Respuesta :

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.

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