Answer
$x^2(2x+3)(7x-1)$
Work Step by Step
Factoring the $GCF=
x^2
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
19x^3-3x^2+14x^4
\\\\=
14x^4+19x^3-3x^2
\\\\=
x^2(14x^2+19x-3)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
x^2(14x^2+19x-3)
\end{array} has $ac=
14(-3)=-42
$ and $b=
19
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
21,-2
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x^2(14x^2+21x-2x-3)
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
x^2[(14x^2+21x)-(2x+3)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x^2[7x(2x+3)-(2x+3)]
.\end{array}
Factoring the $GCF=
(2x+3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
x^2[(2x+3)(7x-1)]
\\\\=
x^2(2x+3)(7x-1)
.\end{array}