Question 1040174
{{{(2x^2-4x+1)^2-14x^2+28x+3=(2x^2-4x+1)^2-14x^2+28x-7+10=(2x^2-4x+1)^2-7(2x^2-4x+1)+10}}} .
Let {{{y}}} be {{{y=2x^2-4x+1}}} .
We can re-write the original expression as
{{{y^2-7y+10=(y-5)(y-2)}}} .
Since {{{y=2x^2-4x+1}}} , {{{system(y-5=2x^2-4x-4,"and",y-2=2x^2-4x-1)}}} , so
{{{highlight((2x^2-4x+1)^2-14x^2+28x+3=(2x^2-4x-4)(2x^2-4x-1))}}} .
That is as far as we can go, unless we use ugly irrational coefficients,
because all the zeros of {{{2x^2-4x-4}}} and {{{2x^2-4x-1}}} are irrational numbers.