How do you factor $x^{2} + 2 x + 3$ $x^2 + 2x +3$ ?

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Answer 1 · 2015-07-30T13:33:21

$(x + 1 + i \sqrt{2}) (x + 1 - i \sqrt{2})$ $(x+1+isqrt(2))(x+1-isqrt(2))$

Solve for the roots. First complete the square:

$x^{2} + 2 x + 3 = {(x + 1)}^{2} + 2 = 0$ $x^2+2x+3=(x+1)^2+2=0$

Solve for $x$ $x$ :

${(x + 1)}^{2} + 2 = 0$ $(x+1)^2+2=0$

=>

${(x + 1)}^{2} = - 2$ $(x+1)^2=-2$

=>

$x + 1 = \pm i \sqrt{2}$ $x+1=+-isqrt(2)$

=>

$x = - 1 \pm i \sqrt{2}$ $x=-1+-isqrt(2)$

Therefore the factorisation is:

$(x + 1 + i \sqrt{2}) (x + 1 - i \sqrt{2})$ $(x+1+isqrt(2))(x+1-isqrt(2))$

Answer 2 · 2015-07-30T17:40:07

Factor $y = x^{2} + 2 x + 3$ $y = x^2 + 2x + 3$

$D = b^{2} - 4 a c = 4 - 12 = - 8 < 0 .$ $D = b^2 - 4ac = 4 - 12 = - 8 < 0.$
Therefor, this function can't be factored.

How do you factor x2+2x+3x^2 + 2x +3?