Use the long division method to find the result when $6x^3 - 25x^2 + 8x + 27$ is divided by $2x - 7$. 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 $6x^3 - 25x^2 + 8x + 27$ by the binomial $2x - 7$ using long division.
2. **Recall the long division formula:** When dividing $f(x)$ by $g(x)$, the result is
$$f(x) = q(x) \cdot g(x) + r(x)$$
where $q(x)$ is the quotient and $r(x)$ is the remainder with degree less than $g(x)$.
3. **Set up the division:**
Divide $6x^3$ by $2x$ to get the first term of the quotient:
$$\frac{6x^3}{2x} = 3x^2$$
4. **Multiply and subtract:**
Multiply $3x^2$ by $2x - 7$:
$$3x^2 \cdot (2x - 7) = 6x^3 - 21x^2$$
Subtract this from the original polynomial:
$$\left(6x^3 - 25x^2 + 8x + 27\right) - \left(6x^3 - 21x^2\right) = -4x^2 + 8x + 27$$
5. **Repeat the process:**
Divide $-4x^2$ by $2x$:
$$\frac{-4x^2}{2x} = -2x$$
Multiply $-2x$ by $2x - 7$:
$$-2x \cdot (2x - 7) = -4x^2 + 14x$$
Subtract:
$$\left(-4x^2 + 8x + 27\right) - \left(-4x^2 + 14x\right) = -6x + 27$$
6. **Continue:**
Divide $-6x$ by $2x$:
$$\frac{-6x}{2x} = -3$$
Multiply $-3$ by $2x - 7$:
$$-3 \cdot (2x - 7) = -6x + 21$$
Subtract:
$$\left(-6x + 27\right) - \left(-6x + 21\right) = 6$$
7. **Write the final answer:**
The quotient is:
$$q(x) = 3x^2 - 2x - 3$$
The remainder is:
$$r(x) = 6$$
The divisor is:
$$b(x) = 2x - 7$$
So the division result is:
$$3x^2 - 2x - 3 + \frac{6}{2x - 7}$$