Question

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

Answer 1

First you can take out the factor `2`

Explanation:

`->y=2(x^2+4x+2)`

This cannot be factored any further without using radical expressions.

Answer 2

`2x^2+8x+4 = 2(x^2+4x+2)`

`= 2(x+2+sqrt(2))(x+2-sqrt(2))`

Explanation:

Given `y = 2x^2+8x+4` first separate out the common scalar factor `2` to get:

`y = 2(x^2+4x+2)`

The quadratic factor is in the form `ax^2+bx+c` with `a=1`, `b=4`, `c=2`. This has zeros given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a) = (-4+-sqrt((-4)^2-(4xx1xx2)))/(2xx1)`

`=(-4+-sqrt(8))/2 = (-4+-sqrt(2^2*2))/2 = (-4+-2sqrt(2))/2 = -2+-sqrt(2)`

These zeros correspond to factors `(x+2+sqrt(2))` and `(x+2-sqrt(2))`

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Another way of finding this is by completing the square, then using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a = (x+2)` and `b = sqrt(2)` as follows:

`y = 2x^2+8x+4`

`=2(x^2+4x+2)`

`=2(x^2+4x+4-2)`

`=2((x+2)^2-2)`

`=2((x+2)^2-(sqrt(2))^2)`

`=2((x+2)-sqrt(2))((x+2)+sqrt(2))`

`=2(x+2-sqrt(2))(x+2+sqrt(2))`