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)#