How do you factor #x^2 + 2x +3#?

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Answer 1 · 2015-07-30T13:33:21

#(x+1+isqrt(2))(x+1-isqrt(2))#

Solve for the roots. First complete the square:

#x^2+2x+3=(x+1)^2+2=0#

Solve for #x#:

#(x+1)^2+2=0#

=>

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

=>

#x+1=+-isqrt(2)#

=>

#x=-1+-isqrt(2)#

Therefore the factorisation is:

#(x+1+isqrt(2))(x+1-isqrt(2))#

Answer 2 · 2015-07-30T17:40:07

Factor #y = x^2 + 2x + 3#

#D = b^2 - 4ac = 4 - 12 = - 8 < 0.#
Therefor, this function can't be factored.