Question: Divide and check.
$$\frac{3x^2 - 28x}{3x - 1}$$
1. **State the problem:** We need to divide the polynomial $$3x^2 - 28x$$ by the binomial $$3x - 1$$ and then check the result.
2. **Recall the formula:** Polynomial long division is similar to numerical long division. We divide the leading term of the dividend by the leading term of the divisor, multiply the divisor by that quotient, subtract, and repeat until the degree of the remainder is less than the divisor.
3. **Set up the division:**
$$\frac{3x^2 - 28x}{3x - 1}$$
4. **Divide the leading terms:**
$$\frac{3x^2}{3x} = x$$
5. **Multiply the divisor by $$x$$:**
$$x(3x - 1) = 3x^2 - x$$
6. **Subtract this from the dividend:**
$$\begin{aligned}
(3x^2 - 28x) - (3x^2 - x) &= 3x^2 - 28x - 3x^2 + x \\
&= -27x
\end{aligned}$$
7. **Bring down the remainder:**
Now we have $$-27x$$.
8. **Divide the leading term of the remainder by the leading term of the divisor:**
$$\frac{-27x}{3x} = -9$$
9. **Multiply the divisor by $$-9$$:**
$$-9(3x - 1) = -27x + 9$$
10. **Subtract this from the current remainder:**
$$\begin{aligned}
(-27x) - (-27x + 9) &= -27x + 27x - 9 \\
&= -9
\end{aligned}$$
11. **Since the degree of the remainder $$-9$$ is less than the degree of the divisor $$3x - 1$$, we stop here.**
12. **Write the quotient and remainder:**
$$\text{Quotient} = x - 9$$
$$\text{Remainder} = -9$$
13. **Express the division result:**
$$\frac{3x^2 - 28x}{3x - 1} = x - 9 + \frac{-9}{3x - 1}$$
14. **Check the result:** Multiply the divisor by the quotient and add the remainder:
$$\begin{aligned}
(3x - 1)(x - 9) + (-9) &= 3x \cdot x - 3x \cdot 9 - 1 \cdot x + 1 \cdot 9 - 9 \\
&= 3x^2 - 27x - x + 9 - 9 \\
&= 3x^2 - 28x
\end{aligned}$$
This matches the original dividend, so the division is correct.
**Final answer:**
$$\frac{3x^2 - 28x}{3x - 1} = x - 9 - \frac{9}{3x - 1}$$