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

**Synthetic Division Setup** For the polynomial \[ f(x)=x^4-16, \] write it in full form (including all degrees): \[ f(x)=1x^4+0x^3+0x^2+0x-16. \] The coefficients are: \[ 1,\quad0,\quad0,\quad0,\quad-16. \] --- **1. Evaluating \( f(3) \) using Synthetic Division** We set up synthetic division with \( x=3 \): - Write the coefficients: \[ \begin{array}{cccccc} \text{Root: }3 & | & 1 & 0 & 0 & 0 & -16 \\ \hline & & 1 & & & & \\ \end{array} \] - **Step 1:** Bring down the first coefficient: \(1\). - **Step 2:** Multiply \(1\) by \(3\) and write the result under the next coefficient: \(1\times3=3\). Now, add \(0+3=3\). - **Step 3:** Multiply the new value \(3\) by \(3\): \(3\times3=9\). Then, add: \(0+9=9\). - **Step 4:** Multiply \(9\) by \(3\): \(9\times3=27\). Add: \(0+27=27\). - **Step 5:** Multiply \(27\) by \(3\): \(27\times3=81\). Finally, add: \(-16+81=65\). The bottom row represents the coefficients of the depressed polynomial with the last number as the remainder, which is \(f(3)\). Thus, \[ f(3)=65. \] --- **2. Evaluating \( f(-3) \) using Synthetic Division** Use the same coefficients and set \( x=-3 \): - Coefficients: \[ 1,\quad 0,\quad 0,\quad 0,\quad -16. \] - **Step 1:** Bring down the first coefficient: \(1\). - **Step 2:** Multiply \(1\) by \(-3\): \(1\times(-3)=-3\). Add: \(0+(-3)=-3\). - **Step 3:** Multiply \(-3\) by \(-3\): \(-3\times(-3)=9\). Add: \(0+9=9\). - **Step 4:** Multiply \(9\) by \(-3\): \(9\times(-3)=-27\). Add: \(0+(-27)=-27\). - **Step 5:** Multiply \(-27\) by \(-3\): \(-27\times(-3)=81\). Add: \(-16+81=65\). Thus, \[ f(-3)=65. \] --- **3. Evaluating \( f(1-\sqrt{6}) \)** We now evaluate \[ f(1-\sqrt{6}) = (1-\sqrt{6})^4 - 16. \] While synthetic division can be used with irrational numbers, it is often simpler to compute using the binomial expansion. Use the formula for \((a-b)^4\) with \( a=1 \) and \( b=\sqrt{6} \): \[ (1-\sqrt{6})^4 = 1^4 - 4(1^3)(\sqrt{6}) + 6(1^2)(\sqrt{6})^2 - 4(1)(\sqrt{6})^3 + (\sqrt{6})^4. \] Now, compute each term: 1. \(1^4 = 1.\) 2. \(- 4(1^3)(\sqrt{6}) = -4\sqrt{6}.\) 3. \(6(1^2)(\sqrt{6})^2 = 6 \times 6 = 36,\) since \((\sqrt{6})^2=6.\) 4. \(- 4(1)(\sqrt{6})^3 = -4(6\sqrt{6}) = -24\sqrt{6},\) because \((\sqrt{6})^3=6\sqrt{6}.\) 5. \((\sqrt{6})^4 = 36.\) Combine the terms: \[ (1-\sqrt{6})^4 = 1 - 4\sqrt{6} + 36 - 24\sqrt{6} + 36. \] Group similar terms: - Constants: \(1 + 36 + 36 = 73.\) - Radical terms: \(-4\sqrt{6} - 24\sqrt{6} = -28\sqrt{6}.\) Thus, \[ (1-\sqrt{6})^4 = 73 - 28\sqrt{6}. \] Subtract 16: \[ f(1-\sqrt{6}) = (73 - 28\sqrt{6}) - 16 = 57 - 28\sqrt{6}. \] --- **Final Answers** \[ \begin{aligned} f(3) &= 65, \\ f(-3) &= 65, \\ f(1-\sqrt{6}) &= 57-28\sqrt{6}. \end{aligned} \] So, the answer for \[ f(3)=\boxed{65}. \]