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

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This is not factorable normally. We can see this since

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

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

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#### Answer:

This quadratic only factors with the help of Complex coefficients:

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

#### Explanation:

Given:

#x^2-2x+2#

This is in the form

It has discriminant

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

Since

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

#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)#

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