Let `f(x)=x^2+b x+c ,w h e r eb ,c in Rdot` If `f(x)` is a factor of both `x^4+6x^2+25a n d3x^4+4x^4+28 x+5` , then the least value of `f(x)` is `2` b. `3` c. `5//2` d. `4`

Options

A. 2
B. 3
C. `(5)/(2)`
D. 4

Answer

To solve the problem, we need to find the least value of the quadratic function \( f(x) = x^2 + bx + c \) given that it is a factor of both polynomials \( x^4 + 6x^2 + 25 \) and \( 3x^4 + 4x^4 + 28x + 5 \).

### Step 1: Identify the Quadratic Function
We start with the quadratic function:
\[ f(x) = x^2 + bx + c \]

### Step 2: Factorization Requirement
Since \( f(x) \) is a factor of both polynomials, we can express each polynomial in terms of \( f(x) \). This means that when we divide the polynomials by \( f(x) \), the remainder should be zero.

### Step 3: Analyze the First Polynomial
Let's first analyze the polynomial:
\[ x^4 + 6x^2 + 25 \]
We can express it as:
\[ x^4 + 0x^3 + 6x^2 + 0x + 25 \]
We will perform polynomial long division of \( x^4 + 6x^2 + 25 \) by \( f(x) \).

### Step 4: Analyze the Second Polynomial
Now, let's analyze the second polynomial:
\[ 3x^4 + 4x^4 + 28x + 5 = 7x^4 + 28x + 5 \]
We will also perform polynomial long division of \( 7x^4 + 28x + 5 \) by \( f(x) \).

### Step 5: Set Up the Long Division
For both polynomials, we will set up the long division and find the coefficients \( b \) and \( c \) such that the remainders are zero.

### Step 6: Find the Roots
Since \( f(x) \) is a quadratic, we can find its roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4c}}{2} \]
The function will have real roots if the discriminant \( b^2 - 4c \geq 0 \).

### Step 7: Calculate \( f(1) \)
To find the least value of \( f(x) \), we will evaluate \( f(1) \):
\[ f(1) = 1^2 + b(1) + c = 1 + b + c \]

### Step 8: Minimize \( f(1) \)
To minimize \( f(1) \), we need to find suitable values of \( b \) and \( c \) that satisfy the conditions from the polynomial divisions.

### Step 9: Substitute Values
After performing the polynomial divisions and solving for \( b \) and \( c \), we find that:
\[ f(x) = x^2 - 2x + 5 \]
Then we calculate:
\[ f(1) = 1^2 - 2(1) + 5 = 1 - 2 + 5 = 4 \]

### Conclusion
Thus, the least value of \( f(x) \) is:
\[ \boxed{4} \]