Question 694376
{{{P(x)=x^4-2x^3+3x^2-2x+1}}}
Maybe you just wanted to factor the polynomial {{{P(x)}}}.
Or maybe you wanted to find its zeros, to solve {{{P(x)=0}}}
I'll try both.
 
FINDING ZEROS:
We can use a trick that I (dimly) remember from high school.
Obviously {{{P(0)=1}}}, so {{{0}}} is not a zero of the polynomial.
We can divide by {{{x}}} (or by {{{x^2}}}) without fear.
When looking for zeros,
{{{x^4-2x^3+3x^2-2x+1=0}}} <--> {{{(x^4-2x^3+3x^2-2x+1)/x^2=0}}}  <--> {{{x^2-2x+3-2(1/x)+1/x^2=0}}}
We can do a change of variable.
If we define {{{y=x+1/x}}}, then {{{y^2=x^2+1/x^2+2}}} <--> {{{x^2+1/x^2=y^2-2}}}
We can re-write {{{x^2-2x+3-2(1/x)+1/x^2=0}}} in terms of {{{y}}} as
{{{y^2-2+2y+3=0}}} <--> {{{y^2+2y+1=0}}} <--> {{{(y+1)^2=0}}} <--> {{{highlight(y=-1)}}}
Going back to {{{x}}}, remembering we defined {{{y=x+1/x}}},
{{{x+1/x=-1}}}, and multiplying by {{{x}}} we get
{{{x^2+1=-x}}} <--> {{{x^2+x+1=0}}}
Applying the quadratic formula we solve to get
{{{x=(1 +- sqrt(1^2-4*1*1))/(2*1)=(1 +- sqrt(-3))/2=(1 +- i*sqrt(3))/2=highlight(1/2 +- i*sqrt(3)/2)}}}

FACTORING (no tricks):
We realize that it is so very symmetrical that if some non-zero {{{a}}} is a zero of {{{P(x)}}}
{{{P(a)=a^4-2a^3+3a^2-2a+1=0}}} --> {{{P(a)/a^4=a^4/a^4-2a^3/a^4+3a^2/a^4-2a/a^4+1/a^4=0}}} --> {{{1-2/a+3/a^2-2/a^3+1/a^4=0}}} --> {{{1-2(1/a)+3(1/a)^2-2(1/a)^3+(1/a)^4=P(1/a)=0}}}
So the 4 complex zeros of the polynomial will come in pairs {{{a}}} with {{{1/a}}} and {{{b}}} with {{{1/b}}},
and the full factoring of the polynomial would be
{{{P(x)=(x-a)(x-1/a)(x-b)(x-1/b)}}}
Multiplying pairs we get
{{{P(x)=(x^2-(a+1/a)x+1)(x^2-(b+1/b)x+1)}}}
Maybe we could find {{{m=a+1/a}}} and {{{n=b+1/b}}} and have a partial factoring with real coefficients.
It would be
{{{P(x)=(x^2-mx+1)(x^2-nx+1)=x^4-(m+n)x^2+(mn+2)x^2-(m+n)x+1}}}
If it must be {{{P(x)=x^4-2x^3+3x^2-2x+1=x^4-(m+n)x^2+(mn+2)x^2-(m+n)x+1}}} for all {{{x}}},
then {{{m+n=2}}} and {{{mn+2=3}}} <--> {{{mn=1}}}
The solution to {{{system (m+n=2,mn=1)}}} is {{{m=n=1}}},
so substituting into {{{P(x)=(x^2-mx+1)(x^2-nx+1)}}}
the factoring with real coefficients is
{{{highlight(P(x)=(x^2-x+1)(x^2-x+1))}}} or {{{highlight(P(x)=(x^2-x+1)^2)}}}
The factoring with real coefficients is done.
To factor further we need to go beyond real numbers.
Since {{{x^2-x+1+0}}} does not have real solutions,
any linear {{{(x-a)}}} factor would have an imaginary {{{a}}} coefficient.