To evaluate $P\left(\frac{1}{3}\right)$ using synthetic division and the Remainder Theorem:

Step 1: Set up synthetic division

The polynomial is $P(x) = 3x^3 - x^2 + 6x + 6$, and we want to evaluate $P\left(\frac{1}{3}\right)$.

• The coefficients of $P(x)$ are: $3, -1, 6, 6$.
• The value of $a = \frac{1}{3}$ is the divisor.

Step 2: Synthetic division process

Write down the coefficients:

$3 \quad -1 \quad 6 \quad 6$

Use $\frac{1}{3}$ as the divisor:

1. Bring down the first coefficient ($3$) unchanged.
2. Multiply $3$ by $\frac{1}{3}$: $3 \times \frac{1}{3} = 1$.
3. Add $-1 + 1 = 0$.
4. Multiply $0$ by $\frac{1}{3}$: $0 \times \frac{1}{3} = 0$.
5. Add $6 + 0 = 6$.
6. Multiply $6$ by $\frac{1}{3}$: $6 \times \frac{1}{3} = 2$.
7. Add $6 + 2 = 8$.

The synthetic division table looks like this:

\frac{1}{3} & 3 & -1 & 6 & 6 \\ & & 1 & 0 & 2 \\ \hline & 3 & 0 & 6 & 8 \\ \end{array}$$ ### Step 3: Interpret the result The remainder is $$8$$. According to the Remainder Theorem, $$P\left(\frac{1}{3}\right) = 8$$. ### Final Answer: $$P\left(\frac{1}{3}\right) = 8$$ --- Would you like more details or examples? ### Related Questions: 1. How does the Remainder Theorem relate to synthetic division? 2. What is the purpose of synthetic division in polynomial evaluation? 3. Can synthetic division be used to factor polynomials? If so, how? 4. What happens if the divisor is not a fraction (e.g., $$a = 2$$)? 5. How does synthetic division differ from long division of polynomials? ### Tip: When using synthetic division with fractional divisors like $$\frac{1}{3}$$, ensure all arithmetic operations with fractions are carefully simplified.