1. **State the problem:** Simplify the expression $$\frac{4p^3 - p^2 + 2p}{3p - 1}$$ using synthetic division and write the result in standard form. 2. **Rewrite divisor:** The divisor is $$3p - 1$$. To use synthetic division, rewrite it as $$p - \frac{1}{3}$$ by factoring out 3. We will divide by $$p - \frac{1}{3}$$ and adjust accordingly. 3. **Set up synthetic division:** The coefficients of the dividend $$4p^3 - p^2 + 2p + 0$$ (note the zero for the constant term) are: $$4, -1, 2, 0$$. 4. **Use synthetic division with root $$\frac{1}{3}$$:** - Bring down 4. - Multiply 4 by $$\frac{1}{3}$$ to get $$\frac{4}{3}$$. - Add to -1: $$-1 + \frac{4}{3} = \frac{-3 + 4}{3} = \frac{1}{3}$$. - Multiply $$\frac{1}{3}$$ by $$\frac{1}{3}$$ to get $$\frac{1}{9}$$. - Add to 2: $$2 + \frac{1}{9} = \frac{18}{9} + \frac{1}{9} = \frac{19}{9}$$. - Multiply $$\frac{19}{9}$$ by $$\frac{1}{3}$$ to get $$\frac{19}{27}$$. - Add to 0: $$0 + \frac{19}{27} = \frac{19}{27}$$ (remainder). 5. **Write quotient and remainder:** The quotient coefficients are $$4, \frac{1}{3}, \frac{19}{9}$$ corresponding to $$4p^2 + \frac{1}{3}p + \frac{19}{9}$$. 6. **Adjust for original divisor:** Since we divided by $$p - \frac{1}{3}$$ instead of $$3p - 1$$, multiply the quotient by 3: $$3 \times \left(4p^2 + \frac{1}{3}p + \frac{19}{9}\right) = 12p^2 + p + \frac{19}{3}$$. The remainder must be divided by the original divisor $$3p - 1$$: $$\frac{19/27}{3p - 1} = \frac{19}{27(3p - 1)}$$. 7. **Final expression:** $$\frac{4p^3 - p^2 + 2p}{3p - 1} = 12p^2 + p + \frac{19}{3} + \frac{19}{27(3p - 1)}$$. **Answer:** $$12p^2 + p + \frac{19}{3} + \frac{19}{27(3p - 1)}$$