# Is it possible to factor y=2x^2+4x+8 ? If so, what are the factors?

Yes it is possible to factorise $2 {x}^{2} + 4 x + 8$
The factors are 2,$\left(x + 1 - i \sqrt{3}\right)$,$\left(x + 1 + i \sqrt{3}\right)$
$2 {x}^{2} + 4 x + 8$ =2(${x}^{2} + 2 x + 4$)
${x}^{2} + 2 x + 4$=$\left(x + 1 - i \sqrt{3}\right)$ $\left(x + 1 + i \sqrt{3}\right)$