Let P(x) = x² +bx+c, where b and c are integer. If P(x) is a factor of both x⁴ +6x² +25 and 

3x⁴ +4x2 +28x+5, find the value of P(1).

$$\begin{gathered} \mathrm{P}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}\text{ is a factor of }{\mathrm{x}}^4+6{\mathrm{x}}^2+25\text{. } \\ {x}^4+6x^2+25=x^4+10x^2-4x^2+25 \\ \Rightarrow \left(x^4+10x^2+25\right)-4x^2={\left(x^2+5\right)}^2-(2x{)}^2 \\ \Rightarrow {\mathrm{x}}^4+6{\mathrm{x}}^2+25=\left({\mathrm{x}}^2+2\mathrm{x}+5\right)\left({\mathrm{x}}^2-2\mathrm{x}+5\right) \end{gathered}$$

$$\begin{gathered} 3x^4+4x^2+28x+5=3x^4-6x^3+15x^2+6x^3-11x^2+28x+5 \\ =3x^2\left(x^2-2x+5\right)+6x^3-12x^2+30x+x^2-2x+5 \\ =3x^2\left(x^2-2x+5\right)+6x\left(x^2-\right.2x+5)+x^2-2x+5 \\ \Rightarrow 3x^4+4x^2+28x+5=\left(x^2-2x+5\right)\left(3x^2+6x+1\right) \\ \text{ So, }P(x)=x^2-2x+5 \end{gathered}$$
 

Hence $\therefore \mathrm{P}\left(1\right)=4$