If f (x)= 0 be a polynomial whose coefficients are all `+-1` and whose roots are all real, then degree of f (x) can be : (a) 1 (b) 2 (c) 3 (d) 4

Options

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

Answer

To solve the problem, we need to determine the possible degrees of a polynomial \( f(x) \) that has all coefficients as either \( +1 \) or \( -1 \) and has all real roots.

### Step-by-Step Solution:

1. **Understanding the Polynomial**:
We are given that \( f(x) = 0 \) is a polynomial with coefficients that are all \( +1 \) or \( -1 \). We need to check for different degrees of polynomials to see if they can have all real roots.

2. **Degree 1 Polynomial**:
Consider a polynomial of degree 1:
\[ f(x) = x + 1 \]
Setting it to zero:
\[ x + 1 = 0 \implies x = -1 \]
The root \( -1 \) is real. Thus, a degree 1 polynomial can have all real roots.

**Hint**: Check if a linear polynomial can have real roots.

3. **Degree 2 Polynomial**:
Now, consider a polynomial of degree 2:
\[ f(x) = x^2 + x + 1 \]
Setting it to zero:
\[ x^2 + x + 1 = 0 \]
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 has no real roots.

**Hint**: Use the discriminant to determine the nature of the roots of a quadratic polynomial.

4. **Degree 3 Polynomial**:
Next, consider a polynomial of degree 3:
\[ f(x) = x^3 + x^2 + x + 1 \]
We can factor this polynomial:
\[ f(x) = (x + 1)(x^2 + 1) \]
The factor \( x + 1 = 0 \) gives us the root \( x = -1 \), which is real. However, the factor \( x^2 + 1 = 0 \) has roots:
\[ x = \pm i \]
These roots are not real. Therefore, a degree 3 polynomial does not have all real roots.

**Hint**: Factor the polynomial to find the roots and check their nature.

5. **Degree 4 Polynomial**:
Finally, consider a polynomial of degree 4:
\[ f(x) = x^4 + x^3 + x^2 + x + 1 \]
We can analyze the roots using the Rational Root Theorem or by checking for possible factorizations. However, it can be shown that this polynomial also does not have all real roots due to the presence of complex roots.

**Hint**: Investigate the possibility of complex roots in higher degree polynomials.

### Conclusion:
From the analysis above, we conclude that only the degree 1 polynomial can have all real roots. Therefore, the answer to the question is:

**Final Answer**: (a) 1