This is just 1 approach.
Lets assume that the plot crosses the x-axis at least once (it does).
Set #y=0=x^4-2x^3+2x-1 #
Have a really good look at what is given. Notice that if we substitute 1 for #x# we have #y=(1)^4-2(1)^3+2(1)-1 = 0#
So #(x-1)(?.......)# is part of the factorisation.
Lets use what we have got to construct the original equation.
The first term is #x^4# so we may do this:
#(x-1)(color(red)(x^3)+......) = x^4-x^3.......#
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But the 2nd term is #-2x^3# so we need an extra #x^3#. Do this:
#(x-1)(x^3color(red)(-x^2)......) = x^4-x^3color(red)(-x^3+x^2).......#
#(x-1)(x^3-x^2......) = x^4-2x^3+x^2.........#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
There is no #x^2# term in the original equation. So we need to 'get rid of it. We do this by turning it into 0 so now we need #-x^2#
#(x-1)(x^3-x^2color(red)(-x)......) = x^4-2x^3+x^2color(red)(-x^2+x.........#
#(x-1)(x^3-x^2-x......) = x^4-2x^3+x.........#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We need to have not just #x# but#+2x#. So we need another #x# from somewhere.
#(x-1)(x^3-x^2-xcolor(red)(+1)) = x^4-2x^3+xcolor(red)(+x-1)#
#color(green)((x-1)(x^3-x^2-x+1) = x^4-2x^3+2x-1)# part answer
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Now we need to solve for #x^3-x^2-x+1=0#
Again notice that #x=1# is a solution
#(+1)^3-(+1)^2-(+1)+1=0#
#color(white)("dd")1color(white)("dddd")-1color(white)("ddddd")-1color(white)("d.")+1=0#
So we can go through the process again.
#(x-1)(color(red)(x^2)..... )=color(red)(x^3-x^2)....#
#(x-1)(x^2color(red)(-1 ) )=x^3-x^2 color(red)(-x+1)#
#color(green)((x-1)(x^2-1 )=x^3-x^2 -x+1)# the other part answer
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together")#
#(x-1)^2(x^2-1)#
But this is the same as
#(x-1)^2(x^2-[1^2])#
#(x-1)^2(x+1)(x-1)#
Which is:
#(x-1)^3(x+1)#
