Given a quadratic expression of the form
[tex]ax^2+b[/tex]We can factor it by:
[tex]ax^2+b=0[/tex]And solve for x:
[tex]x=\pm\sqrt{-\frac{b}{a}}[/tex]But, if we try to apply this in the given equation:
[tex]2x^2+8=0[/tex][tex]\begin{gathered} 2x^2=-8 \\ . \\ x^2=-\frac{8}{2} \\ . \\ x=\pm\sqrt{-4} \end{gathered}[/tex]Which is not a real number. Thus, the expression is prime.