A polynomial of degree `nge1` can have at most n real zeroes. A quadratic polynomial can have at most two real zeroes .
Find `p(1)`, if `p(x)=x^(3)-22x^(2)+141x-120` .

Options

A. `-1`
B. `-12`
C. `0`
D. `9`

Answer

To find \( p(1) \) for the polynomial \( p(x) = x^3 - 22x^2 + 141x - 120 \), we will substitute \( x = 1 \) into the polynomial and simplify.

### Step-by-step Solution:

1. **Substitute \( x = 1 \) into the polynomial:**
\[ p(1) = 1^3 - 22 \cdot 1^2 + 141 \cdot 1 - 120 \]

2. **Calculate each term:**
- \( 1^3 = 1 \)
- \( 22 \cdot 1^2 = 22 \)
- \( 141 \cdot 1 = 141 \)

So, substituting these values back into the equation:
\[ p(1) = 1 - 22 + 141 - 120 \]

3. **Combine the terms:**
- Start with \( 1 - 22 = -21 \)
- Then, add \( 141 \):
\[ -21 + 141 = 120 \]
- Finally, subtract \( 120 \):
\[ 120 - 120 = 0 \]

4. **Final result:**
\[ p(1) = 0 \]

### Conclusion:
The value of \( p(1) \) is \( 0 \).