How do you factor # x^2-2x+2# ?

Given:

#x^2-2x+2#

This is in the form #ax^2+bx+c# with #a=1#, #b=-2# and #c=2#

It has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (color(blue)(-2))^2-4(color(blue)(1))(color(blue)(2)) = 4 - 8 = -4#

Since #Delta < 0#, this quadratic has no Real zeros and no linear factors with Real coefficients.

We can still factor it, but we need to use Complex coefficients.

The difference of squares identity can be written:

#A^2-B^2 = (A-B)(A+B)#

To factor our quadratic, we can complete the square and use the difference of squares identity with #A=(x-1)# and #B=i# as follows:

#x^2-2x+2 = x^2-2x+1+1#

#color(white)(x^2-2x+2) = (x-1)^2+1#

#color(white)(x^2-2x+2) = (x-1)^2-i^2#

#color(white)(x^2-2x+2) = ((x-1)-i)((x-1)+i)#

#color(white)(x^2-2x+2) = (x-1-i)(x-1+i)#

This is not factorable normally. We can see this since #x^2-2x+1# is a perfect square, so it touches the x-axis at a single root. Adding 1 to this produces #x^2-2x+2#, and raises the graph of #y = x^2-2x+1# one upwards, meaning it no longer touches the x-axis, so it has no real roots.

However, we can find its imaginary roots like this:

#x^2-2x+2#

#x^2-2x+1+1#

#(x-1)^2 - i^2 color(white)"XXX"# (since #i^2 = -1#)

This is of the form #a^2-b^2#, so it factors into #(a+b)(a-b)#.

#(x-1+i)(x-1-i)#

Final Answer