#### Answer

$10(x^2+1)(x+1)(x-1)$

#### Work Step by Step

Factoring the $GCF=
10
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
10x^4-10
\\\\=
10(x^4-1)
.\end{array}
The expressions $
x^4
$ and $
1
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^4-1
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
10(x^4-1)
\\\\=
10[(x^2)^2-(1)^2]
\\\\=
10(x^2+1)(x^2-1)
.\end{array}
The expressions $
x^2
$ and $
1
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^2-1
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
10(x^2+1)(x^2-1)
\\\\=
10(x^2+1)[(x)^2-(1)^2]
\\\\=
10(x^2+1)[(x+1)(x-1)]
\\\\=
10(x^2+1)(x+1)(x-1)
.\end{array}