Let $\text{P}\left(\text{x}\right)={\text{x}}^2+\text{bx}+\text{c}$, where $\text{b}$ and $\text{c}$ are integers. If $\mathrm{P}\left(\mathrm{x}\right)$ is a factor of both ${\text{x}}^4+6{\text{x}}^2+25$ and $3{\text{x}}^4+4{\text{x}}^2+28\text{x}+5$, then the value of $\mathrm{P}\left(1\right)$ is

If $\text{x} = \alpha$ is the common root of both the equations, then

${\alpha }^4+6{\alpha }^2+25=0\Rightarrow {\alpha }^4=-25-6{\alpha }^2$

Using this value of ${\alpha }^4$ in the second equation

$\Rightarrow 3\left(-25-6{\alpha }^2\right)+4{\alpha }^2+28\alpha +5=0$

$\Rightarrow -75-18{\alpha }^2+4{\alpha }^2+28\alpha +5=0$

$\Rightarrow 2{\alpha }^2-4\alpha +10=0$

$\Rightarrow {\alpha }^2-2\alpha +5=0$

on comparing with ${\text{x}}^2+\text{bx}+\text{c}$, we get

$\text{b}=-2, \text{c}=5$

So, $\text{P}\left(\text{x}\right)={\text{x}}^2-\text{2x}+\text{5}$

$\Rightarrow \mathrm{P}\left(1\right)=\left(1\right)-2+5$$=4$