Answer
$x=\{ 0,1 \}$
Work Step by Step
The factored form of the given expression, $
2x^4-2x^3+2x^2=2x
,$ is
\begin{align*}
2x^4-2x^3+2x^2-2x&=0
\\\\
\dfrac{2x^4-2x^3+2x^2-2x}{2}&=\dfrac{0}{2}
\\\\
x^4-x^3+x^2-x&=0
\\
x(x^3-x^2+x-1)&=0
&\text{ (factor $GCF=x$)}
\\
x[(x^3-x^2)+(x-1)]&=0
&\text{ (use factoring by grouping)}
\\
x[x^2(x-1)+(x-1)]&=0
&\text{ (factor $GCF$ in each group)}
\\
x[(x-1)(x^2+1)]&=0
&\text{ (factor $GCF=(x-1)$)}
\\
x(x-1)(x^2+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) results to
\begin{align*}
x=0
\\\\\text{ OR }\\\\
x-1&=0
\\\\\text{ OR }\\\\
x^2+1&=0
.\end{align*}
Solving each of the equations above results to
\begin{align*}
x=0
\\\\\text{ OR }\\\\
x-1&=0
\\
x&=1
\\\\\text{ OR }\\\\
x^2+1&=0
\\
x^2&=-1
\\
x&=\pm\sqrt{-1}
&\text{ (not a real number)}
.\end{align*}
Hence, the real solutions are $
x=\{ 0,1 \}
$.
