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

Responda:

This quadratic only factors with the help of Complex coefficients:

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

Explicação:

Dado:

#x^2-2x+2#

Isto está na forma #ax^2+bx+c# com #a=1#, #b=-2# e #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#

Desde #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.

A diferença da identidade dos quadrados pode ser escrita:

#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)# e #B=i# como se segue:

#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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