If f(x) = 0 is a polynomial whose coefficients all `pm 1` and whose roots are all real, then the degree of f(x) can be equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to determine the degree of the polynomial \( f(x) \) such that all coefficients are either +1 or -1, and all roots are real.

### Step-by-Step Solution:

1. **Understanding the Polynomial**:
We know that \( f(x) \) is a polynomial with coefficients of either +1 or -1. This means that the polynomial can be expressed in the form:
\[ f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \]
where each \( a_i \) is either +1 or -1.

2. **Testing Degree 1 Polynomial**:
Let's consider a polynomial of degree 1:
\[ f(x) = x + 1 \]
Setting \( f(x) = 0 \):
\[ x + 1 = 0 \implies x = -1 \]
This polynomial has a real root (-1).

3. **Testing Degree 2 Polynomial**:
Now, consider a polynomial of degree 2:
\[ f(x) = x^2 + x + 1 \]
Setting \( f(x) = 0 \):
\[ x^2 + x + 1 = 0 \]
The discriminant \( D \) is given by:
\[ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, this polynomial does not have real roots.

4. **Testing Degree 3 Polynomial**:
Now, consider a polynomial of degree 3:
\[ f(x) = x^3 + x^2 + x + 1 \]
Setting \( f(x) = 0 \):
\[ x^3 + x^2 + x + 1 = 0 \]
The discriminant for cubic polynomials is more complex, but we can check for real roots using the Rational Root Theorem or numerical methods. However, it can be shown that this polynomial does not have all real roots.

5. **Testing Degree 4 Polynomial**:
Finally, consider a polynomial of degree 4:
\[ f(x) = x^4 + x^3 + x^2 + x + 1 \]
Again, using the discriminant or numerical methods, we find that this polynomial does not have all real roots either.

### Conclusion:
From the analysis, we find that the only polynomial degree that allows for all real roots is degree 1. Therefore, the degree of \( f(x) \) can only be equal to 1.

### Final Answer:
The degree of \( f(x) \) can be equal to **1**.