If f(x) is a polynomial of degree n with rational coefficients and `1 +2 i ,2 - sqrt(3) ` and 5 are roots of f(x) =0 then the least value of n is

Options

A. 5
B. 4
C. 3
D. 6

Answer

To find the least value of \( n \) for the polynomial \( f(x) \) with rational coefficients, given the roots \( 1 + 2i \), \( 2 - \sqrt{3} \), and \( 5 \), we need to consider the properties of polynomials with rational coefficients.

### Step 1: Identify the given roots
The roots provided are:
1. \( 1 + 2i \)
2. \( 2 - \sqrt{3} \)
3. \( 5 \)

### Step 2: Determine the necessary conjugate roots
Since the polynomial has rational coefficients, any complex root must have its conjugate also as a root. Thus, the conjugate of \( 1 + 2i \) is:
- \( 1 - 2i \)

Also, since \( 2 - \sqrt{3} \) is an irrational root, its conjugate (which is also a root) is:
- \( 2 + \sqrt{3} \)

### Step 3: List all the roots
Now we can list all the roots of the polynomial:
1. \( 1 + 2i \)
2. \( 1 - 2i \) (conjugate of \( 1 + 2i \))
3. \( 2 - \sqrt{3} \)
4. \( 2 + \sqrt{3} \) (conjugate of \( 2 - \sqrt{3} \))
5. \( 5 \)

### Step 4: Count the total number of roots
Now we count the total number of roots:
- \( 1 + 2i \)
- \( 1 - 2i \)
- \( 2 - \sqrt{3} \)
- \( 2 + \sqrt{3} \)
- \( 5 \)

This gives us a total of 5 roots.

### Step 5: Determine the degree of the polynomial
The degree \( n \) of the polynomial \( f(x) \) is equal to the number of roots (counting multiplicities). Since we have identified 5 roots, the least value of \( n \) is:
\[ n = 5 \]

### Conclusion
Thus, the least value of \( n \) is \( 5 \).

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