Question

How do you factor `2x^2+8x+4`?

Answer 1

It is `2x^2+8x+4=2*(x^2+4x+4)-4=2*(x+2)^2-2^2=(sqrt2(x+2))^2-2^2=(sqrt2(x+2)-2)*(sqrt2(x+2)+2)`

We used the formula `a^2-b^2=(a-b)*(a+b)`

Answer 2

`2x^2+8x+4`
`color(white)("XXX")=color(green)(2)color(red)((x+2+sqrt(2)))color(blue)((x+2-sqrt(2)))`

Explanation:

Extract the obvious constant factor of `2`
`color(white)("XXX")color(green)(2)color(orange)((x^2+4x+2))`

Factor the second term using the quadratic formula for roots
`color(white)("XXX")r=(-b+-sqrt(b^2-4ac))/(2a)`

In this case:
`color(white)("XXX")r=(-4+-sqrt(4^2-4(1)(2)))/(2(1))`

`color(white)("XXX")= (-4+-sqrt(8))/2`

`color(white)("XXX")=-2+-sqrt(2)`

If `r` is a root then `(x-r)` is a factor
So
`color(white)("XXX")color(orange)((x^2+4x+2))`
factors as
`color(white)("XXX")=color(red)((x+2+sqrt(2)))color(blue)((x+2-sqrt(2)))`

Giving the final factoring:
`color(white)("XXX")2x^2+8x+4`
`color(white)("XXX")=color(green)(2)color(red)((x+2+sqrt(2)))color(blue)((x+2-sqrt(2)))`