Question

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

Answer 1

Not sure about my solution. Have a look!!

Explanation:

Certainly at the moment I can not think of an alternative to:

Factors of `x^2` can only be `+-(x " and " x)`
Factors of 2 can only be `+-(1 " and " 2)`. You could have `sqrt(2) times sqrt(2)color(white)(x)` I suppose!

The problem is that `(x times 1)+(x times 2) != 4x`

Suppose we had instead `(x times 2)+(x times 2) = 4x`
This works for the coefficient (number in front) of `x` but fails for the constant of 2 at the end of the expression.

`2 times 2 =4` this is 2 too much. We can get over this by subtracting the amount that is too much.

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

The problem is that when asked to `color(green)("factorise completely")` I expect that I am able to do so. Consequently, `color(blue)("either I have gone wrong or the question is not quite correct.")`

Answer 2

Complete the square, then use the difference of squares identity to find:

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

Explanation:

The difference of squares identity can be written:

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

In our example:

`x^2+4x+2`

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

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

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

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

In general, given `ax^2+bx+c`, you can express it in terms of a square and a constant term as follows:

`ax^2+bx+c`

`= a(x+b/(2a))^2+(c-b^2/(4a))`

`= a((x+b/(2a))^2-(b^2/(4a^2)-c/a))`

`= a((x+b/(2a))^2-(sqrt(b^2/(4a^2)-c/a))^2)`

`=a(x+b/(2a)- sqrt(b^2/(4a^2)-c/a))(x+b/(2a)+sqrt(b^2/(4a^2)-c/a))`

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