Answer
$(x^{n}+4)(x^{n}+2)$
Work Step by Step
$x^{2n}+6x^{n}+8$
We can write it as
$(x^{n})^{2}+6x^{n}+8$
Since the last term of $(x^{n})^{2}+6x^{n}+8$ is 8. We need to find two factors of 8 that have a sum of 6.
4 and 2 are factors of 8. Also, $4\times2=8$ and $4+2=6$
Therefore, $(x^{n})^{2}+6x^{n}+8$ can be written as
$(x^{n})^{2}+4x^{n}+2x^{n}+8$
Group the terms.
$((x^{n})^{2}+4x^{n})+(2x^{n}+8)$
Factor out common factors.
$=x^{n}(x^{n}+4)+2(x^{n}+4)$
Factor out common factor $(x^{n}+4)$
$=(x^{n}+4)(x^{n}+2)$
$x^{2n}+6x^{n}+8 =(x^{n}+4)(x^{n}+2)$
