How do you factor #x^2+1#?

1 Answer
Gió
May 5, 2015

You can do it using Complex Numbers! A complex number is given as #a+ib# where #i=sqrt(-1)# represents the Immaginary Unit (notice that #sqrt(-1)# cannot be evaluated as a real number so you call it simply #i#).
In your case you have that, to get #x^2+1#, you need to use two complex numbers:
#(x+i)(x-i)#
Try to multiply them remembering that #i=sqrt(-1)#
You get:
#x^2-ix+ix-i^2=#
but #i^2=[sqrt(-1)]^2=-1# so
#=x^2cancel(-ix)+cancel(ix)-i^2=x^2+1#

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