Use synthetic division to find the function values. \( f(x)=x^{4}-16 \); find \( f(3), f(-3) \), and \( f(1-\sqrt{6}) \). \( f(3)=65 \) (Type an exact answer, using radicals as needed.) \( f(-3)=\square \) (Type an exact answer, using radicals as needed.)

To find the function values \( f(-3) \) and \( f(1 - \sqrt{6}) \) using synthetic division, we will first evaluate \( f(-3) \) using synthetic division. ### Step 1: Synthetic Division for \( f(-3) \) The polynomial is \( f(x) = x^4 - 16 \). We can rewrite it in standard form as: \[ f(x) = 1x^4 + 0x^3 + 0x^2 + 0x - 16 \] We will perform synthetic division with \( -3 \). #### Synthetic Division Setup - Coefficients: \( [1, 0, 0, 0, -16] \) - Divisor: \( -3 \) #### Synthetic Division Process 1. Write down the coefficients: \( 1, 0, 0, 0, -16 \) 2. Bring down the leading coefficient (1). 3. Multiply by the divisor (-3) and add to the next coefficient. Let's perform the calculations: - Bring down \( 1 \). - Multiply \( 1 \) by \( -3 \) to get \( -3 \), and add to \( 0 \) to get \( -3 \). - Multiply \( -3 \) by \( -3 \) to get \( 9 \), and add to \( 0 \) to get \( 9 \). - Multiply \( 9 \) by \( -3 \) to get \( -27 \), and add to \( 0 \) to get \( -27 \). - Multiply \( -27 \) by \( -3 \) to get \( 81 \), and add to \( -16 \) to get \( 65 \). The synthetic division gives us: \[ \begin{array}{r|rrrrr} -3 & 1 & 0 & 0 & 0 & -16 \\ & & -3 & 9 & -27 & 81 \\ \hline & 1 & -3 & 9 & -27 & 65 \\ \end{array} \] The remainder is \( 65 \), so: \[ f(-3) = 65 \] ### Step 2: Synthetic Division for \( f(1 - \sqrt{6}) \) Now we will evaluate \( f(1 - \sqrt{6}) \) using synthetic division. However, since \( 1 - \sqrt{6} \) is not a simple rational number, we will directly substitute it into the polynomial instead of using synthetic division. #### Direct Substitution \[ f(1 - \sqrt{6}) = (1 - \sqrt{6})^4 - 16 \] We will first calculate \( (1 - \sqrt{6})^4 \). Using the binomial theorem: \[ (1 - \sqrt{6})^4 = \sum_{k=0}^{4} \binom{4}{k} (1)^{4-k} (-\sqrt{6})^k \] Calculating each term: - For \( k = 0 \): \( \binom{4}{0} (1)^4 (-\sqrt{6})^0 = 1 \) - For \( k = 1 \): \( \binom{4}{1} (1)^3 (-\sqrt{6})^1 = -4\sqrt{6} \) - For \( k = 2 \): \( \binom{4}{2} (1)^2 (-\sqrt{6})^2 = 6 \cdot 6 = 36 \) - For \( k = 3 \): \( \binom{4}{3} (1)^1 (-\sqrt{6})^3 = -4 \cdot 6\sqrt{6} = -24\sqrt{6} \) - For \( k = 4 \): \( \binom{4}{4} (1)^0 (-\sqrt{6})^4 = 36 \) Now, summing these terms: \[ (1 - \sqrt{6})^4 = 1 - 4\sqrt{6} + 36 - 24\sqrt{6} + 36 = 73 - 28\sqrt{6} \] Now substituting back into \( f(1 - \sqrt{6}) \): \[ f(1 - \sqrt{6}) = (73 - 28\sqrt{6}) - 16 = 57 - 28\sqrt{6} \] ### Final Answers - \( f(3) = 65 \) - \( f(-3) = 65 \) - \( f(1 - \sqrt{6}) = 57 - 28\sqrt{6} \) Thus, the final answers are: - \( f(-3) = 65 \) - \( f(1 - \sqrt{6}) = 57 - 28\sqrt{6} \)