#### Answer

$(x+3)(x^2+1)(x+1)(x-1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Group the terms of the given expression, $
x^5+3x^4-x-3
,$ such that the factored form of the groupings will result to a factor that is common to the entire expression. Then, factor the $GCF$ in each group. and factor the $GCF$ of the entire expression. Finally, factor the resulting difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^5+3x^4)-(x+3)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x^4(x+3)-(x+3)
.\end{array}
Factoring the $GCF=
(x+3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(x+3)(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}
(x+3)[(x^2)^2-(1)^2]
\\\\=
(x+3)[(x^2+1)(x^2-1)]
\\\\=
(x+3)(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}
(x+3)(x^2+1)[(x)^2-(1)^2]
\\\\=
(x+3)(x^2+1)[(x+1)(x-1)]
\\\\=
(x+3)(x^2+1)(x+1)(x-1)
.\end{array}