Polynomially Yours!

Easy to see that $x^4+6x^2+25=\left(x^2-2x+5\right)\left(x^2+2x+5\right)$ . So $f(x)$ must be one of these factors. We get $f(1)=4$ from the first and $f(1)=8$ from the second, none of which is an option.

See that if $f(x)$ divides $1^{st}$ equation and it also divides $2^{nd}$ equation, then we can write

$x^4 + 6x^2 + 25 = f(x) \cdot g(x)$ ............. where $g(x)$ is a monic quadratic polynomial.

Similarly,

$3x^4 + 4x^2+28x + 5 = f(x) \cdot h(x)$ ......... where $h(x)$ is a quadratic polynomial with leading co-efficient 3.

Thus , always , $3 (x^4 + 6x^2 + 25 ) - (3x^4 + 4x^2+28x + 5) = f(x) (3 \times g(x)-h(x) )$

Hence $f(x) (3\times g(x) - h(x) = 14x^2-28x + 70$

Hence $x^2+bx+c$ divides $14x^2-28x+70=14(x^2-2x+5)$ And because the other factor is constant, we can say that $x^2+bx+c= x^2-2x+5$ giving $f(1) = 1^2-1\times 2 + 5 = \boxed{4}$ which is not in options. (Answer None of these)

${ x }^{ 4 }+{ 6 }x^{ 2 }+25\quad \quad \quad \quad \quad ------>\left( 1 \right) \\ { 3x }^{ 4 }+{ 4x }^{ 2 }+28x+5\quad ------>(2)\\$

Then, $3[1]-[2]=14{ x }^{ 2 }-28x+70=14({ x }^{ 2 }-2x+5)$ will also be divisible by ${ x }^{ 2 }+bx+c$

Therefor, $f(x)={ x }^{ 2 }-2x+5\\ f(1)=1-2+5=\boxed{4}$ .