SOLUTION: x^4-2x^3+3x^2-2x+1 How to solve without using trial and error method ??

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Question 694376: x^4-2x^3+3x^2-2x+1
How to solve without using trial and error method ??
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
P%28x%29=x%5E4-2x%5E3%2B3x%5E2-2x%2B1
Maybe you just wanted to factor the polynomial P%28x%29.
Or maybe you wanted to find its zeros, to solve P%28x%29=0
I'll try both.

FINDING ZEROS:
We can use a trick that I (dimly) remember from high school.
Obviously P%280%29=1, so 0 is not a zero of the polynomial.
We can divide by x (or by x%5E2) without fear.
When looking for zeros,
x%5E4-2x%5E3%2B3x%5E2-2x%2B1=0 <--> %28x%5E4-2x%5E3%2B3x%5E2-2x%2B1%29%2Fx%5E2=0 <--> x%5E2-2x%2B3-2%281%2Fx%29%2B1%2Fx%5E2=0
We can do a change of variable.
If we define y=x%2B1%2Fx, then y%5E2=x%5E2%2B1%2Fx%5E2%2B2 <--> x%5E2%2B1%2Fx%5E2=y%5E2-2
We can re-write x%5E2-2x%2B3-2%281%2Fx%29%2B1%2Fx%5E2=0 in terms of y as
y%5E2-2%2B2y%2B3=0 <--> y%5E2%2B2y%2B1=0 <--> %28y%2B1%29%5E2=0 <--> highlight%28y=-1%29
Going back to x, remembering we defined y=x%2B1%2Fx,
x%2B1%2Fx=-1, and multiplying by x we get
x%5E2%2B1=-x <--> x%5E2%2Bx%2B1=0
Applying the quadratic formula we solve to get

FACTORING (no tricks):
We realize that it is so very symmetrical that if some non-zero a is a zero of P%28x%29
P%28a%29=a%5E4-2a%5E3%2B3a%5E2-2a%2B1=0 --> P%28a%29%2Fa%5E4=a%5E4%2Fa%5E4-2a%5E3%2Fa%5E4%2B3a%5E2%2Fa%5E4-2a%2Fa%5E4%2B1%2Fa%5E4=0 --> 1-2%2Fa%2B3%2Fa%5E2-2%2Fa%5E3%2B1%2Fa%5E4=0 --> 1-2%281%2Fa%29%2B3%281%2Fa%29%5E2-2%281%2Fa%29%5E3%2B%281%2Fa%29%5E4=P%281%2Fa%29=0
So the 4 complex zeros of the polynomial will come in pairs a with 1%2Fa and b with 1%2Fb,
and the full factoring of the polynomial would be
P%28x%29=%28x-a%29%28x-1%2Fa%29%28x-b%29%28x-1%2Fb%29
Multiplying pairs we get
P%28x%29=%28x%5E2-%28a%2B1%2Fa%29x%2B1%29%28x%5E2-%28b%2B1%2Fb%29x%2B1%29
Maybe we could find m=a%2B1%2Fa and n=b%2B1%2Fb and have a partial factoring with real coefficients.
It would be

If it must be for all x,
then m%2Bn=2 and mn%2B2=3 <--> mn=1
The solution to system+%28m%2Bn=2%2Cmn=1%29 is m=n=1,
so substituting into P%28x%29=%28x%5E2-mx%2B1%29%28x%5E2-nx%2B1%29
the factoring with real coefficients is
highlight%28P%28x%29=%28x%5E2-x%2B1%29%28x%5E2-x%2B1%29%29 or highlight%28P%28x%29=%28x%5E2-x%2B1%29%5E2%29
The factoring with real coefficients is done.
To factor further we need to go beyond real numbers.
Since x%5E2-x%2B1%2B0 does not have real solutions,
any linear %28x-a%29 factor would have an imaginary a coefficient.