How do you factor $2 x^{2} + 4 x + 6$ $2x^2 +4x +6$ ?

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Answer 1 · 2015-06-07T06:51:44

$2 x^{2} + 4 x + 6 = 2 (x^{2} + 2 x + 3)$ $2x^2+4x+6 = 2(x^2+2x+3)$

The expression $x^{2} + 2 x + 3$ $x^2+2x+3$ is of the form $a x^{2} + b x + c$ $ax^2+bx+c$ , with $a = 1$ $a=1$ , $b = 2$ $b=2$ and $c = 3$ $c=3$

This has discriminant given by the formula:

$Delta = b^2-4ac = 2^2-(4xx1xx3) = 4 - 12 = -8$

Since $Delta < 0$ the equation $x^2+2x+3=0$ has no real roots,
so there are no linear factors with real coefficients.

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