Answer
$(3-x) \left( 6+2x-x^2 \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Get the $GCF$ of the given expression, $
4(3-x)^2-(3-x)^3+3(3-x)
.$ Divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient.
$\bf{\text{Solution Details:}}$
The $GCF$ of the constants of the terms $\{
4,-1,3
\}$ is $
1
.$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $\{
(3-x)^2,(3-x)^3,(3-x)
\}$ is $
(3-x)
.$ Hence, the entire expression has $GCF=
(3-x)
.$
Factoring the $GCF=
(3-x)
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
(3-x) \left( \dfrac{4(3-x)^2}{(3-x)}-\dfrac{(3-x)^3}{(3-x)}+\dfrac{3(3-x)}{(3-x)} \right)
.\end{array}
Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
(3-x) \left( 4(3-x)^{2-1}-(3-x)^{3-1}+3(3-x)^{1-1} \right)
\\\\=
(3-x) \left( 4(3-x)^{1}-(3-x)^{2}+3(3-x)^{0} \right)
\\\\=
(3-x) \left( 4(3-x)-(3-x)^{2}+3(1) \right)
\\\\=
(3-x) \left( 4(3-x)-(3-x)^{2}+3 \right)
.\end{array}
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the product of the expression above is
\begin{array}{l}\require{cancel}
(3-x) \left( 4(3-x)-(9-6x+x^2)+3 \right)
.\end{array}
Using $a(b+c)=ab+ac$ or the Distributive Property and then combining like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(3-x) \left( 12-4x-9+6x-x^2+3 \right)
\\\\=
(3-x) \left( (12-9+3)+(-4x+6x)-x^2 \right)
\\\\=
(3-x) \left( 6+2x-x^2 \right)
.\end{array}