Find factor $f(x)$ conmmon to two quartic equation

Question

Let $f(x) = x^2 + bx + c$, where $b, c ∈ R$. If f(x) is a factor of both $x^4 + 6x^2 + 25$ and $3x^4 + 4x^2 + 28x + 5$, then find $f(x)$

My approach ,on dividing both quartic equation by $f(x)$ remainder is zero, but not getting the answer

Answer 1

Hint :

$$\begin{align}x^4 + 6x^2 + 25 &= x^4 + 10x^2 + 25 -4x^2 \\ &= (x^2+5)^2 - (2x)^2\\&=(x^2+2x+5)(x^2-2x+5) \\ &\end{align}$$

$$\begin{align}3x^4 + 4x^2 + 28x + 5 &= (x^2-2x+5)(3x^2+6x+1)\end{align} $$

Answer 2

Hint: $$ x^4+6x^2+25 = (x^2+3)^2+16 $$ $$= (x^2+3+4i)(x^2+3-4i)$$

So, the roots of the first functions are$$ \pm 1\pm 2i$$

Answer 3

If is a factor of $x^4 + 6 x^2 + 25$ and $3 x^4 + 4 x^2 + 28 x + 5$

then it is a factor of $$ 3 (x^4 + 6 x^2 + 25) - (3 x^4 + 4 x^2 + 28 x + 5)=14 (5 - 2 x + x^2) $$