How do you factor the trinomial #x^2 +1 - x#?

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Answer 1 · 2017-08-17T11:03:55

# (x-1/2+isqrt3/2)(x-1/2-isqrt3/2).#

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

Since, #x^2-x=x^2-2(1/2)x,# to make it a perfect square, we need the

last term #(1/2)^2=1/4.#

Hence, #x^2-x+1=x^2-x+1/4+3/4,#

#=(x-1/2)^2-(-3/4),#

#=(x-1/2)^2-{i^2*(sqrt3/2)^2},#

#=(x-1/2)^2-(isqrt3/2)^2,#

#=(x-1/2+isqrt3/2)(x-1/2-isqrt3/2),# is the desired

factorisation.