\[ P(x)=x^{3}+2 x^{2}-6 x-4, \quad c=2 \] Evaluate \( D(x) \) for the given value of c, (Simplify your answers completely,) \[ \begin{aligned} P(2) & =(2)^{3}+2(\square)-6(2)-4 \\ & =8+2(\square)-6(2)-4 \\ & =8+(\square)=12-4 \\ & =\square \square \square) \end{aligned} \] Since? \( \square \) is 3 zero of \( P(x), x=2 \) \( \square \) is a factor of \( P(x) \),

We are given   P(x) = x³ + 2x² – 6x – 4      and  c = 2. Step 1. Evaluate P(2):   P(2) = (2)³ + 2(2)² – 6(2) – 4      = 8 + 2(4) – 12 – 4      = 8 + 8 – 12 – 4      = 0 Thus, P(2) = 0. Step 2. Since P(2)=0, by the Factor Theorem, x = 2 is a zero of P(x) and (x – 2) is a factor. Step 3. To find D(x) (also called the depressed polynomial), perform synthetic division of P(x) by x – 2. Set up synthetic division using the coefficients of P(x):   Coefficients: 1  2  –6  –4   c = 2     Bring down the 1.     Multiply 1 by 2 -> 2. Add to 2 -> 4.     Multiply 4 by 2 -> 8. Add to –6 -> 2.     Multiply 2 by 2 -> 4. Add to –4 -> 0. The synthetic division steps:     2 |  1   2  –6  –4         2   4   2      –––––––––––––––       1   4   2   0 The quotient (depressed polynomial) is given by the coefficients 1, 4, and 2, which corresponds to:   D(x) = x² + 4x + 2 Summary of Fill-in-the-Blanks:   P(2) = (2)³ + 2(2)² – 6(2) – 4       = 8 + 2(4) – 6(2) – 4       = 8 + 8 – 12 – 4       = 0 Thus, since 0 was obtained, x = 2 is a zero of P(x) and hence (x – 2) is a factor. Final Answers:   P(2) = 0,   D(x) = x² + 4x + 2.