If `p(y)=3y^(4)-5y^(3)+y^(2)+8`, then p(-1) will be

Options

A. 2
B. 15
C. 17
D. `-17`

Answer

To find \( p(-1) \) for the polynomial \( p(y) = 3y^4 - 5y^3 + y^2 + 8 \), we will substitute \( y = -1 \) into the polynomial and simplify.

### Step-by-Step Solution:

1. **Write the polynomial**:
\[ p(y) = 3y^4 - 5y^3 + y^2 + 8 \]

2. **Substitute \( y = -1 \)**:
\[ p(-1) = 3(-1)^4 - 5(-1)^3 + (-1)^2 + 8 \]

3. **Calculate each term**:
- Calculate \( (-1)^4 \):
\[ (-1)^4 = 1 \quad \text{(since any even power of -1 is 1)} \]
- Calculate \( (-1)^3 \):
\[ (-1)^3 = -1 \quad \text{(since any odd power of -1 is -1)} \]
- Calculate \( (-1)^2 \):
\[ (-1)^2 = 1 \quad \text{(since any even power of -1 is 1)} \]

4. **Substitute these values back into the equation**:
\[ p(-1) = 3(1) - 5(-1) + 1 + 8 \]

5. **Simplify the expression**:
\[ p(-1) = 3 + 5 + 1 + 8 \]

6. **Combine the terms**:
\[ p(-1) = 3 + 5 = 8 \]
\[ p(-1) = 8 + 1 = 9 \]
\[ p(-1) = 9 + 8 = 17 \]

### Final Answer:
\[ p(-1) = 17 \]