How to factorize the above expression?

Question

How to factorize $x^4+x+7$? I am trying this question by adding and subtracting $x^2$. So, the expression is becoming $(x^4+x^2+x+7)-x^2$. Now I am trying to factorize $x^2+x+7$ first. So, therefore $x^2+x+7=x^2+2\times x \times (1/2)+(1/2)^2+7-(1/2)^2$. So, finally I am getting $x^2+x+7=(x+1/2)^2+27/4$. So, the actual expression is becoming $x^4+(x+1/2)^2+(27/4)-x^2$. But I can't approach further. Please help me out with this factorization.

Answer 1

$$\left(x^4+x+\frac{a^4}{4}-\frac{1}{4a^2}\right)=\left(x^2+a x +\frac{a^2}{2}-\frac{1}{2a}\right)\left(x^2-a x +\frac{a^2}{2}+\frac{1}{2a}\right)$$

Now find a real number $a$ such that $$7=\frac{a^4}{4}-\frac{1}{4a^2}$$

This is a cubic in $a^2$

EDIT: going further, based on two comments by @Will Jagy

Let $t=a^2$ and note that the resulting cubic $t^3-28t-1=0$ has a real root $t>0$

Also, the factorization can be written in the form of a difference of two squares as follows

$$\left(x^4+x+\frac{a^4}{4}-\frac{1}{4a^2}\right)=\left(x^2+\frac{a^2}{2}\right)^2-\left(a x -\frac{1}{2a}\right)^2$$